Class 12 RD Sharma Solutions - Chapter 8 Solution of Simultaneous Linear Equations - Exercise 8.1 | Set 1

Last Updated : 23 Jul, 2025

Chapter 8 of RD Sharma's Class 12 textbook focuses on solving the simultaneous linear equations a fundamental topic in algebra. This chapter provides the methods to find the values of variables that satisfy the multiple linear equations simultaneously. Mastery of these techniques is crucial for solving real-world problems involving constraints and optimization. Exercise 8.1 | Set 1 specifically helps students practice these methods through a variety of problems.

Simultaneous Linear Equations

Simultaneous linear equations involve solving two or more linear equations with multiple variables to find a common solution. The equations are solved simultaneously meaning the solution must satisfy all equations in the system. Methods such as substitution, elimination, and matrix approaches are used to find the intersection point(s) of the equations on a graph. The goal is to determine the values of the variables that make all the equations true at once.

Question 1. Solve the following system of equations by matrix method:

(i) 5x + 7y + 2 = 0

4x + 6y + 3 = 0

Solution:

The given system of equations can be written in matrix form as,
\begin{bmatrix}5 & 7 \\ 4 & 6\end{bmatrix} \binom{x}{y} = \binom{ - 2}{ - 3}
AX = B
Here,
A = \begin{bmatrix}5 & 7 \\ 4 & 6\end{bmatrix} , X = \binom{x}{y} and B = \binom{ - 2}{ - 3}
Now,
|A| = \begin{bmatrix}5 & 7 \\ 4 & 6\end{bmatrix}
= 30 - 28
= 2
The given system has a unique solution given by, X = A^-1 B.
Let C_ij be the cofactor of the elements a_ijin A.
C₁₁ = (-1)¹⁺¹(6) = 6, C₁₂ = (-1)¹⁺² (4) = -4, C₂₁ = -1²⁺¹ (7) = -7 and C₂₂ = (-1)²⁺² (5) = 5
adj A = \begin{bmatrix}6 & - 4 \\ - 7 & 5\end{bmatrix}^T
= \begin{bmatrix}6 & - 7 \\ - 4 & 5\end{bmatrix}
A^-1 = \frac{1}{\left| A \right|}adj A
A^-1 = \frac{1}{2}\begin{bmatrix}6 & - 7 \\ - 4 & 5\end{bmatrix}\binom{ - 2}{ - 3}
So, X = A^-1 B
= \frac{1}{2}\begin{bmatrix}6 & - 7 \\ - 4 & 5\end{bmatrix}
= \frac{1}{2}\binom{ - 12 + 21}{8 - 15}
=> \binom{x}{y} = \binom{\frac{9}{2}}{\frac{- 7}{2}}
Therefore, x = 9/2 and y = -7/2.

(ii) 5x + 2y = 3

3x + 2y = 5

Solution:

The given system of equations can be written in matrix form as,
\begin{bmatrix}5 & 2 \\ 3 & 2\end{bmatrix} \binom{x}{y} = \binom{3}{5}
AX = B
Here,
A = \begin{bmatrix}5 & 2 \\ 3 & 2\end{bmatrix} , X = \binom{x}{y} and B = \binom{3}{5}
Now,
|A| = \begin{bmatrix}5 & 2 \\ 3 & 2\end{bmatrix}
= 10 - 6
= 4
The given system has a unique solution given by, X = A^-1 B
Let C_ij be the cofactor of the elements a_ij in A.
C₁₁ = -1¹⁺¹ (2) = 2, C₁₂ = (-1)¹⁺² (3) = - 3, C₂₁= (-1)²⁺¹ (2) = - 2 and C₂₂ = (-1)²⁺² (5) = 5
adj A = \begin{bmatrix}2 & - 3 \\ - 2 & 5\end{bmatrix}^T
= \begin{bmatrix}2 & - 2 \\ - 3 & 5\end{bmatrix}
A^-1 = \frac{1}{\left| A \right|}adj A
= \frac{1}{4}\begin{bmatrix}2 & - 2 \\ - 3 & 5\end{bmatrix}
Now, X = A^-1 B
= \frac{1}{4}\begin{bmatrix}2 & - 2 \\ - 3 & 5\end{bmatrix}\binom{3}{5}
= \frac{1}{4}\binom{6 - 10}{ - 9 + 25}
=> \binom{x}{y} = \binom{\frac{- 4}{4}}{\frac{16}{4}}
Therefore, x = - 1 and y = 4.

(iii) 3x + 4y − 5 = 0

x − y + 3 = 0

Solution:

The given system of equations can be written in matrix form as,
\begin{bmatrix}3 & 4 \\ 1 & - 1\end{bmatrix} \binom{x}{y} = \binom{5}{ - 3}
AX = B
Here,
A = \begin{bmatrix}3 & 4 \\ 1 & - 1\end{bmatrix} , X = \binom{x}{y} and B = \binom{5}{ - 3}
Now,
|A| = \begin{bmatrix}3 & 4 \\ 1 & - 1\end{bmatrix}
= - 3 - 4
= -7
So, the given system has a unique solution given by, X = A^-1 B
Let C_ij be the cofactors of the elements a_ijin A.
C₁₁ = (-1)¹⁺¹ (1) = -1, C₁₂ = (-1)¹⁺² (-1) = 1, C₂₁ = (-1)²⁺¹ (4) = -4 and C₂₂ = (-1)²⁺² (3) = 3
adj A = \begin{bmatrix}- 1 & - 1 \\ - 4 & 3\end{bmatrix}^T = \begin{bmatrix}- 1 & - 4 \\ - 1 & 3\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{- 7}\begin{bmatrix}- 1 & - 4 \\ - 1 & 3\end{bmatrix}
Now, X = A^-1 B
= \frac{1}{- 7}\begin{bmatrix}- 1 & - 4 \\ - 1 & 3\end{bmatrix}\binom{5}{ - 3}
= \frac{1}{- 7}\binom{ - 5 + 12}{ - 5 - 9}
=> \binom{x}{y} = \binom{\frac{7}{- 7}}{\frac{- 14}{- 7}}
Therefore, x = -1 and y = 2.

(iv) 3x + y = 19

3x − y = 23

Solution:

The given system of equations can be written in matrix form as,
\begin{bmatrix}3 & 1 \\ 3 & - 1\end{bmatrix} \binom{x}{y} = \binom{19}{23}
AX = B
Here,
A = \begin{bmatrix}3 & 1 \\ 3 & - 1\end{bmatrix} , X = \binom{x}{y} and B = \binom{19}{23}
Now,
|A| = \begin{bmatrix}3 & 1 \\ 3 & - 1\end{bmatrix}
= - 3 - 3
= -6
So, the given system has a unique solution given by X = A^-1 B.
Let C_ij be the cofactors of the elements a_ij in A.
C₁₁ = (-1)¹⁺¹ (-1) = -1, C₁₂ = (-1)¹⁺² (3) = -3, C₂₁ = (-1)²⁺¹ (1) = -4 and C₂₂ = (-1)²⁺² (3) = 3
adj A = \begin{bmatrix}- 1 & - 3 \\ - 1 & 3\end{bmatrix}^T
= \begin{bmatrix}- 1 & - 1 \\ - 3 & 3\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{- 6}\begin{bmatrix}- 1 & - 1 \\ - 3 & 3\end{bmatrix}
Now, X = A^-1 B
= \frac{1}{- 6}\begin{bmatrix}- 1 & - 1 \\ - 3 & 3\end{bmatrix}\binom{19}{23}
= \frac{1}{- 6}\binom{ - 19 - 23}{ - 57 + 69}
=> \binom{x}{y} = \binom{\frac{- 42}{- 6}}{\frac{12}{- 6}}
Therefore, x = 7 and y = -2.

(v) 3x + 7y = 4

x + 2y = −1

Solution:

The given system of equations can be written in matrix form as,
\begin{bmatrix}3 & 7 \\ 1 & 2\end{bmatrix} \binom{x}{y} = \binom{4}{ - 1}
AX = B
Here,
A = \begin{bmatrix}3 & 7 \\ 1 & 2\end{bmatrix} , X = \binom{x}{y} and B = \binom{4}{ - 1}
Now,
|A| = \begin{bmatrix}3 & 7 \\ 1 & 2\end{bmatrix}
= 6 - 7
= -1
So, the given system has a unique solution given by X = A^-1 B.
Let C_ij be the cofactors of the elements a_ijin A.
C₁₁ = (-1)¹⁺¹ (2) = 2, C₁₂ = (-1)¹⁺² (1) = -1, C₂₁ = (-1)²⁺¹ (7) = -7 and C₂₂ = (-1)²⁺² (3) = 3
adj A = \begin{bmatrix}2 & - 1 \\ - 7 & 3\end{bmatrix}^T
= \begin{bmatrix}2 & - 7 \\ - 1 & 3\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{- 1}\begin{bmatrix}2 & - 7 \\ - 1 & 3\end{bmatrix}
X = A^-1 B
= \frac{1}{- 1}\begin{bmatrix}2 & - 7 \\ - 1 & 3\end{bmatrix}\binom{4}{ - 1}
= \frac{1}{- 1}\binom{8 + 7}{ - 4 - 3}
=> \binom{x}{y} = \binom{\frac{15}{- 1}}{\frac{- 7}{- 1}}
Therefore x = - 15 and y = 7.

