Practice Problems on Determinant using Elementary Operations

Using the properties of determinants and without expanding in Exercise 1 to 7, prove that:

Question 1. \begin{vmatrix}x & a & x+a\\y & b & y+b\\z &c & z+c\end{vmatrix}=0

Solution:

L.H.S.= \begin{vmatrix}x & a & x+a\\y & b & y+b\\z &c & z+c\end{vmatrix}

C ₁ →C ₁ +C ₂

= \begin{vmatrix}x+a & a & x+a\\y+b & b & y+b\\z+c &c & z+c\end{vmatrix}

According to Properties of Determinant

=0                                [∵  C ₁ & C ₃ are identical]

Now, L.H.S.=R.H.S.

Hence Proved

Question 2. \begin{vmatrix}a-b & b-c & c-a\\b-c & c-a & a-b\\c-a & a-b & b-c\end{vmatrix}=0

Solution:

L.H.S.= \begin{vmatrix}a-b & b-c & c-a\\b-c & c-a & a-b\\c-a & a-b & b-c\end{vmatrix}

\begin{array}{l} \text { } \qquad\mathrm{C}{1} \rightarrow \mathrm{C}{1}+\mathrm{C}{2}+\mathrm{C}{3} \\ =\left|\begin{array}{lll} a-b+b-c+c-a & b-c & c-a \\ b-c+c-a+a-b & c-a & a-b \\ c-a+a-b+b-c & a-b & b-c \end{array}\right| \\ =\left|\begin{array}{lll} 0 & b-c & c -a \\ 0 & c-a & a-b \\ 0 & a-b & b-c \end{array}\right|=0 \end{array}

=0                        [∵ Every element of C ₁ are 0]

Now,  L.H.S.=R.H.S.

Hence Proved

Question 3. \begin{vmatrix}2 & 7 & 65\\3 & 8 & 75\\5 & 9 & 86\end{vmatrix}=0

Solution:

L.H.S.= \begin{vmatrix}2 & 7 & 65\\3 & 8 & 75\\5 & 9 & 86\end{vmatrix}

C ₃ →C ₃ -C ₁

= \begin{vmatrix}2 & 7 & 63\\3 & 8 & 72\\5 & 9 & 81\end{vmatrix}

= 2\begin{vmatrix}2 & 7 & 7\\3 & 8 & 8\\5& 9 & 9\end{vmatrix}

=9 ×0=0          [∵C ₂ & C ₃ are identical]

Now,  L.H.S.=R.H.S.

Hence Proved

Question 4. \begin{vmatrix}1 & bc & a(b+c)\\1 & ca & b(c+a)\\1 & ab & c(a+b)\end{vmatrix}=0

Solution:

L.H.S.= \begin{vmatrix}1 & bc & a(b+c)\\1 & ca & b(c+a)\\1 & ab & c(a+b)\end{vmatrix}

= \text {  } \mathrm{C}{3} \rightarrow \mathrm{C}{3}+\mathrm{C}_{2}\\\\ \text {=  }\begin{vmatrix}1 & bc & ab+ac\\1 & ca & bc+ba\\1 & ab & ca+cb\end{vmatrix}\\\\ \text {=  }\begin{vmatrix}1 & bc & ab+ac+bc\\1 & ca & bc+ba+ca\\1 & ab & ca+cb+ab\end{vmatrix}\\\\ \text {=  }(ab+ac+bc)\begin{vmatrix}1 & bc & 1\\1 & ca &1\\1 & ab & 1\end{vmatrix}\\\\ \text {=  }(ab+ac+bc)0\\\\ \text {= 0 }\qquad\qquad[∵Two \:columns\: are\: identical]

Now,  L.H.S.=R.H.S.

