Determinant of 4x4 Matrix (Practice Questions)

Last Updated : 12 Sep, 2026

A 4 × 4 matrix determinant is a single number calculated from a square matrix containing 4 rows and 4 columns.

Solved Examples

Example 1: Calculate the determinant of the matrix A = \begin{bmatrix} 2 & 1 & 0 & 3 \\ 4 & -1 & 2 & 0 \\ -3 & 2 & 1 & 5 \\ 1 & 0 & -2 & 3 \end{bmatrix}

Solution:

First Expand along the first row:

\text{det}(A) = 2 \cdot \text{det}(A_{11}) - 1 \cdot \text{det}(A_{12}) + 0 \cdot \text{det}(A_{13}) - 3 \cdot \text{det}(A_{14})

Now, compute the determinant of each 3x3 submatrix.

For (A11):

A_{11} = \begin{bmatrix} -1 & 2 & 0 \\ 2 & 1 & 5 \\ 0 & -2 & 3 \end{bmatrix}

\text{det}(A_{11}) = (-1) \cdot \text{det}\left(\begin{bmatrix} 1 & 5 \\ -2 & 3 \end{bmatrix}\right) - 2 \cdot \text{det}\left(\begin{bmatrix} 2 & 5 \\ 0 & 3 \end{bmatrix}\right) + 0 \cdot \text{det}\left(\begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix}\right)

= (-1)((1)(3)-(5)(-2)) - 2((2)(3)-(5)(0)) + 0((2)(-2)-(1)(0))

= (-1)((3)+(10)) - 2((6)-(0)) + 0((-4)-(0))

= (-1)(13) - 2(6) + 0(-4)

= -13 - 12

= -25

For (A12):

A_{12} = \begin{bmatrix} 4 & 2 & 0 \\ -3 & 1 & 5 \\ 1 & 2 & 3 \end{bmatrix}

\text{det}(A_{12}) = (4) \cdot \text{det}\left(\begin{bmatrix} 1 & 5 \\ 2 & 3 \end{bmatrix}\right) - (2) \cdot \text{det}\left(\begin{bmatrix} -3 & 5 \\ 1 & 3 \end{bmatrix}\right) + (0) \cdot \text{det}\left(\begin{bmatrix} -3 & 1 \\ 1 & 2 \end{bmatrix}\right)

= (4)((1)(3)-(5)(2)) - (2)((-3)(3)-(5)(1)) + (0)((-3)(2)-(1)(1))

= (4)((3)-(10)) - (2)((-9)-(5)) + 0

= 4 (-7) - 2(- 14 )

= =28 + 28 = 0

For (A13):

Since a13 = 0, the contribution of C13​ will be zero. (Computation unnecessary.)

For (A14):

A_{14} = \begin{bmatrix} 4 & -1 & 2 \\ -3 & 2 & 1 \\ 1 & 0 & -2 \end{bmatrix}

\text{det}(A_{14}) = (4) \cdot \text{det}\left(\begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix}\right) - (-1) \cdot \text{det}\left(\begin{bmatrix} -3 & 1 \\ 1 & -2 \end{bmatrix}\right) + (2) \cdot \text{det}\left(\begin{bmatrix} -3 & 2 \\ 1 & 0 \end{bmatrix}\right)

=4[(2)(−2)−(1)(0)]−(−1)[(−3)(−2)−(1)(1)]+2[(−3)(0)−(2)(1)]

=4 [− 4 − 0] + 1[6 − 1] + 2[0 − 2]

= 4(-4) + 5 + (- 4)

=− 16 + 5 − 4

= - 15

Now, substitute the determinants of the 3x3 submatrices into the expansion formula:

det(A) = 2 ( - 25 ) - 1( 80 ) - 3 ( - 15)

= -50 - 80 + 45 + 0

= - 130 + 45

= - 85

So, the determinant of matrix (A) is - 85 .

Example 2: Calculate the determinant of the matrix A = \begin{bmatrix} 2 & 1 & -3 & 4 \\ -1 & 0 & 2 & 5 \\ 3 & 2 & 1 & 0 \\ 4 & -2 & 3 & 1 \end{bmatrix}

Solution:

To find the determinant of the matrix ( A ), we'll use the expansion by minors method along the first row:

\text{det}(A) = 2 \cdot \begin{vmatrix} 0 & 2 & 5 \\ 2 & 1 & 0 \\ -2 & 3 & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} -1 & 2 & 5 \\ 3 & 1 & 0 \\ 4 & 3 & 1 \end{vmatrix} - 3 \cdot \begin{vmatrix} -1 & 0 & 5 \\ 3 & 2 & 0 \\ 4 & -2 & 1\end{vmatrix} + 4 \cdot \begin{vmatrix} -1 & 0 & 2 \\ 3 & 2 & 1 \\ 4 & -2 & 3 \end{vmatrix}

Now, let's compute the determinants of the 3x3 submatrices:

\text{det}\left( \begin{vmatrix} 0 & 2 & 5 \\ 2 & 1 & 0 \\ -2 & 3 & 1 \end{vmatrix} \right) = (0 \cdot (1 \cdot 1 - 0 \cdot 3) - 2 \cdot (2 \cdot 1 - 0 \cdot (-2)) + 5 \cdot (2 \cdot 3 - 2 \cdot (-2)))

= 0 - 2 ( 2 - 0) + 5 ( 6 + 2)

= -4 + 5(8) = - 4 + 40 = 36

\text{det}\left( \begin{vmatrix} -1 & 2 & 5 \\ 3 & 1 & 0 \\ 4 & 3 & 1 \end{vmatrix} \right) = -1 \cdot ((1 \cdot 1 - 0 \cdot 3) - 2 \cdot (3 \cdot 1 - 0 \cdot 4) + 5 \cdot (3 \cdot 3 - 1 \cdot 4))

= -1 · ( 1 - 0) - 2 ( 3- 0) + 5(9 - 4)

= - 1 - 6 + 5(5)

= - 7 + 25 = 18

\text{det}\left( \begin{vmatrix} -1 & 0 & 5 \\ 3 & 2 & 0 \\ 4 & -2 & 3 \end{vmatrix} \right) = -1 \cdot ((2 \cdot 3 - 0 \cdot (-2)) - 0 \cdot (3 \cdot 5 - 0 \cdot 4) + 5 \cdot (3 \cdot (-2) - 2 \cdot 4))

= -1( 2- 0) - 0 + 5 (-6 - 8 )

= - 72

\text{det}\left( \begin{vmatrix} -1 & 0 & 2 \\ 3 & 2 & 1 \\ 4 & -2 & 3 \end{vmatrix} \right) = -1 \cdot ((2 \cdot 3 - 1 \cdot (-2)) - 0 \cdot (2 \cdot 3 - 1 \cdot 4) + 2 \cdot (3 \cdot (-2) - 2 \cdot 4))

= -1 · (6 + 2) - 0 + 2( - 6 - 8)

= (- 1)(8) + 2 (-14) = -8 - 28 = - 36

Now, substitute these determinants back into the expansion formula:

det(A) = 72 - 18 + 216 + 144 = 54 + 360 = 414

So, the determinant of matrix ( A ) is det(A) = 414.

Example 3: Let C = \begin{bmatrix}3 & 0 & -2 & 1 \\0 & 4 & 1 & -3 \\2 & -1 & 3 & 0 \\1 & 2 & 0 & 5\end{bmatrix} find its determinants using the cofactor expansion method.

Solution:

C_{11} =\begin{vmatrix}4 & 1 & -3\\-1 & 3 & 0\\2 & 0 & 5\end{vmatrix}

Expand along the first row [3,  0,  2,  1]. The expansion is

det⁡(C) = 3  C11 + 0  C12 + 2  C13 + 1C14

where C1j = (−1)1+j det⁡(M1j) (cofactor).

=4(15) - 1(-5) - 3(-6)
= 60 + 5 + 18
= 83.

C_{13} =\begin{vmatrix}0 & 4 & -3\\2 & -1 & 0\\1 & 2 & 5\end{vmatrix}
= 0 - 4(10) - 3(5)
= -40 - 15
= -55

C_{14} =\begin{vmatrix}0 & 4 & 1\\2 & -1 & 3\\1 & 2 & 0\end{vmatrix}

=0 - 4(-3) + 1(5)
=12 + 5
= 17.

Now substitute:

|C| = 3(83) - 2(-55) + (-17)

|C| = 249 + 110 - 17

|C| = 342

Practice Questions

Question 1: Calculate the determinant of the following 4×4 matrix: A = \begin{bmatrix} 2 & 0 & 1 & 3 \\ -1 & 2 & 2 & 0 \\ 3 & -2 & 0 & 1 \\ 1 & 1 & 2 & -1 \\ \end{bmatrix}

Question 2: Find the determinant of the matrix: B = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 2 & 3 & 4 & 5 \\ \end{bmatrix}

Question 3: Calculate the determinant of the following 4×4 matrix: C = \begin{bmatrix} 2 & 1 & 0 & -1 \\ 3 & 2 & -1 & 0 \\ 0 & -3 & 2 & 1 \\ 1 & 0 & 3 & -2 \\ \end{bmatrix}

Question 4: Determine the determinant of the matrix: D = \begin{bmatrix} 4 & 2 & 1 & 0 \\ -1 & 3 & 0 & 2 \\ 0 & 2 & 1 & -3 \\ 2 & 0 & -1 & 4 \\ \end{bmatrix}

Question 5: Find the determinant of the matrix: E = \begin{bmatrix} 3 & 1 & -2 & 0 \\ 2 & 0 & 1 & 1 \\ -1 & 2 & 3 & -2 \\ 0 & 3 & -1 & 1 \\ \end{bmatrix}

Question 6: Find the determinant of the following lower triangular matrix: B = \begin{bmatrix}5 & 0 & 0 & 0 \\-2 & 3 & 0 & 0 \\4 & 1 & 7 & 0 \\6 & -3 & 2 & 9\end{bmatrix}

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