(vi) 3x + y = 7

5x + 3y = 12

Solution:

The given system of equations can be written in matrix form as,
\begin{bmatrix}3 & 1 \\ 5 & 3\end{bmatrix} \binom{x}{y} = \binom{7}{12}
AX = B
Here,
A = \begin{bmatrix}3 & 1 \\ 5 & 3\end{bmatrix} , X = \binom{x}{y} and B = \binom{7}{12}
Now,
|A| = \begin{bmatrix}3 & 1 \\ 5 & 3\end{bmatrix}
= 9 - 5
= 4
So, the given system has a unique solution given by X = A^-1 B.
Let C_ij be the cofactors of the elements a_ijin A.
C₁₁ = (-1)¹⁺¹ (3) = 3, C₁₂ = (-1)¹⁺² (5) = -5, C₂₁ = (-1)²⁺¹ (1) = -1 and C₂₂ = (-1)²⁺² (3) = 3
adj A = \begin{bmatrix}3 & - 5 \\ - 1 & 3\end{bmatrix}^T
= \begin{bmatrix}3 & - 1 \\ - 5 & 3\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{4}\begin{bmatrix}3 & - 1 \\ - 5 & 3\end{bmatrix}
X = A^-1B
= \frac{1}{4}\begin{bmatrix}3 & - 1 \\ - 5 & 3\end{bmatrix}\binom{7}{12}
= \frac{1}{4}\binom{21 - 12}{ - 35 + 36}
=> \binom{x}{y} = \binom{\frac{9}{4}}{\frac{1}{4}}
Therefore x = 9/4 and y = 1/4.

Question 2. Solve the following system of equations by matrix method:

(i) x + y − z = 3

2x + 3y + z = 10

3x − y − 7z = 1

Solution:

A = \begin{bmatrix}1 & 1 & - 1 \\ 2 & 3 & 1 \\ 3 & - 1 & - 7\end{bmatrix}
|A| = \begin{vmatrix}1 & 1 & - 1 \\ 2 & 3 & 1 \\ 3 & - 1 & - 7\end{vmatrix}
= 1 (- 21 + 1) - 1(-14 - 3) - 1(-2 - 9)
= - 20 + 17 + 11
= 8
So, the given system has a unique solution given by X = A^-1 B.
Let C_ij be the cofactor of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}3 & 1 \\ - 1 & - 7\end{vmatrix} = - 20 , C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 1 \\ 3 & - 7\end{vmatrix} = 17, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & 3 \\ 3 & - 1\end{vmatrix} = - 11
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & - 1 \\ - 1 & - 7\end{vmatrix} = 8 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & - 1 \\ 3 & - 7\end{vmatrix} = - 4, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 3 & - 1\end{vmatrix} = 4
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & - 1 \\ 3 & 1\end{vmatrix} = 4 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & - 1 \\ 2 & 1\end{vmatrix} = - 3, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 2 & 3\end{vmatrix} = 1
adj A = \begin{bmatrix}- 20 & 17 & - 11 \\ 8 & - 4 & 4 \\ 4 & - 3 & 1\end{bmatrix}^T
= \begin{bmatrix}- 20 & 8 & 4 \\ 17 & - 4 & - 3 \\ - 11 & 4 & 1\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{8}\begin{bmatrix}- 20 & 8 & 4 \\ 17 & - 4 & - 3 \\ - 11 & 4 & 1\end{bmatrix}
X = A^-1 B
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{8}\begin{bmatrix}- 20 & 8 & 4 \\ 17 & - 4 & - 3 \\ - 11 & 4 & 1\end{bmatrix}\begin{bmatrix}3 \\ 10 \\ 1\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{8}\begin{bmatrix}- 60 + 80 + 4 \\ 51 - 40 - 3 \\ - 33 + 40 + 1\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{8}\begin{bmatrix}24 \\ 8 \\ 8\end{bmatrix}
=> x = 24/8, y = 8/8 and z = 8/8
Therefore, x = 3, y = 1 and z = 1.

(ii) x + y + z = 3

2x − y + z = − 1

2x + y − 3z = − 9

Solution:

A = \begin{bmatrix}1 & 1 & 1 \\ 2 & - 1 & 1 \\ 2 & 1 & - 3\end{bmatrix}
|A| = \begin{vmatrix}1 & 1 & 1 \\ 2 & - 1 & 1 \\ 2 & 1 & - 3\end{vmatrix}
= 1 (3 - 1) - 1 (-6 - 2) + 1 (2 + 2)
= 2 + 8 + 4
= 14
So, the given system has a unique solution given by X = A^-1 B.
Let C_ij be the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 1 & 1 \\ 1 & - 3\end{vmatrix} = 2 , C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 1 \\ 2 & - 3\end{vmatrix} = 8, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & - 1 \\ 2 & 1\end{vmatrix} = 4
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 1 & - 3\end{vmatrix} = 4, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 2 & - 3\end{vmatrix} = - 5, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = 1
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ - 1 & 1\end{vmatrix} = 2 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = 1, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 2 & - 1\end{vmatrix} = - 3
adj A = \begin{bmatrix}2 & 8 & 4 \\ 4 & - 5 & 1 \\ 2 & 1 & - 3\end{bmatrix}^T
= \begin{bmatrix}2 & 4 & 2 \\ 8 & - 5 & 1 \\ 4 & 1 & - 3\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{14}\begin{bmatrix}2 & 4 & 2 \\ 8 & - 5 & 1 \\ 4 & 1 & - 3\end{bmatrix}
Now, X = A^-1 B
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{14}\begin{bmatrix}2 & 4 & 2 \\ 8 & - 5 & 1 \\ 4 & 1 & - 3\end{bmatrix}\begin{bmatrix}3 \\ - 1 \\ - 9\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{14}\begin{bmatrix}6 - 4 - 18 \\ 24 + 5 - 9 \\ 12 - 1 + 27\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{14}\begin{bmatrix}- 16 \\ 20 \\ 38\end{bmatrix}
=> x = -16/14, y = 20/14 and z = 38/14
Therefore, x = -8/7, y = 10/7 and z = 19/7.

(iii) 6x − 12y + 25z = 4

4x + 15y − 20z = 3

2x + 18y + 15z = 10

Solution:

A = \begin{bmatrix}6 & - 12 & 25 \\ 4 & 15 & - 20 \\ 2 & 18 & 15\end{bmatrix}
|A| = \begin{vmatrix}6 & - 12 & 25 \\ 4 & 15 & - 20 \\ 2 & 18 & 15\end{vmatrix}
= 6 (225 + 360) + 12 (60 + 40) + 25 (72 - 30)
= 3510 + 1200 + 1050
= 5760
So, the given system has a unique solution given by X = A^-1 B.
Let C_ij be the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}15 & - 20 \\ 18 & 15\end{vmatrix} = 585, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}4 & - 20 \\ 2 & 15\end{vmatrix} = - 100 , C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}4 & 15 \\ 2 & 18\end{vmatrix} = 42
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 12 & 25 \\ 18 & 15\end{vmatrix} = 630, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}6 & 25 \\ 2 & 15\end{vmatrix} = 40, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}6 & - 12 \\ 2 & 18\end{vmatrix} = - 132
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 12 & 25 \\ 15 & - 20\end{vmatrix} = - 135, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}6 & 25 \\ 4 & - 20\end{vmatrix} = 220, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}6 & - 12 \\ 4 & 15\end{vmatrix} = 138
adj A = \begin{bmatrix}585 & - 100 & 42 \\ 630 & 40 & - 132 \\ - 135 & 220 & 138\end{bmatrix}^T
= \begin{bmatrix}585 & 630 & - 135 \\ - 100 & 40 & 220 \\ 42 & - 132 & 138\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{5760}\begin{bmatrix}585 & 630 & - 135 \\ - 100 & 40 & 220 \\ 42 & - 132 & 138\end{bmatrix}
Now, X = A^-1 B
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{5760}\begin{bmatrix}585 & 630 & - 135 \\ - 100 & 40 & 220 \\ 42 & - 132 & 138\end{bmatrix}\begin{bmatrix}4 \\ 3 \\ 10\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{5760}\begin{bmatrix}2340 + 1890 - 1350 \\ - 400 + 120 + 2200 \\ 168 - 396 + 1380\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{5760}\begin{bmatrix}2880 \\ 1920 \\ 1152\end{bmatrix}
=> x = 2880/5760, y = 1920/5760 and z = 1152/5760
Therefore x = 1/2, y = 1/3 and z = 1/5.