Hence Proved

Question 5. \begin{vmatrix}b+c & q+r & y+z\\c+a & r+p & z+x\\a+b & p+q & x+y\end{vmatrix}=2\begin{vmatrix}a& p & x\\b & q & y\\c & r & z\end{vmatrix}

Solution:

L.H.S.= \begin{vmatrix}b+c & q+r & y+z\\c+a & r+p & z+x\\a+b & p+q & x+y\end{vmatrix}

\begin{array}{l} \text {  } \qquad\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3} \\ =\left|\begin{array}{lll} b+c+c+a+a+b & q+r+r+p+p+q & y+z+z+x+x+y \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right|\\\\ =\left|\begin{array}{lll} 2(a+b+c) & 2(p+q+r) & 2(x+y+z) \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right| \\\\ =\left|\begin{array}{lll} (a+b+c) & (p+q+r) & (x+y+z) \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right| \end{array}

\begin{aligned} &\text {  } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}\\ &=2\left|\begin{array}{lcc} b & q & y \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right|\\ &\text {  } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\\ &=2\left|\begin{array}{ccc} b & q & y \\ c+a & r+p & z+x \\ a & p & x \end{array}\right|\\ &\text {  } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\\ &=2\left|\begin{array}{lll} b & q & y \\ c & r & z \\ a & p & x \end{array}\right|\\ &\text {Interchange  } \mathrm{R}_{2} \text {  and } \mathrm{R}_{3} \text {  }\\ &=-2\left|\begin{array}{ccc} b & q & y \\ a & p & x \\ c & r & z \end{array}\right| \end{aligned}

=-(-2)\begin{vmatrix}a & p & q\\b & q & y\\c & r & z\end{vmatrix}\\\\ =2\begin{vmatrix}a & p & q\\b & q & y\\c & r & z\end{vmatrix}\\\\

Now, L.H.S.=R.H.S.

Hence Proved

Question 6. \begin{vmatrix}0& a & -b\\-a & 0 & -c\\b & c & 0\end{vmatrix}=0

Solution:

Let Δ= \begin{vmatrix}0& a & -b\\-a & 0 & -c\\b & c & 0\end{vmatrix}

Taking (-1) common from every row

Δ=(-1) ³ \begin{vmatrix}0& -a & b\\a & 0 & c\\-b & -c & 0\end{vmatrix}

Interchange rows and columns

Δ=- \begin{vmatrix}0& a & -b\\-a & 0 & -c\\b & c & 0\end{vmatrix}

Now, Δ=-Δ

Δ+Δ=0

2Δ=0

Δ=0

Now,  L.H.S.=R.H.S.

Hence Proved

Question 7. \begin{vmatrix}-a^{2}& ab& ac\\ba & -b^{2} & bc\\ca & cb & -c^{2}\end{vmatrix}=4a^{2}b^{2}c^{2}

Solution:

L.H.S.= \begin{vmatrix}-a^{2}& ab& ac\\ba & -b^{2} & bc\\ca & cb & -c^{2}\end{vmatrix}

Taking common a from Row 1,

b from Row 2,

c from Row 3, we have

\begin{array}{l} =a b c\left|\begin{array}{ccc} -a & b & c \\ a & -b & b \\ a & b & -c \end{array}\right| \\ \\\text { } R_{1} \rightarrow R_{1}+R_{2} \\\\ =a b c\left|\begin{array}{ccc} 0 & 0 & 2 c \\ a & -b & c \\ a & b & -c \end{array}\right| \\\\ =a b c \cdot 2 c\left|\begin{array}{cc} a & -b \\ a & b \end{array}\right| \\\\ =a b c \cdot 2 c(a b+a b) \\\\ =a b c 2 c .2 a b=4 a^{2} b^{2} c^{2} \end{array}

Now,  L.H.S.=R.H.S.

