Determinants of Matrix 4×4 with Examples

Her you will learn how to find determinants of matrix 4×4 with example.

Let’s begin –

Determinants of Matrix 4×4

To evaluate the determinant of a square matrix of order 4 we follow the same procedure as discussed in previous post in evaluating the determinant of a square matrix of order 3.

If A = $\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix}$ is a square matrix of order 4,

then | A | = $a_{11}\begin{vmatrix} a_{22} & a_{23} & a_{24} \\ a_{32} & a_{33} & a_{34} \\  a_{42} & a_{43} & a_{44} \end{vmatrix}$ – $a_{12}\begin{vmatrix} a_{21} & a_{23} & a_{24} \\ a_{31} & a_{33} & a_{34} \\  a_{41} & a_{43} & a_{44} \end{vmatrix}$ + $a_{13}\begin{vmatrix} a_{21} & a_{22} & a_{24} \\ a_{31} & a_{32} & a_{34} \\  a_{41} & a_{42} & a_{44} \end{vmatrix}$ – $a_{14}\begin{vmatrix} a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\  a_{41} & a_{42} & a_{43} \end{vmatrix}$

Example :

Find the determinant of A = $\begin{bmatrix} 1 & 2 & -1  &  3 \\ 2 & 1 & -2  & 3\\ 3 & 1 & 2 & 1 \\ 1 & -1 & 0 & 2 \end{bmatrix}$.

Solution : | A | = $\begin{vmatrix} 1 & 2 & -1  &  3 \\ 2 & 1 & -2  & 3\\ 3 & 1 & 2 & 1 \\ 1 & -1 & 0 & 2 \end{vmatrix}$

$\implies$ | A | = $1\begin{vmatrix} 1 & -2 & 3 \\ 1 & 2 & 1 \\  -1 & 0 & 2 \end{vmatrix}$ – $2\begin{vmatrix} 2 & -2 & 3 \\ 3 & 2 & 1 \\  1 & 0 & 2 \end{vmatrix}$ + $(-1)\begin{vmatrix} 2 & 1 & 3 \\ 3 & 1 & 1 \\  1 & -1 & 2 \end{vmatrix}$ – $3\begin{vmatrix} 2 & 1 & -2 \\ 3 & 1 & 2 \\  1 & -1 & 0 \end{vmatrix}$

| A | =  (1){$(1)\begin{vmatrix} 2 & 1 \\  0 & 2 \end{vmatrix}$ – $(-2)\begin{vmatrix} 1 & 1 \\ -1 & 2 \end{vmatrix}$ + $(3)\begin{vmatrix} 1 & 2 \\ -1 & 0 \end{vmatrix}$}

– (2){$(2)\begin{vmatrix} 2 & 1 \\  0 & 2 \end{vmatrix}$ – $(-2)\begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix}$ + $(3)\begin{vmatrix} 3 & 2 \\ 1 & 0 \end{vmatrix}$}

+ (-1){$(2)\begin{vmatrix} 1 & 1 \\  -1 & 2 \end{vmatrix}$ – $(1)\begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix}$ + $(3)\begin{vmatrix} 3 & 1 \\ 1 & -1 \end{vmatrix}$}

– (3){$(2)\begin{vmatrix} 1 & 2 \\  -1 & 0 \end{vmatrix}$ – $(1)\begin{vmatrix} 3 & 2 \\ 1 & 0 \end{vmatrix}$ + $(-2)\begin{vmatrix} 3 & 1 \\ 1 & -1 \end{vmatrix}$}

$\implies$ | A | = 1{(1)(4 – 0) – (-2)(2 + 1) + (3)(0 + 2)} – 2{(2)(4 – 0) – (-2)(6 – 1) + (3)(0 – 2)} – (-1){(2)(2 + 1) – (1)(6 – 1) + (3)(-3 – 1)} – 3{(2)(0 + 2) – (1)(0 – 2) + (-2)(-3 – 1)}

$\implies$ | A | = 1(16) – 2(12) + (-1)(-11) – 3(14) = -39