(iv) 3x + 4y + 7z = 14

2x − y + 3z = 4

x + 2y − 3z = 0

Solution:

A = \begin{bmatrix}3 & 4 & 7 \\ 2 & - 1 & 3 \\ 2 & 1 & - 3\end{bmatrix}
|A| = \begin{vmatrix}3 & 4 & 7 \\ 2 & - 1 & 3 \\ 2 & 1 & - 3\end{vmatrix}
= 3 (3 - 3) - 4 (- 6 - 6) + 7 (2 + 2)
= 0 + 48 + 28
= 76
So, the given system has a unique solution given by X = A^-1 B.
Let C_ij be the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 1 & 3 \\ 1 & - 3\end{vmatrix} = 0, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 3 \\ 2 & - 3\end{vmatrix} = 12, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & - 1 \\ 2 & 1\end{vmatrix} = 4
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}4 & 7 \\ 1 & - 3\end{vmatrix} = 19 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}3 & 7 \\ 2 & - 3\end{vmatrix} = - 23 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}3 & 4 \\ 2 & 1\end{vmatrix} = 5
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}4 & 7 \\ - 1 & 3\end{vmatrix} = 19, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}3 & 7 \\ 2 & 3\end{vmatrix} = 5, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}3 & 4 \\ 2 & - 1\end{vmatrix} = - 11
adj A = \begin{bmatrix}0 & 12 & 4 \\ 19 & - 23 & 5 \\ 19 & 5 & - 11\end{bmatrix}^T
= \begin{bmatrix}0 & 19 & 19 \\ 12 & - 23 & 5 \\ 4 & 5 & - 11\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{76}\begin{bmatrix}0 & 19 & 19 \\ 12 & - 23 & 5 \\ 4 & 5 & - 11\end{bmatrix}
Now, X = A^-1 B
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{76}\begin{bmatrix}0 & 19 & 19 \\ 12 & - 23 & 5 \\ 4 & 5 & - 11\end{bmatrix}\begin{bmatrix}14 \\ 4 \\ 0\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{76}\begin{bmatrix}0 + 76 + 0 \\ 168 - 92 + 0 \\ 56 + 20 + 0\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{76}\begin{bmatrix}76 \\ 76 \\ 76\end{bmatrix}
=> x = 76/76, y = 76/76 and z = 76/76
Therefore x = 1, y = 1 and z = 1.

(v) \frac{2}{x} - \frac{3}{y} + \frac{3}{z} = 10\\ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 10\\ \frac{3}{x} - \frac{1}{y} + \frac{2}{z} = 13

Solution:

Let 1/x be a, 1/y be b and 1/z be c.
Here,
A = \begin{bmatrix}2 & - 3 & 3 \\ 1 & 1 & 1 \\ 3 & - 1 & 2\end{bmatrix}
|A| = \begin{vmatrix}2 & - 3 & 3 \\ 1 & 1 & 1 \\ 3 & - 1 & 2\end{vmatrix}
= 2 (2 + 1) + 3 (2 - 3) + 3 (-1 - 3)
= 6 - 3 - 12
= -9
So, the given system has a unique solution given by X = A^-1 B.
Let C_ij be the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}1 & 1 \\ - 1 & 2\end{vmatrix} = 3, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & 1 \\ 3 & 2\end{vmatrix} = 1 , C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & 1 \\ 3 & - 1\end{vmatrix} = - 4
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 3 & 3 \\ - 1 & 2\end{vmatrix} = 3 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & 3 \\ 3 & 2\end{vmatrix} = - 5, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & - 3 \\ 3 & - 1\end{vmatrix} = -7
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 3 & 3 \\ 1 & 1\end{vmatrix} = - 6 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & 3 \\ 1 & 1\end{vmatrix} = 1, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & - 3 \\ 1 & 1\end{vmatrix} = 5
adj A = \begin{bmatrix}3 & 1 & - 4 \\ 3 & - 5 & - 7 \\ - 6 & 1 & 5\end{bmatrix}^T
= \begin{bmatrix}3 & 3 & - 6 \\ 1 & - 5 & 1 \\ - 4 & - 7 & 5\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{- 9}\begin{bmatrix}3 & 3 & - 6 \\ 1 & - 5 & 1 \\ - 4 & - 7 & 5\end{bmatrix}
X = A^-1 B
\begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{- 9}\begin{bmatrix}3 & 3 & - 6 \\ 1 & - 5 & 1 \\ - 4 & - 7 & 5\end{bmatrix}\begin{bmatrix}10 \\ 10 \\ 13\end{bmatrix}
\begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{- 9}\begin{bmatrix}30 + 30 - 78 \\ 10 - 50 + 13 \\ - 40 - 70 + 65\end{bmatrix}
\begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{- 9}\begin{bmatrix}- 18 \\ - 27 \\ - 45\end{bmatrix}
=> x = 1/a = - 9/-18, y = 1/b = - 9/- 27 and z = 1/c = -9/-45
Therefore x = 1/2, y = 1/3 and z = 1/5.

(vi) 5x + 3y + z = 16

2x + y + 3z = 19

x + 2y + 4z = 25

Solution:

A = \begin{bmatrix}5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4\end{bmatrix}
|A| = \begin{vmatrix}5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4\end{vmatrix}
= 5 (4 - 6) - 3 (8 - 3) + 1 (4 - 1)
= -10 - 15 + 3
= - 22
So, the given system has a unique solution given by X = A^-1 B.
Let C_ij be the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}1 & 3 \\ 2 & 4\end{vmatrix} = - 2, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 3 \\ 1 & 4\end{vmatrix} = - 5 , C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & 1 \\ 1 & 2\end{vmatrix} = 3
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}3 & 1 \\ 2 & 4\end{vmatrix} = - 10 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}5 & 1 \\ 1 & 4\end{vmatrix} = 19, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}5 & 3 \\ 1 & 2\end{vmatrix} = -7
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}3 & 1 \\ 1 & 3\end{vmatrix} = 8 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}5 & 1 \\ 2 & 3\end{vmatrix} = - 13 , C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}5 & 3 \\ 2 & 1\end{vmatrix} = - 1
adj A = \begin{bmatrix}- 2 & - 5 & 3 \\ - 10 & 19 & - 7 \\ 8 & - 13 & - 1\end{bmatrix}^T
= \begin{bmatrix}- 2 & - 10 & 8 \\ - 5 & 19 & - 13 \\ 3 & - 7 & - 1\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{- 22}\begin{bmatrix}- 2 & - 10 & 8 \\ - 5 & 19 & - 13 \\ 3 & - 7 & - 1\end{bmatrix}
X = A^-1 B
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 22}\begin{bmatrix}- 2 & - 10 & 8 \\ - 5 & 19 & - 13 \\ 3 & - 7 & - 1\end{bmatrix}\begin{bmatrix}16 \\ 19 \\ 25\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 22}\begin{bmatrix}- 32 - 190 + 200 \\ - 80 + 361 - 325 \\ 48 - 133 - 25\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 22}\begin{bmatrix}- 22 \\ - 44 \\ - 110\end{bmatrix}
=> x = - 22/- 22, y = - 44/- 22 and z = -110/-22
Therefore x = 1, y = 2 and z = 5.

(vii) 3x + 4y + 2z = 8

2y − 3z = 3

x − 2y + 6z = −2

Solution:

A = \begin{bmatrix}3 & 4 & 2 \\ 0 & 2 & - 3 \\ 1 & - 2 & 6\end{bmatrix}
|A| = \begin{vmatrix}3 & 4 & 2 \\ 0 & 2 & - 3 \\ 1 & - 2 & 6\end{vmatrix}
= 3 (12 - 6) - 4 (0 + 3) + 2 (0 - 2)
= 18 - 12 - 4
= 2
Let C_ij be the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & - 3 \\ - 2 & 6\end{vmatrix} = 6, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}0 & - 3 \\ 1 & 6\end{vmatrix} = - 3 , C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}0 & 2 \\ 1 & - 2\end{vmatrix} = - 2
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}4 & 2 \\ - 2 & 6\end{vmatrix} = - 28 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}3 & 2 \\ 1 & 6\end{vmatrix} = 16 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}3 & 4 \\ 1 & - 2\end{vmatrix} = 10
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}4 & 2 \\ 2 & - 3\end{vmatrix} = - 16, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}3 & 2 \\ 0 & - 3\end{vmatrix} = 9, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}3 & 4 \\ 0 & 2\end{vmatrix} = 6
adj A = \begin{bmatrix}6 & - 3 & - 2 \\ - 28 & 16 & 10 \\ - 16 & 9 & 6\end{bmatrix}^T
= \begin{bmatrix}6 & - 28 & - 16 \\ - 3 & 16 & 9 \\ - 2 & 10 & 6\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{2}\begin{bmatrix}6 & - 28 & - 16 \\ - 3 & 16 & 9 \\ - 2 & 10 & 6\end{bmatrix}
Now X = A^-1 B
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{2}\begin{bmatrix}6 & - 28 & - 16 \\ - 3 & 16 & 9 \\ - 2 & 10 & 6\end{bmatrix}\begin{bmatrix}8 \\ 3 \\ - 2\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{2}\begin{bmatrix}48 - 84 + 32 \\ - 24 + 48 - 18 \\ - 16 + 30 - 12\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{2}\begin{bmatrix}- 4 \\ 6 \\ 2\end{bmatrix}
=> x = -4/2, y = 6/2 and z = 2/2
Therefore x = -2, y = 3 and z = 1.