Hence Proved

By using properties of determinants, in Exercises 8 to 14, show that:

Question 8(i). \begin{vmatrix}1 & a & a^{2}\\1 & b & b^{2}\\1 & c & c^{2}\end{vmatrix}=(a-b)(b-c)(c-a)

(ii) \begin{vmatrix}1 & 1 & 1\\a & b & c\\a^{3} &b^{3}  & c^{3}\end{vmatrix}=(a-b)(b-c)(c-a)(a+b+c)

Solution:

(i) L.H.S.= \begin{vmatrix}1 & a & a^{2}\\1 & b & b^{2}\\1 & c & c^{2}\end{vmatrix}

\begin{aligned} &R _{2} \rightarrow R _{2}- R _{1} \text { and } R _{3} \rightarrow R _{3}- R _{1}\\ &=\left|\begin{array}{ccc} 1 & a & a^{2} \\ 0 & b-a & b^{2}-a^{2} \\ 1 & c-a & c^{2}-a^{2} \end{array}\right|\\ &\left|\begin{array}{ccc} 1 & a & a^{2} \\ 0 & b-a & (b-a)(b+a) \\ 0 & c-a & (c-a)(c+a) \end{array}\right|\\ &\text { Takina (b-a) and (c-a) common from } R _{2} \text { and } R _{3} \text { respective }\\ &=(b-a)(c-a)\left|\begin{array}{ccc} 1 & a & a^{2} \\ 0 & 1 & (b+a) \\ 0 & 1 & (c+a) \end{array}\right|\\ &R_{2} \rightarrow R_{2}-R_{3}\\ &=(b-a)(c-a)\left|\begin{array}{ccc} 1 & a & a^{2} \\ 0 & 0 & (b-c) \\ 0 & 1 & (c+a) \end{array}\right| \end{aligned}

\begin{aligned} &\text { Expanding along } 1^{\text {st }} \text { column }\\ &=(b-a)(c-a)  \begin{array}{ll} =\left|\begin{array}{lll} 0 & (b+a) \\ 1 & (c+a) \end{array}\right| \end{array}\\ &=(b-a)(c-a)(0-(b-c))\\ &=(b-a)(c-a)(c-b)\\ &=(a-b)(b-c)(c-a)\text {   }\\ &=0 \end{aligned}

Now,  L.H.S.=R.H.S.

Hence Proved

(ii) L.H.S.= \begin{vmatrix}1 & 1 & 1\\a & b & c\\a^{3} &b^{3}  & c^{3}\end{vmatrix}

\begin{aligned} &\text { operating } C _{2} \rightarrow C _{2}- C _{1} \text { and } C _{3} \rightarrow C _{3}- C _{3}\\ &=\left|\begin{array}{ccc} 1 & 0 & 0 \\ a & b-a & c-a \\ a^{3} & b^{3}-a^{3} & c^{3}-a^{3} \end{array}\right|\\ &=1\left|\begin{array}{cc} b-a & c-a \\ b^{3}-a^{3} & c^{3}-a^{3} \end{array}\right|\\ &=1\left|\begin{array}{cc} b-a & c-a \\ (b-a)\left(b^{2}+a^{2}+a b\right) & (c-a)\left(c^{2}+a^{2}+a c\right) \end{array}\right|\\ &=(b-a)(c-a) \mid\left(b^{2}+a^{2}+a b\right) \quad\left(c^{2}+a^{2}+a c\right)\\ &=(b-a)(c-a)\left(c^{2}+a^{2}+a c-b^{2}-a^{2}-a b\right)\\ &=(b-a)(c-a)\left(c^{2}-b^{2}+a c-a b\right)\\ &=(b-a)(c-a)[(c-b)(c+b)+a(c-b)]\\ &=(b-a)(c-a)(c-b)(c+b+a)\\ &=-(a-b)(c-a)[-(b-c)(c+b+a)]\\ &=(a-b)(b-c)(c-a)(a+b+c) \end{aligned}

Now,  L.H.S.=R.H.S.

