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