(viii) 2x + y + z = 2

x + 3y − z = 5

3x + y − 2z = 6

Solution:

Here,
A = \begin{bmatrix}2 & 1 & 1 \\ 1 & 3 & - 1 \\ 3 & 1 & - 2\end{bmatrix}
|A| = \begin{vmatrix}2 & 1 & 1 \\ 1 & 3 & - 1 \\ 3 & 1 & - 2\end{vmatrix}
= - 10 - 1 - 8
= -19
Let C_ij be the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}3 & - 1 \\ 1 & - 2\end{vmatrix} = - 5, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & - 1 \\ 3 & - 2\end{vmatrix} = - 1, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & 3 \\ 3 & 1\end{vmatrix} = - 8
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 1 & - 2\end{vmatrix} = 3, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & - 1 \\ 3 & - 2\end{vmatrix} = - 7, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & 1 \\ 3 & 1\end{vmatrix} = 1
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 3 & - 1\end{vmatrix} = - 4, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & 1 \\ 1 & - 1\end{vmatrix} = 3, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & 1 \\ 1 & 3\end{vmatrix} = 5
adj A = \begin{bmatrix}- 5 & - 1 & - 8 \\ 3 & - 7 & 1 \\ - 4 & 3 & 5\end{bmatrix}^T
= \begin{bmatrix}- 5 & 3 & - 4 \\ - 1 & - 7 & 3 \\ - 8 & 1 & 5\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{- 19}\begin{bmatrix}- 5 & 3 & - 4 \\ - 1 & - 7 & 3 \\ - 8 & 1 & 5\end{bmatrix}
X = A^-1 B
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 19}\begin{bmatrix}- 5 & 3 & - 4 \\ - 1 & - 7 & 3 \\ - 8 & 1 & 5\end{bmatrix}\begin{bmatrix}2 \\ 5 \\ 6\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 19}\begin{bmatrix}- 10 + 15 - 24 \\ - 2 - 35 + 18 \\ - 16 + 5 + 30\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 19}\begin{bmatrix}- 19 \\ -19 \\ 19\end{bmatrix}
x = -19/-19, y = -19/-19 and z = 19/-19
Therefore x = 1, y = 1 and z = - 1.

(ix) 2x + 6y = 2

3x − z = −8

2x − y + z = −3

Solution:

A = \begin{bmatrix}2 & 6 & 0 \\ 3 & 0 & - 1 \\ 2 & - 1 & 1\end{bmatrix}
|A| = \begin{vmatrix}2 & 6 & 0 \\ 3 & 0 & - 1 \\ 2 & - 1 & 1\end{vmatrix}
= 2 (0 - 1) - 6 (3 + 2) + 0 (-3 + 0)
= -2 - 30
= - 32
Let C_ij be the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}0 & - 1 \\ - 1 & 1\end{vmatrix} = - 1 , C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}3 & - 1 \\ 2 & 1\end{vmatrix} = - 5 , C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}3 & 0 \\ 2 & - 1\end{vmatrix} = - 3
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}6 & 0 \\ - 1 & 1\end{vmatrix} = - 6 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & 0 \\ 2 & 1\end{vmatrix} = 2 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & 6 \\ 2 & - 1\end{vmatrix} = 14
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}6 & 0 \\ 0 & - 1\end{vmatrix} = - 6 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & 0 \\ 3 & - 1\end{vmatrix} = 2 , C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & 6 \\ 3 & 0\end{vmatrix} = - 18
adj A = \begin{bmatrix}- 1 & - 5 & - 3 \\ - 6 & 2 & 14 \\ - 6 & 2 & - 18\end{bmatrix}^T
= \begin{bmatrix}- 1 & - 6 & - 6 \\ - 5 & 2 & 2 \\ - 3 & 14 & - 18\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{- 32}\begin{bmatrix}- 1 & - 6 & - 6 \\ - 5 & 2 & 2 \\ - 3 & 14 & - 18\end{bmatrix}
X = A^-1 B
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 32}\begin{bmatrix}- 1 & - 6 & - 6 \\ - 5 & 2 & 2 \\ - 3 & 14 & - 18\end{bmatrix}\begin{bmatrix}2 \\ - 8 \\ - 3\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 32}\begin{bmatrix}- 2 + 48 + 18 \\ - 10 - 16 - 6 \\ - 6 - 112 + 54\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 32}\begin{bmatrix}64 \\ - 32 \\ - 64\end{bmatrix}
=> x = 64/-32, y = -32/-32 and z = -64/-32
Therefore x = - 2, y = 1 and z = 2.

(x) x − y + z = 2

2x − y = 0

2y − z = 1

Solution:

A = \begin{bmatrix}1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 0 & 2 & - 1\end{bmatrix}
|A| = \begin{vmatrix}1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 0 & 2 & - 1\end{vmatrix}
= 1 (1 - 0) + 1 (-2 - 0) + 1(4 - 0)
= 1 - 2 + 4
= 3
Let C_ij be the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 1 & 0 \\ 2 & - 1\end{vmatrix} = 1, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 0 \\ 0 & - 1\end{vmatrix} = 2, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & - 1 \\ 0 & 2\end{vmatrix} = 4
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 1 & 1 \\ 2 & - 1\end{vmatrix} = 1, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 0 & - 1\end{vmatrix} = - 1, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & - 1 \\ 0 & 2\end{vmatrix} = - 2
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 1 & 1 \\ - 1 & 0\end{vmatrix} = 1, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 2 & 0\end{vmatrix} = 2, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & - 1 \\ 2 & - 1\end{vmatrix} = 1
adj A = \begin{bmatrix}1 & 2 & 4 \\ 1 & - 1 & - 2 \\ 1 & 2 & 1\end{bmatrix}^T
= \begin{bmatrix}1 & 1 & 1 \\ 2 & - 1 & 2 \\ 4 & - 2 & 1\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{1}\begin{bmatrix}1 & 1 & 1 \\ 2 & - 1 & 2 \\ 4 & - 2 & 1\end{bmatrix}
X = A^-1 B
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{3}\begin{bmatrix}1 & 1 & 1 \\ 2 & - 1 & 2 \\ 4 & - 2 & 1\end{bmatrix}\begin{bmatrix}2 \\ 0 \\ 1\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{3}\begin{bmatrix}2 + 1 \\ 4 + 2 \\ 8 + 1\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{1}\begin{bmatrix}3 \\ 6 \\ 9\end{bmatrix}
=> x = 3/3, y = 6/3 and z = 9/3
Therefore x = 1, y = 2 and z = 3.

(xi) 8x + 4y + 3z = 18

2x + y +z = 5

x + 2y + z = 5

Solution:

A = \begin{bmatrix}8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 1\end{bmatrix}
|A| = \begin{vmatrix}8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 1\end{vmatrix}
= 8 (1 - 2) - 4 (2 - 1) + 3(4 - 1)
= - 8 - 4 + 9
= -3
Let C_ij be the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = - 1, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 1 \\ 1 & 1\end{vmatrix} = - 1, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & 1 \\ 1 & 2\end{vmatrix} = 3
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}4 & 3 \\ 2 & 1\end{vmatrix} = 2, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}8 & 3 \\ 1 & 1\end{vmatrix} = 5, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}8 & 4 \\ 1 & 2\end{vmatrix} = - 12
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}4 & 3 \\ 1 & 1\end{vmatrix} = 1, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}8 & 3 \\ 2 & 1\end{vmatrix} = - 2 , C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}8 & 4 \\ 2 & 1\end{vmatrix} = 0
adj A = \begin{bmatrix}- 1 & - 1 & 3 \\ 2 & 5 & - 12 \\ 1 & - 2 & 0\end{bmatrix}^T
= \begin{bmatrix}- 1 & 2 & 1 \\ - 1 & 5 & - 2 \\ 3 & - 12 & 0\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{- 3}\begin{bmatrix}- 1 & 2 & 1 \\ - 1 & 5 & - 2 \\ 3 & - 12 & 0\end{bmatrix}
X = A^-1 B
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 3}\begin{bmatrix}- 1 & 2 & 1 \\ - 1 & 5 & - 2 \\ 3 & - 12 & 0\end{bmatrix}\begin{bmatrix}18 \\ 5 \\ 5\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 3}\begin{bmatrix}- 18 + 10 + 5 \\ - 18 + 25 - 10 \\ 54 - 60\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 3}\begin{bmatrix}- 3 \\ - 3 \\ - 6\end{bmatrix}
=> x = -3/-3, y = -3/-3 and z = -6/-3
Therefore x = 1, y = 1 and z = 2.