Hence Proved

Question 9. \begin{vmatrix}x & x^{2} & yz\\y & y^{2} & zx\\z & z^{2} & xy\end{vmatrix}=(x-y)(y-z)(z-x)(xy+yz+zx)

Solution:

L.H.S.= \begin{vmatrix}x & x^{2} & yz\\y & y^{2} & zx\\z & z^{2} & xy\end{vmatrix}

R_{1} \rightarrow R_{1}-R_{2}

=\left|\begin{array}{ccc} (x-y) & (x+y)(x-y) & z(y-x) \\ y^{2} & y^{2} & z x \\ z &z^{2}& x y \end{array}\right|

\begin{aligned} &=(x-y)\left|\begin{array}{ccc} 1 & (x+y) & -2 \\ y^{2} & y^{2} x & 2 x \\ z & z^{2} & x y \end{array}\right|\\ &\begin{array}{l} \qquad R_{2} \rightarrow R_{2}-R_{3} \\ =(x-y)\left|\begin{array}{lll} 1 & (x+y) & -2 \\ y-2 & y^{2}-2^{2} & 2 x-x y \\ 2 & 2^{2} & x y \end{array}\right| \end{array}\\ &=(x-y)(y-2)\left|\begin{array}{ccc} 1 & (x+y) & -2 \\ 1 & (y+2) & 2 x-x y \\ 2 & z^{2} & x y \end{array}\right|\\\\ &=(x-y)(y-2)\left|\begin{array}{ccc} 1 & (x+y) & -z \\ 1 & (y+z) & -x \\ z & \left(z^{2}\right) & x y \end{array}\right|\\\\ &\begin{array}{l} R_{1} \rightarrow R_{1}-R_{2} \\ \end{array} \end{aligned}

=(x-y)(y-z)\left|\begin{array}{ccc} 0 & x-2 & -z+x \\ 1 & y+2 & -x \\ z & z^{2} & x y \end{array}\right|

=(x-y)(y-z)(z-x)\left|\begin{array}{ccc} 0 & -1 & -1 \\ 1 & y+z & -x \\ z & z^{2} & x y \end{array}\right|

=(x-y)(y-z)(z-x)\left|\begin{array}{ccc} 0 & 0 & -1 \\ 1 & x+y+z & -x \\ z & z^{2}-xy & x y \end{array}\right|\\\\ =(x-y)(y-z)(z-x)[-1(z^{2}-xy-zx-zy-z^{2}]\\\\ =(x-y)(y-z)(z-x)(xy+zx+zy)

Now, L.H.S.=R.H.S.

Hence Proved

Question 10.(i) \begin{vmatrix}x+4& 2x & 2x\\2x & x+4 & 2x\\2x & 2x & x+4\end{vmatrix}=(5x+4)(4-x)^{2}

(ii) \begin{vmatrix}y+k & y & y\\y & y+k & y\\y & y & y+k\end{vmatrix}=k^{2}(3y+k)

Solution:

(i) L.H.S.= \begin{vmatrix}x+4& 2x & 2x\\2x & x+4 & 2x\\2x & 2x & x+4\end{vmatrix}

\begin{aligned} &R_{1} \rightarrow R_{1}+R_{2}+R_{3}\\ &=\left|\begin{array}{ccc} 5 x+4 & 5 x+4 & 5 x+4 \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4 \end{array}\right|\\ &\text { Taking } (5 x+4) \text { common from } R_{1}\\ &=(5 x+4)\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4 \end{array} \right|.\\ &\text {} C _{2} \rightarrow C _{2}- C _{1} \text { and } \left. C _{3} \rightarrow C _{3}- C _{1}\right]\\ &=(5 x+4)\left|\begin{array}{ccc} 1 & 0 & 0 \\ 2 x & 4-x & 0 \\ 2 x & 0 & 4-x \end{array}\right|\\ &=(5 x+4) \cdot 1\left|\begin{array}{cc} 4-x & 0 \\ 0 & 4-x \end{array}\right|\\ &=(5 x+4)(4- x^{2}) \end{aligned}

Now,  L.H.S.=R.H.S.

Hence Proved

(ii) L.H.S.= \begin{vmatrix}y+k & y & y\\y & y+k & y\\y & y & y+k\end{vmatrix}

\begin{aligned} & C _{1} \rightarrow C _{1}+ C _{2}+ C _{3}\\ &=\left|\begin{array}{ccc} 3 y+k & y & y \\ 3 y+k & y+k & y \\ 3 y+k & y & y+k \end{array}\right|\\ &\text { Taking } 3 y+k \text { common from } C_{1}\\ &=(3 y+k)\left|\begin{array}{ccc} 1 & y & y \\ 1 & y+k & y \\ 1 & y & y+k \end{array}\right|\\ &\text {  } C _{2} \rightarrow C _{2}- C _{1} \text { and } C _{3} \rightarrow C _{3}-C_{1} \end{aligned}

\begin{array}{l} =(3 y+k)\left|\begin{array}{lll} 1 & y & y \\ 0 & k & 0 \\ 0 & 0 & k \end{array}\right| \\ =(3 y+k) \cdot 1\left|\begin{array}{ll} k & 0 \\ 0 & k \end{array}\right| \\ =(3 y+k) k^{2}=k^{2}(3 y+k) \end{array}

Now,  L.H.S.=R.H.S.

Hence Proved

Question 11. (i)\begin{array}{ccc} \left|\begin{array}{lll} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right| \end{array} \mid=(a+b+c)^{3}

(ii)\left|\begin{array}{ccc} x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y \end{array}\right|=2(x+y+z)^{3}

Solution:

(i) L.H.S.=\begin{array}{ccc} \left|\begin{array}{lll} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right| \end{array} \mid

\begin{aligned} &R_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} a+b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|\\ &\text { Taking } a+b+c \text { common from } R_{1}\\ &=(a+b+c)\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|\\ &[C_{2} \rightarrow C_{2}-C_{1} \text { and } C_{3} \rightarrow C_{3}-C_{1} \end{aligned}

\begin{array}{l} =(a+b+c)\left|\begin{array}{ccc} 1 & 0 & 0 \\ 2 b & -b-c-a & 0 \\ 2 c & 0 & -c-a-b \end{array}\right| \\ =(a+b+c) \cdot 1  \begin{array}{cc} =\left|\begin{array}{lll} -b-c-a & 0 \\ 0 & -c-a-b \mid \end{array}\right| \end{array} \\ =(a+b+c)\{-(b+c+a)\}\{-(c+a+b)\} \\ =  |a+b+c|^{3} \text {  } \end{array}

∵ L.H.S.=R.H.S

Hence proved

(ii) \left|\begin{array}{ccc} x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y \end{array}\right|

\begin{aligned} & C _{1} \rightarrow C _{1}+ C _{2}+ C _{3}\\ &=\left|\begin{array}{ccc} 2(x+y+z) & x & y \\ 2(x+y+z) & y+z+2 x & y \\ 2(x+y+z) & x & z+x+2 y \end{array}\right|\\ &\text { Taking } 2(x+y+z) \text { common from } C_{1}\\ &=2(x+y+z)\left|\begin{array}{ccc} 1 & x & y \\ 1 & y+z+2 x & y \\ 1 & x & z+x+2 y \end{array}\right|\\ & R _{2} \rightarrow R _{2}- R _{1} \text { and } R _{3} \rightarrow R _{3}- R _{1} \end{aligned}

\begin{array}{l} =2(x+y+z)\left|\begin{array}{ccc} 1 & x & y \\ 0 & x+y+z & 0 \\ 0 & 0 & x+y+z \end{array}\right| \\ =2(x+y+z) \cdot 1 \cdot\left|\begin{array}{cc} x+y+z & 0 \\ 0 & x+y+z \end{array}\right| \\ =2(x+y+z)\left[(x+y+z)^{2}-0\right] \\ =2(x+y+z)^{3} \text {  } \end{array}