(xii) x + y + z = 6

x + 2z = 7

3x + y + z = 12

Solution:

A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1\end{bmatrix}
|A| = \begin{vmatrix}1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1\end{vmatrix}
= 1 (0 - 2) - 1 (1 - 6) + 1(1 - 0)
= - 2 + 5 + 1
= 4
Let C_ij be the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}0 & 2 \\ 1 & 1\end{vmatrix} = - 2, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & 2 \\ 3 & 1\end{vmatrix} = 5, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & 0 \\ 3 & 1\end{vmatrix} = 1
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 1 & 1\end{vmatrix} = 0, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 3 & 1\end{vmatrix} = - 2, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 3 & 1\end{vmatrix} = 2
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 0 & 2\end{vmatrix} = 2, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 1 & 2\end{vmatrix} = - 1, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 1 & 0\end{vmatrix} = - 1
adj A = \begin{bmatrix}- 2 & 5 & 1 \\ 0 & - 2 & 2 \\ 2 & - 1 & - 1\end{bmatrix}^T
= \begin{bmatrix}- 2 & 0 & 2 \\ 5 & - 2 & - 1 \\ 1 & 2 & - 1\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{4}\begin{bmatrix}- 2 & 0 & 2 \\ 5 & - 2 & - 1 \\ 1 & 2 & - 1\end{bmatrix}
X = A^-1 B
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{4}\begin{bmatrix}- 2 & 0 & 2 \\ 5 & - 2 & - 1 \\ 1 & 2 & - 1\end{bmatrix}\begin{bmatrix}6 \\ 7 \\ 12\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{4}\begin{bmatrix}- 12 + 0 + 24 \\ 30 - 14 - 12 \\ 6 - 14 - 12\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{4}\begin{bmatrix}12 \\ 4 \\ - 20\end{bmatrix}
=> x = 12/4, y = 4/4 and z = -20/4
Therefore x = 3, y = 1 and z = - 5.

(xiii) \frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4\\\frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1\\\frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2 , x, y, z ≠ 0

Solution:

Let 1/x be a, 1/y be b and 1/z be c.
Here,
A = \begin{bmatrix}2 & 3 & 10 \\ 4 & - 6 & 5 \\ 6 & 9 & - 20\end{bmatrix}
|A| = \begin{vmatrix}2 & 3 & 10 \\ 4 & - 6 & 5 \\ 6 & 9 & - 20\end{vmatrix}
= 2 (120 - 45) - 3 (-80 - 30) + 10 (36 + 36)
= 150 + 330 + 720
= 1200
Let C_ij be the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 6 & 5 \\ 9 & - 20\end{vmatrix} = 75, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}4 & 5 \\ 6 & - 20\end{vmatrix} = 110, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}4 & - 6 \\ 6 & 9\end{vmatrix} = 72
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}3 & 10 \\ 9 & - 20\end{vmatrix} = 150, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & 10 \\ 6 & - 20\end{vmatrix} = - 100, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & 3 \\ 6 & 9\end{vmatrix} = 0
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}3 & 10 \\ - 6 & 5\end{vmatrix} = 75, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & 10 \\ 4 & 5\end{vmatrix} = 30, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & 3 \\ 4 & - 6\end{vmatrix} = - 24
adj A = \begin{bmatrix}75 & 110 & 72 \\ 150 & - 100 & 0 \\ 75 & 30 & - 24\end{bmatrix}^T
= \begin{bmatrix}75 & 150 & 75 \\ 110 & - 100 & 30 \\ 72 & 0 & - 24\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{1200} \begin{bmatrix}75 & 150 & 75 \\ 110 & - 100 & 30 \\ 72 & 0 & - 24\end{bmatrix}
X = A^-1 B
\begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{1200} \begin{bmatrix}75 & 150 & 75 \\ 110 & - 100 & 30 \\ 72 & 0 & - 24\end{bmatrix}\begin{bmatrix}4 \\ 1 \\ 2\end{bmatrix}
\begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{1200}\begin{bmatrix}300 + 150 + 150 \\ 440 - 100 + 60 \\ 288 - 48\end{bmatrix}
\begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{1200}\begin{bmatrix}600 \\ 400 \\ 240\end{bmatrix}
=> x = 1/a = 1200/600, y = 1/b = 1200/400 and z = 1/c = 1200/240
Therefore x = 2, y = 3 and z = 5.

(xiv) x − y + 2z = 7

3x + 4y − 5z = −5

2x − y + 3z = 12

Solution:

A = \begin{bmatrix}1 & - 1 & 2 \\ 3 & 4 & - 5 \\ 2 & - 1 & 3\end{bmatrix}
|A| = \begin{vmatrix}1 & - 1 & 2 \\ 3 & 4 & - 5 \\ 2 & - 1 & 3\end{vmatrix}
= 1 (12 - 5) + 1 (9 + 10) + 2 (-3 - 8)
= 7 + 19 - 22
= 4
Let C_ij be the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}4 & - 5 \\ - 1 & 3\end{vmatrix} = 7, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}3 & - 5 \\ 2 & 3\end{vmatrix} = - 19, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}3 & 4 \\ 2 & - 1\end{vmatrix} = - 11
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 1 & 2 \\ - 1 & 3\end{vmatrix} = 1 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 2 \\ 2 & 3\end{vmatrix} = - 1 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & - 1 \\ 2 & - 1\end{vmatrix} = - 1
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 1 & 2 \\ 4 & - 5\end{vmatrix} = - 3, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 2 \\ 3 & - 5\end{vmatrix} = 11, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & - 1 \\ 3 & 4\end{vmatrix} = 7
adj A = \begin{bmatrix}7 & - 19 & - 11 \\ 1 & - 1 & - 1 \\ - 3 & 11 & 7\end{bmatrix}^T
= \begin{bmatrix}7 & 1 & - 3 \\ - 19 & - 1 & 11 \\ - 11 & - 1 & 7\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{4}\begin{bmatrix}7 & 1 & - 3 \\ - 19 & - 1 & 11 \\ - 11 & - 1 & 7\end{bmatrix}
X = A^-1 B
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{4}\begin{bmatrix}7 & 1 & - 3 \\ - 19 & - 1 & 11 \\ - 11 & - 1 & 7\end{bmatrix}\begin{bmatrix}7 \\ - 5 \\ 12\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{4}\begin{bmatrix}49 - 5 - 36 \\ - 133 + 5 + 132 \\ - 77 + 5 + 84\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{4}\begin{bmatrix}8 \\ 4 \\ 12\end{bmatrix}
=> x = 8/4, y = 4/4 and z = 12/4
Therefore x = 2, y = 1 and z = 3.

Question 3. Show that the following systems of linear equations is consistent:

(i) 6x + 4y = 2

9x + 6y = 3

Solution:

Here,
6x + 4y = 2
9x + 6y = 3
We know, AX = B
A = \begin{bmatrix}6 & 4 \\ 9 & 6\end{bmatrix} , X = \binom{x}{y} and B = \binom{2}{3}
\begin{bmatrix}6 & 4 \\ 9 & 6\end{bmatrix}\binom{x}{y} = \binom{2}{3}
|A| = \begin{vmatrix}6 & 4 \\ 9 & 6\end{vmatrix}
= 36 - 36
= 0
Let C_ij be the cofactors of the elements a_ij in A.
C₁₁ = 6, C₁₂ = -9, C₂₁ = -4 and C₂₂ = 6
adj A = \begin{bmatrix}6 & - 9 \\ - 4 & 6\end{bmatrix}^T
= \begin{bmatrix}6 & - 4 \\ - 9 & 6\end{bmatrix}
(adj A) B = \begin{bmatrix}6 & - 4 \\ - 9 & 6\end{bmatrix}\binom{2}{3}
= \binom{12 - 12}{ - 18 + 18}
= \binom{0}{0}
Therefore, the system is consistent and has infinitely many solutions.
Hence proved.