∵ L.H.S.=R.H.S

Hence proved

Question 12. \begin{aligned} &\text {  }\left|\begin{array}{lll} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right|\\ \end{aligned}

Solution:

\begin{aligned} &\text { L.H.S. }=\left|\begin{array}{lll} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right|\\ &R_{1} \rightarrow R_{1}+R_{2}+R_{3}\\ &=\left|\begin{array}{ccc} 1+x+x^{2} & 1+x+x^{2} & 1+x+x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right| \\ &\text { Taking } 1+x+x^{2} \text { common from } R_{1}\\ &=\left(1+x+x^{2}\right)\left|\begin{array}{ccc} 1 & 1 & 1 \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right|\\ &\left[C_{2} \rightarrow C_{2}-C_{1}\right. \& C_{3} \rightarrow C_{3}-C_{1}\\ &=\left|1+x+x^{2}\right| \begin{array}{ccc} =\left|\begin{array}{lll} 1 & 0 & 0 \\ x^{2} & 1-x^{2} & x-x^{2} \\ x & x-x & 1-x \end{array}\right| \end{array} \mid \end{aligned}

\begin{array}{l} =\left(1+x+x^{2}\right) \cdot 1 \cdot\left|\begin{array}{ll} 1-x^{2} & x-x^{2} \\ x^{2}-x & 1-x \end{array}\right| \\ =\left(1+x+x^{2}\right) \cdot 1 \cdot \mid \begin{array}{ll} (1-x)(1+x) & x(1-x) \\ -x(1-x) & 1-x \end{array} \\ =\left(1+x+x^{2}\right)\left[(1-x)^{2}(1+x)+x^{2}(1-x)^{2}\right] \\ =\left(1+x+x^{2}\right)(1-x)^{2}\left(1+x+x^{2}\right) \\ =\left(1+x+x^{2}\right)^{2}(1-x)^{2} \\ =\left[\left(1+x+x^{2}\right)(1-x)\right]^{2} \\ =\left(1-x+x-x^{2}+x^{2}-x^{2}\right)^{2} \\ =\left(1-x^{3}\right)^{2} \quad \text {  } \end{array}

∵ L.H.S.=R.H.S

Hence proved

Question 13. \left|\begin{array}{ccc} 1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2} \end{array}\right|=\left(1+a^{2}+b^{2}\right)^{3}

Solution:

L.H.S.= \left|\begin{array}{ccc} 1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2} \end{array}\right|

\begin{array}{l} C _{1} \rightarrow C _{1}-b C _{3} \text { and } \left. C _{2} \rightarrow C _{2}+ aC _{3}\right] \\ =\begin{array}{ccc} \left|\begin{array}{lll} 1+a^{2}+b^{2} & 0 & -2 b \\ 0 & 1+a^{2}+b^{2} & 2 a \\ b\left(1+a^{2}+b^{2}\right) & -a\left(1+a^{2}+b^{2}\right) & 1-a^{2}-b^{2} \end{array}\right| \end{array}  \\ =\left(1+a^{2}+b^{2}\right)\left|\begin{array}{ccc} 1 & 0 & -2 b \\ 0 & 1 & 2 a \\ b & -a & 1-a^{2}-b^{2} \end{array}\right| \\ {\left[ R _{1} \rightarrow R _{1}-bR_{1} \right]} \end{array}

\begin{array}{l} =\left(1+a^{2}+b^{2}\right)^{2}\left|\begin{array}{ccc} 1 & 0 & -2 b \\ 0 & 1 & 2 a \\ 0 & -a & 1-a^{2}+b^{2} \end{array}\right| \\ =\left(1+a^{2}+b^{2}\right)^{2}\left|\begin{array}{cc} 1 & 2 a \\ -a & 1-a^{2}+b^{2} \end{array}\right| \\ =\left(1+a^{2}+b^{2}\right)^{2}\left(1-a^{2}+b^{2}+2 a^{2}\right) \\ =\left(1+a^{2}-b^{2}\right)^{3} \end{array}