(ii) 2x + 3y = 5

6x + 9y = 15

Solution:

Here,
2x + 3y = 5
6x + 9y = 15
We know, AX = B
A = \begin{bmatrix}2 & 3 \\ 6 & 9\end{bmatrix} , X = \binom{x}{y} and B = \binom{5}{15}
\begin{bmatrix}2 & 3 \\ 6 & 9\end{bmatrix}\binom{x}{y} = \binom{5}{15}
|A| = \begin{vmatrix}2 & 3 \\ 6 & 9\end{vmatrix}
= 18 - 18
= 0
Let C_ijbe the cofactors of the elements a_ij in A.
C₁₁ = 9, C₁₂ = -6, C₂₁ = -3 and C₂₂ = 2
adj A = \begin{bmatrix}9 & - 6 \\ - 3 & 2\end{bmatrix}^T
= \begin{bmatrix}2 & - 3 \\ - 6 & 9\end{bmatrix}
(adj A) B = \begin{bmatrix}9 & - 3 \\ - 6 & 2\end{bmatrix}\binom{5}{15}
= \binom{45 - 45}{ - 30 + 30}
= \binom{0}{0}
Therefore, the system is consistent and has infinitely many solutions.
Hence proved.

(iii) 5x + 3y + 7z = 4

3x + 26y + 2z = 9

7x + 2y + 10z = 5

Solution:

Here,
5x + 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5
We know, AX = B
A = \begin{bmatrix}5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10\end{bmatrix} , X = \begin{bmatrix}x \\ y \\ z\end{bmatrix} and B = \begin{bmatrix}4 \\ 9 \\ 5\end{bmatrix}
\begin{bmatrix}5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}4 \\ 9 \\ 5\end{bmatrix}
|A| = \begin{vmatrix}5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10\end{vmatrix}
= 1280 - 48 - 1232\]
= 0
Let C_ijbe the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}26 & 2 \\ 2 & 10\end{vmatrix} = 256 , C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}3 & 2 \\ 7 & 10\end{vmatrix} = - 16, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}3 & 26 \\ 7 & 2\end{vmatrix} = - 176
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}3 & 7 \\ 2 & 10\end{vmatrix} = - 16 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}5 & 7 \\ 7 & 10\end{vmatrix} = 1, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}5 & 3 \\ 7 & 2\end{vmatrix} = 11
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}3 & 7 \\ 26 & 2\end{vmatrix} = - 176, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}5 & 7 \\ 3 & 2\end{vmatrix} = 11, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}5 & 3 \\ 3 & 26\end{vmatrix} = 121
adj A = \begin{bmatrix}256 & - 16 & - 176 \\ - 16 & 1 & 11 \\ - 176 & 11 & 121\end{bmatrix}^T
= \begin{bmatrix}256 & - 16 & - 176 \\ - 16 & 1 & 11 \\ - 176 & 11 & 121\end{bmatrix}
(adj A)B = \begin{bmatrix}256 & - 16 & - 176 \\ - 16 & 1 & 11 \\ - 176 & 11 & 121\end{bmatrix}\begin{bmatrix}4 \\ 9 \\ 5\end{bmatrix}
= \begin{bmatrix}1024 - 144 - 880 \\ - 64 + 9 + 55 \\ - 704 + 99 + 605\end{bmatrix}
= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}
Therefore, the system is consistent and has infinitely many solutions.
Hence proved.

(iv) x − y + z = 3

2x + y − z = 2

−x −2y + 2z = 1

Solution:

Here,
x − y + z = 3
2x + y − z = 2
−x −2y + 2z = 1
We know, AX = B
A = \begin{bmatrix}1 & - 1 & 1 \\ 2 & 1 & - 1 \\ - 1 & - 2 & 2\end{bmatrix} , X = \begin{bmatrix}x \\ y \\ z\end{bmatrix} and B = \begin{bmatrix}3 \\ 2 \\ 1\end{bmatrix}
\begin{bmatrix}1 & - 1 & 1 \\ 2 & 1 & - 1 \\ - 1 & - 2 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}3 \\ 2 \\ 1\end{bmatrix}
|A| = \begin{vmatrix}1 & - 1 & 1 \\ 2 & 1 & - 1 \\ - 1 & - 2 & 2\end{vmatrix}
= 1\left( 2 - 2 \right) + 1\left( 4 - 1 \right) + 1( - 4 + 1)\]
= 0 + 3 - 3
= 0
Let C_ij be the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}1 & - 1 \\ - 2 & 2\end{vmatrix} = 0, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & - 1 \\ - 1 & 2\end{vmatrix} = - 3, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & 1 \\ - 1 & - 2\end{vmatrix} = - 3
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 1 & 1 \\ - 2 & 2\end{vmatrix} = 0, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ - 1 & 2\end{vmatrix} = 3, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & - 1 \\ - 1 & - 2\end{vmatrix} = 3
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 1 & 1 \\ 1 & - 1\end{vmatrix} = 0, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 2 & - 1\end{vmatrix} = 3 , C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & - 1 \\ 2 & 1\end{vmatrix} = 3
adj A = \begin{bmatrix}0 & - 3 & - 3 \\ 0 & 3 & 3 \\ 0 & 3 & 3\end{bmatrix}^T
= \begin{bmatrix}0 & 0 & 0 \\ - 3 & 3 & 3 \\ - 3 & 3 & 3\end{bmatrix}
(adj A) B = \begin{bmatrix}0 & 0 & 0 \\ - 3 & 3 & 3 \\ - 3 & 3 & 3\end{bmatrix}\begin{bmatrix}3 \\ 2 \\ 1\end{bmatrix}
= \begin{bmatrix}0 \\ - 9 + 6 + 3 \\ - 9 + 6 + 3\end{bmatrix}
= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}
Therefore, the system is consistent and has infinitely many solutions.
Hence proved.

(v) x + y + z = 6

x + 2y + 3z = 14

x + 4y + 7z = 30

Solution:

Here,
x + y + z = 6
x + 2y + 3z = 14
x + 4y + 7z = 30
A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7\end{bmatrix} , X = \begin{bmatrix}x \\ y \\ z\end{bmatrix} and B = \begin{bmatrix}6 \\ 14 \\ 30\end{bmatrix}
\begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}6 \\ 14 \\ 30\end{bmatrix}
|A| = \begin{vmatrix}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7\end{vmatrix}
= 2 - 4 + 2
= 0
Let C_ijbe the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & 3 \\ 4 & 7\end{vmatrix} = 2, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & 3 \\ 1 & 7\end{vmatrix} = - 4 , C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & 2 \\ 1 & 4\end{vmatrix} = 2
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 4 & 7\end{vmatrix} = - 3, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 1 & 7\end{vmatrix} = 6, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 1 & 4\end{vmatrix} = - 3
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 2 & 3\end{vmatrix} = 1 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 1 & 3\end{vmatrix} = - 2 , C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 1 & 2\end{vmatrix} = 1
adj A = \begin{bmatrix}2 & - 4 & 2 \\ - 3 & 6 & - 3 \\ 1 & - 2 & 1\end{bmatrix}^T
= \begin{bmatrix}2 & - 3 & 1 \\ - 4 & 6 & - 2 \\ 2 & - 3 & 1\end{bmatrix}
\left( adj A \right)B = \begin{bmatrix}2 & - 3 & 1 \\ - 4 & 6 & - 2 \\ 2 & - 3 & 1\end{bmatrix}\begin{bmatrix}6 \\ 14 \\ 30\end{bmatrix}
= \begin{bmatrix}12 - 42 + 30 \\ - 24 + 84 - 60 \\ 12 - 42 + 30\end{bmatrix}
= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}
Therefore, the system is consistent and has infinitely many solutions.
Hence proved.

(vi) 2x + 2y − 2z = 1

4x + 4y − z = 2

6x + 6y + 2z = 3

Solution:

Here,
2x + 2y − 2z = 1
4x + 4y − z = 2
6x + 6y + 2z = 3
A = \begin{bmatrix}2 & 2 & - 2 \\ 4 & 4 & - 1 \\ 6 & 6 & 2\end{bmatrix} , X = \begin{bmatrix}x \\ y \\ z\end{bmatrix} and B = \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}
\begin{bmatrix}2 & 2 & - 2 \\ 4 & 4 & - 1 \\ 6 & 6 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}
|A| = \begin{vmatrix}2 & 2 & - 2 \\ 4 & 4 & - 1 \\ 6 & 6 & 2\end{vmatrix}
= 2 (8 + 6) - 2 (8 + 6) - 2 (24 - 24)
= 28 - 28 - 0
= 0
Let C_ij be the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}4 & - 1 \\ 6 & 2\end{vmatrix} = 14, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}4 & - 1 \\ 6 & 2\end{vmatrix} = - 14, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}4 & 4 \\ 6 & 6\end{vmatrix} = 0
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}2 & - 2 \\ 6 & 2\end{vmatrix} = - 16, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & - 2 \\ 6 & 2\end{vmatrix} = 16, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & 2 \\ 6 & 6\end{vmatrix} = 0
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}2 & - 2 \\ 4 & - 1\end{vmatrix} = 6, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & - 2 \\ 4 & - 1\end{vmatrix} = - 6, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & 2 \\ 4 & 4\end{vmatrix} = 0
adj A = \begin{bmatrix}14 & - 14 & 0 \\ - 16 & 16 & 0 \\ 6 & - 6 & 0\end{bmatrix}^T
= \begin{bmatrix}14 & - 16 & 6 \\ - 14 & 16 & - 6 \\ 0 & 0 & 0\end{bmatrix}
(adj A) B = \begin{bmatrix}14 & - 16 & 6 \\ - 14 & 16 & - 6 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}
= \begin{bmatrix}14 - 32 + 18 \\ - 14 + 32 - 18 \\ 0\end{bmatrix}
= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}
Therefore, the system is consistent and has infinitely many solutions.
Hence proved.

Question 4. Show that each one of the following systems of linear equation is inconsistent:

(i) 2x + 5y = 7

6x + 15y = 13

Solution:

The given system of equations can be expressed as follows:
AX = B
Here,
A = \begin{bmatrix}2 & 5 \\ 6 & 15\end{bmatrix} , X = \binom{x}{y} and B = \binom{7}{13}
Now,
|A| = \begin{vmatrix}2 & 5 \\ 6 & 15\end{vmatrix}
= 30 - 30
= 0
Let C_ij be the cofactors of the elements a_ij in A.
C₁₁ = 15, C₁₂ = -6, C₂₁ = -5 and C₂₂ = 2
adj A = \begin{bmatrix}15 & - 6 \\ - 5 & 2\end{bmatrix}^T
= \begin{bmatrix}15 & - 5 \\ - 6 & 2\end{bmatrix}
(adj A) B = \begin{bmatrix}15 & - 5 \\ - 6 & 2\end{bmatrix}\binom{7}{13}
= \binom{105 - 65}{ - 42 + 26}
= \binom{40}{ - 16} ≠ 0
Therefore, the given system of equations is inconsistent.
Hence proved.

(ii) 2x + 3y = 5

6x + 9y = 10

Solution:

The given system of equations can be expressed as follows:
AX = B
Here,
A = \begin{bmatrix}2 & 3 \\ 6 & 9\end{bmatrix} , X = \binom{x}{y} and B = \binom{5}{10}
|A| = \begin{vmatrix}2 & 3 \\ 6 & 9\end{vmatrix}
= 18 - 18
= 0
Let C_ij be the cofactors of the elements a_ijin A.
C₁₁ = 9, C₁₂ = -6, C₂₁ = -3 and C₂₂ = 2
adj A = \begin{bmatrix}9 & - 6 \\ - 3 & 2\end{bmatrix}^T
= \begin{bmatrix}9 & - 3 \\ - 6 & 2\end{bmatrix}
(adj A) B = \begin{bmatrix}9 & - 3 \\ - 6 & 2\end{bmatrix}\binom{5}{10}
= \binom{45 - 30}{ - 30 + 20}
= \binom{15}{ - 10} ≠ 0
Therefore, the given system of equations is inconsistent.
Hence proved.

(iii) 4x − 2y = 3

6x − 3y = 5

Solution:

The given system of equations can be expressed as,
AX = B
Here,
A = \begin{bmatrix}4 & - 2 \\ 6 & - 3\end{bmatrix} , X = \binom{x}{y} and B = \binom{3}{5}
|A| = \begin{vmatrix}4 & - 2 \\ 6 & - 3\end{vmatrix}
= 12 - 12
= 0
Let C_ij be the cofactors of the elements a_ij in A.
C₁₁ = -3, C₁₂ = -6, C₂₁ = 2 and C₂₂ = 4
adj A = \begin{bmatrix}- 3 & - 6 \\ 2 & 4\end{bmatrix}^T
= \begin{bmatrix}- 3 & 2 \\ - 6 & 4\end{bmatrix}
(adj A) B = \begin{bmatrix}- 3 & 2 \\ - 6 & 4\end{bmatrix}\binom{3}{5}
= \binom{ - 9 + 10}{ - 18 + 20}
= \binom{1}{2} ≠ 0
Therefore, the given system of equations is inconsistent.
Hence proved.

(iv) 4x − 5y − 2z = 2

5x − 4y + 2z = −2

2x + 2y + 8z = −1

Solution:

The given system of equations can be written as,
AX = B
Here,
A = \begin{bmatrix}4 & - 5 & - 2 \\ 5 & - 4 & 2 \\ 2 & 2 & 8\end{bmatrix} , X = \begin{bmatrix}x \\ y \\ z\end{bmatrix} and B = \begin{bmatrix}2 \\ - 2 \\ - 1\end{bmatrix}
|A| = \begin{vmatrix}4 & - 5 & - 2 \\ 5 & - 4 & 2 \\ 2 & 2 & 8\end{vmatrix}
= -144 + 180 - 36
= 0
Let C_ij be the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 4 & 2 \\ 2 & 8\end{vmatrix} = 28, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}5 & 2 \\ 2 & 8\end{vmatrix} = - 36, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}5 & - 4 \\ 2 & 2\end{vmatrix} = 18
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 5 & - 2 \\ 2 & 8\end{vmatrix} = 36 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}4 & - 2 \\ 2 & 8\end{vmatrix} = 36 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}4 & - 5 \\ 2 & 2\end{vmatrix} = - 18
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 5 & - 2 \\ - 4 & 2\end{vmatrix} = - 18, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}4 & - 2 \\ 5 & 2\end{vmatrix} = - 18, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}4 & - 5 \\ 5 & - 4\end{vmatrix} = 9
adj A = \begin{bmatrix}28 & - 36 & 18 \\ 36 & 36 & - 18 \\ - 18 & - 18 & 9\end{bmatrix}^T
= \begin{bmatrix}28 & 36 & - 18 \\ - 36 & 36 & - 18 \\ 18 & - 18 & 9\end{bmatrix}
(adj A) B = \begin{bmatrix}28 & 36 & - 18 \\ - 36 & 36 & - 18 \\ 18 & - 18 & 9\end{bmatrix}\begin{bmatrix}2 \\ - 2 \\ - 1\end{bmatrix}
= \begin{bmatrix}56 - 72 + 18 \\ - 72 - 72 + 18 \\ 36 + 36 - 9\end{bmatrix}
= \begin{bmatrix}2 \\ - 126 \\ 63\end{bmatrix} ≠ 0
Therefore, the given system of equations is inconsistent.
Hence proved.

(v) 3x − y − 2z = 2

2y − z = −1

3x − 5y = 3

Solution:

The given system of equations can be written as,
AX = B
Here,
A = \begin{bmatrix}3 & - 1 & - 2 \\ 0 & 2 & - 1 \\ 3 & - 5 & 0\end{bmatrix} , X = \begin{bmatrix}x \\ y \\ z\end{bmatrix} and B = \begin{bmatrix}2 \\ - 1 \\ 3\end{bmatrix}
|A| = \begin{vmatrix}3 & - 1 & - 2 \\ 0 & 2 & - 1 \\ 3 & - 5 & 0\end{vmatrix}
= -15 + 3 + 12
= 0
Let C_ijbe the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & - 1 \\ - 5 & 0\end{vmatrix} = - 5, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}0 & - 1 \\ 3 & 0\end{vmatrix} = - 3, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}0 & 2 \\ 3 & - 5\end{vmatrix} = - 6
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 1 & - 2 \\ - 5 & 0\end{vmatrix} = 10, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}3 & - 2 \\ 3 & 0\end{vmatrix} = 6, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}3 & - 1 \\ 3 & - 5\end{vmatrix} = 12
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 1 & - 2 \\ 2 & - 1\end{vmatrix} = 5, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}3 & - 2 \\ 0 & - 1\end{vmatrix} = 3, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}3 & - 1 \\ 0 & 2\end{vmatrix} = 6
adj A = \begin{bmatrix}- 5 & - 3 & - 6 \\ 10 & 6 & 12 \\ 5 & 3 & 6\end{bmatrix}^T
= \begin{bmatrix}- 5 & 10 & 5 \\ - 3 & 6 & 3 \\ - 6 & 12 & 6\end{bmatrix}
(adj A) B = \begin{bmatrix}- 5 & 10 & 5 \\ - 3 & 6 & 3 \\ - 6 & 12 & 6\end{bmatrix}\begin{bmatrix}2 \\ - 1 \\ 3\end{bmatrix}
= \begin{bmatrix}- 10 - 10 + 15 \\ - 6 - 6 + 9 \\ - 12 - 12 + 18\end{bmatrix}
= \begin{bmatrix}- 5 \\ - 3 \\ - 6\end{bmatrix} ≠ 0
Therefore, the given system of equations is inconsistent.
Hence proved.