∵ L.H.S.=R.H.S

Hence proved

Question 14. \left|\begin{array}{ccc} a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1 \end{array}\right|=1+a^{2}+b^{2}+c^{2}

Solution:

L.H.S= \left|\begin{array}{ccc} a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1 \end{array}\right|

Multiplying column1,column2,column3  by  a,b,c respectively and then dividing the determinant by  a,b,c .

\begin{aligned} &=\frac{1}{a b c}\left|\begin{array}{ccc} a\left(a^{2}+1\right) & a b^{2} & a c^{2} \\ a^{2} b & b\left(b^{2}+1\right) & b c^{2} \\ a^{2} c & b^{2} c & c\left(c^{2}+1\right) \end{array}\right|\\ &\text { Taking a } b \text { and } c \text { common from } R_{1}, R_{2} \text { and } R_{3} \text { respectively }\\ &=\frac{a b c}{a b c}  \begin{array}{ccc} =\left|\begin{array}{lll} a^{2}+1 & b^{2} & c^{2} \\ a^{2} & b^{2}+1 & c^{2} \\ a^{2} & b^{2} & c^{2}+1 \end{array}\right| \end{array}\\ &\left[ C _{1} \rightarrow C _{1}- C _{2}+ C _{3}\right]\\ &=\frac{a b c}{a b c}\left|\begin{array}{ccc} 1+a^{2}+b^{2}+c^{2} & b^{2} & c^{2} \\ 1+a^{2}+b^{2}+c^{2} & b^{2}-1 & c^{2} \\ 1-a^{2}+b^{2}-c^{2} & b^{2} & c^{2}+1 \end{array}\right| \end{aligned}

\begin{aligned} &=\left(1+a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc} 1 & b^{2} & c^{2} \\ 1 & b^{2}+1 & c^{2} \\ 1 & b^{2} & c^{2}+1 \end{array}\right|\\ &\left[ R _{2} \rightarrow R _{2}- R _{1} \text { and } R _{3} \rightarrow R _{3}- R _{1}\right]\\ &=\left(1+a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc} 1 & b^{2} & c^{2} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right|\\ &=\left(1+a^{2}+b^{2}+c^{2}\right)(1)(1-0)\\ &=1+a^{2}+b^{2}+c^{2}\\ &\text { } \end{aligned}

∵ L.H.S.=R.H.S

Hence proved

Choose the correct answer in Exercises 15 and 16.

Question 15. Let A be a square matrix of order 3 x 3, then|A| is equal to:

(A) k|A|

(B) k ² |A|

(C) k ³ |A|

(D) 3k|A|

Solution:

Let A =\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\\7 & 8 & 9\end{bmatrix}      be a square matrix of order 3 x 3   ….(1)

Now, kA=\begin{bmatrix}k1 & k2 & k3\\k4 & k5 & k6\\k7 & k8 & k9\end{bmatrix}

⇒|kA|=\begin{vmatrix}k1 & k2 & k3\\k4 & k5 & k6\\k7 & k8 & k9\end{vmatrix}

⇒|kA|=k ³ \begin{vmatrix}1 & 2 & 3\\4 & 5 & 6\\7 & 8 & 9\end{vmatrix}

∴ |kA|= k ³ |A|            [ from eq ⁿ .(1) ]

∴ option (C) is correct.

Question 16.Which of the following is correct

(A) Determinant is a square matrix.

(B) Determinant is a number associated to a matrix.

(C) Determinant is a number associated to a square matrix.

(D) None of these

Solution:

Since, Determinant is a number which is always associated to a square matrix.

∴ option (C) is correct.

somesh_barthwal

Follow

Improve

Article Tags :

• Mathematics
• School Learning
• Class 12
• Matrix and Determinant
• Algebra
• Maths-Class-12

+2 More