(vi) x + y − 2z = 5

x − 2y + z = −2

−2x + y + z = 4

Solution:

The given system of equations can be written as,
AX = B
Here,
A = \begin{bmatrix}1 & 1 & - 2 \\ 1 & - 2 & 1 \\ - 2 & 1 & 1\end{bmatrix} , X = \begin{bmatrix}x \\ y \\ z\end{bmatrix} and B = \begin{bmatrix}5 \\ - 2 \\ 4\end{bmatrix}
|A| = \begin{vmatrix}1 & 1 & - 2 \\ 1 & - 2 & 1 \\ - 2 & 1 & 1\end{vmatrix}
= - 3 - 3 + 6
= 0
Let C_ijbe the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 2 & 1 \\ 1 & 1\end{vmatrix} = - 3 , C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & 1 \\ - 2 & 1\end{vmatrix} = - 3, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & - 2 \\ - 2 & 1\end{vmatrix} = - 3
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & - 2 \\ 1 & 1\end{vmatrix} = - 3, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & - 2 \\ - 2 & 1\end{vmatrix} = - 3 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ - 2 & 1\end{vmatrix} = - 3
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & - 2 \\ - 2 & 1\end{vmatrix} = - 3, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & - 2 \\ 1 & 1\end{vmatrix} = - 3, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 1 & - 2\end{vmatrix} = - 3
adj A = \begin{bmatrix}- 3 & - 3 & - 3 \\ - 3 & - 3 & - 3 \\ - 3 & - 3 & - 3\end{bmatrix}^T
= \begin{bmatrix}- 3 & - 3 & - 3 \\ - 3 & - 3 & - 3 \\ - 3 & - 3 & - 3\end{bmatrix}
(adj A) B = \begin{bmatrix}- 3 & - 3 & - 3 \\ - 3 & - 3 & - 3 \\ - 3 & - 3 & - 3\end{bmatrix}\begin{bmatrix}5 \\ - 2 \\ 4\end{bmatrix}
= \begin{bmatrix}- 15 + 6 - 12 \\ - 15 + 6 - 12 \\ - 15 + 6 - 12\end{bmatrix}
= \begin{bmatrix}- 21 \\ - 21 \\ - 21\end{bmatrix} ≠ 0
Therefore, the given system of equations is inconsistent.
Hence proved.

Question 5. If A = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix} and B = \begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix} are two square matrices, find AB and hence solve the system of linear equations: x − y = 3, 2x + 3y + 4z = 17, y + 2z = 7.

Solution:

Here,
A = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix} and B = \begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}
Now,
AB = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix} \begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}
AB = \begin{bmatrix}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{bmatrix}
AB = 6\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
AB = 6I
\frac{1}{6}AB = I
A^{- 1} = \frac{1}{6}B
A^{- 1} = \frac{1}{6}\begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}
X = A^-1 B
X = \frac{1}{6}\begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}\begin{bmatrix}3 \\ 17 \\ 7\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{6}\begin{bmatrix}6 + 34 - 28 \\ - 12 + 34 - 28 \\ 6 - 17 + 35\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{6}\begin{bmatrix}12 \\ - 6 \\ 24\end{bmatrix}
Therefore x = 2, y = -1 and z = 4.

Question 6. If A = \begin{bmatrix}2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2\end{bmatrix} , find A⁻¹ and hence solve the system of linear equations 2x − 3y + 5z = 11, 3x + 2y − 4z = −5, x + y + 2z = −3.

Solution:

Here,
A = \begin{bmatrix}2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2\end{bmatrix}
|A| = \begin{vmatrix}2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2\end{vmatrix}
= 0 - 6 + 5
= -1
Let C_ijbe the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & - 4 \\ 1 & - 2\end{vmatrix} = 0, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}3 & - 4 \\ 1 & - 2\end{vmatrix} = 2, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}3 & 2 \\ 1 & 1\end{vmatrix} = 1
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 3 & 5 \\ 1 & - 2\end{vmatrix} = - 1, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & 5 \\ 1 & - 2\end{vmatrix} = - 9, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & - 3 \\ 1 & 1\end{vmatrix} = - 5
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 3 & 5 \\ 2 & - 4\end{vmatrix} = 2 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & 5 \\ 3 & - 4\end{vmatrix} = 23, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & - 3 \\ 3 & 2\end{vmatrix} = 13
adj A = \begin{bmatrix}0 & 2 & 1 \\ - 1 & - 9 & - 5 \\ 2 & 23 & 13\end{bmatrix}^T
= \begin{bmatrix}0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{- 1}\begin{bmatrix}0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13\end{bmatrix}
\begin{bmatrix}2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}11 \\ - 5 \\ - 3\end{bmatrix}
X = A^-1 B
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 1}\begin{bmatrix}0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13\end{bmatrix}\begin{bmatrix}11 \\ - 5 \\ - 3\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 1}\begin{bmatrix}0 + 5 - 6 \\ 22 + 45 - 69 \\ 11 + 25 - 39\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 1}\begin{bmatrix}- 1 \\ - 2 \\ - 3\end{bmatrix}
=> x = - 1/- 1, y = -2/-1\ and z = -3/-1
Therefore x = 1, y = 2 and z = 3.

Question 7. Find A⁻¹, if A = \begin{bmatrix}1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1\end{bmatrix} . Hence solve the following system of linear equations: x + 2y + 5z = 10, x − y − z = −2, 2x + 3y − z = −11.

Solution:

A = \begin{bmatrix}1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1\end{bmatrix}
|A| = \begin{vmatrix}1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1\end{vmatrix}
= 4 - 2 + 25
= 27
Let C_ij be the cofactors of the elements a_ij in A.
C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 1 & - 1 \\ 3 & - 1\end{vmatrix} = 4, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & - 1 \\ 2 & - 1\end{vmatrix} = - 1, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & - 1 \\ - 2 & 3\end{vmatrix} = 5
C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}2 & 5 \\ 3 & - 1\end{vmatrix} = 17, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 5 \\ 2 & - 1\end{vmatrix} = - 11, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 2 \\ 2 & 3\end{vmatrix} = 1
C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}2 & 5 \\ - 1 & - 1\end{vmatrix} = 3, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 5 \\ 1 & - 1\end{vmatrix} = 6, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 2 \\ 1 & - 1\end{vmatrix} = - 3
adj A = \begin{bmatrix}4 & - 1 & 5 \\ 17 & - 11 & 1 \\ 3 & 6 & - 3\end{bmatrix}^T
= \begin{bmatrix}4 & 17 & 3 \\ - 1 & - 11 & 6 \\ 5 & 1 & - 3\end{bmatrix}
A^{- 1} = \frac{1}{\left| A \right|}adj A
= \frac{1}{27}\begin{bmatrix}4 & 17 & 3 \\ - 1 & - 11 & 6 \\ 5 & 1 & - 3\end{bmatrix}
\begin{bmatrix}1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}10 \\ - 2 \\ - 11\end{bmatrix}
X = A^-1 B
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{27}\begin{bmatrix}4 & 17 & 3 \\ - 1 & - 11 & 6 \\ 5 & 1 & - 3\end{bmatrix}\begin{bmatrix}10 \\ - 2 \\ - 11\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{27}\begin{bmatrix}40 - 34 - 33 \\ - 10 + 22 - 66 \\ 50 - 2 + 33\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{27}\begin{bmatrix}- 27 \\ - 54 \\ 81\end{bmatrix}
=> x = -27/27, y = -54/27 and z = 81/27
Therefore, x = - 1, y = -2 and z = 3.

Practice Questions

1).Solve the following system of equations using matrix method:

2x + 3y = 8

4x - y = 1

2).Use Cramer's rule to solve:

x + 2y = 5

3x - y = 2

3).For what value of k will the following system have no solution?

2x + 3y = 7

6x + ky = 21

4).Solve the system of equations:

ax + by = p

cx + dy = q

Where a, b, c, d, p, and q are constants and ad - bc ≠ 0

5).Determine the nature of solutions for:

2x - 3y = 4

4x - 6y = 8

Summary

Exercise 8.1 in Set 1 of Chapter 8 typically focuses on solving systems of linear equations using matrix methods. Key topics usually include:

Expressing systems of linear equations in matrix form
Solving systems using inverse matrix method
Applying Cramer's rule to solve systems of equations
Determining the nature of solutions (unique, infinite, or no solution)
Solving problems involving parameters in the equations

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