Question

How do I calculate the determinant of a $4 \times 4$ matrix?

Answer 1

The determinant of a square matrix A is the integer obtained through a range of methods using the elements of the matrix. Only square matrices have a determinant, for non-square ones it’s not defined. In this question, we have $4 \times 4$ matrix. And here we have non-zero entries, so it doesn’t matter what row or column to pick to perform so-called Laplace expansion. The general formula for calculating determinants which work for determinants of any size:
$\Delta = \sum\limits_{i,j = 1}^n {{{( - 1)}^{i + j}}{a_{ij}}{M_{ij}}}$

Complete step by step solution:
Let us take a $4 \times 4$matrix.
A = \left( {\begin{array}{*{20}{c}} {a11}&{a12}&{a13}&{a14} \\\ {a21}&{a22}&{a23}&{a24} \\\ {a31}&{a32}&{a33}&{a34} \\\ {a41}&{a42}&{a43}&{a44} \end{array}} \right)
Let us apply the formula.
$\Delta = \sum\limits_{i,j = 1}^n {{{( - 1)}^{i + j}}{a_{ij}}{M_{ij}}}$
Here, $4 \times 4$matrix is given. So, the value of n is equal to 4.
For j=1:
The index i is changing from 1 to 4 and j is equal to 1. Put the values into the above equation.
$\Delta = \sum\limits_{i,j = 1}^4 {{{( - 1)}^{i + j}}{a_{i1}}{M_{i1}}}$
That is equal to,
$\Delta = {\left( { - 1} \right)^{1 + 1}}{a_{11}}{M_{11}} + {\left( { - 1} \right)^{2 + 1}}{a_{21}}{M_{21}} + {\left( { - 1} \right)^{3 + 1}}{a_{31}}{M_{31}} + {\left( { - 1} \right)^{4 + 1}}{a_{41}}{M_{41}}$
Now, we already know that if the base is negative and the exponent is odd then the answer is negative and if the base is negative and the exponent is even then the answer is positive.
So,
$\Delta = \left( 1 \right){a_{11}}{M_{11}} + \left( { - 1} \right){a_{21}}{M_{21}} + \left( 1 \right){a_{31}}{M_{31}} + \left( { - 1} \right){a_{41}}{M_{41}}$
That is equal to,
$\Delta = {a_{11}}{M_{11}} - {a_{21}}{M_{21}} + {a_{31}}{M_{31}} - {a_{41}}{M_{41}}$ ...(1)
To find the value of $\Delta$ we need to calculate minors ${M_{11}}$, ${M_{21}}$, ${M_{31}}$, ${M_{41}}$. These are determinants of order $3 \times 3$:
The value of ${M_{11}}$ is:
{M_{11}} = \left| {\begin{array}{*{20}{c}} {{a_{22}}}&{{a_{23}}}&{{a_{24}}} \\\ {{a_{32}}}&{{a_{33}}}&{{a_{34}}} \\\ {{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right|
${M_{11}} = {a_{22}}\left( {{a_{33}}{a_{44}} - {a_{34}}{a_{43}}} \right) - {a_{23}}\left( {{a_{32}}{a_{44}} - {a_{34}}{a_{42}}} \right) + {a_{24}}\left( {{a_{32}}{a_{43}} - {a_{42}}{a_{33}}} \right)$
The value of ${M_{21}}$ is:
{M_{21}} = \left| {\begin{array}{*{20}{c}} {{a_{12}}}&{{a_{13}}}&{{a_{14}}} \\\ {{a_{32}}}&{{a_{33}}}&{{a_{34}}} \\\ {{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right|
${M_{21}} = {a_{12}}\left( {{a_{33}}{a_{44}} - {a_{34}}{a_{43}}} \right) - {a_{13}}\left( {{a_{31}}{a_{44}} - {a_{34}}{a_{41}}} \right) + {a_{14}}\left( {{a_{32}}{a_{43}} - {a_{42}}{a_{33}}} \right)$
The value of ${M_{31}}$ is:
{M_{31}} = \left| {\begin{array}{*{20}{c}} {{a_{12}}}&{{a_{13}}}&{{a_{14}}} \\\ {{a_{22}}}&{{a_{23}}}&{{a_{24}}} \\\ {{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right|
So,
${M_{31}} = {a_{12}}\left( {{a_{23}}{a_{44}} - {a_{24}}{a_{42}}} \right) - {a_{13}}\left( {{a_{22}}{a_{44}} - {a_{24}}{a_{42}}} \right) + {a_{14}}\left( {{a_{22}}{a_{43}} - {a_{23}}{a_{42}}} \right)$
The value of ${M_{41}}$ is:
{M_{41}} = \left| {\begin{array}{*{20}{c}} {{a_{12}}}&{{a_{13}}}&{{a_{14}}} \\\ {{a_{22}}}&{{a_{23}}}&{{a_{24}}} \\\ {{a_{32}}}&{{a_{33}}}&{{a_{34}}} \end{array}} \right|
So,
${M_{41}} = {a_{12}}\left( {{a_{23}}{a_{34}} - {a_{32}}{a_{24}}} \right) - {a_{13}}\left( {{a_{22}}{a_{34}} - {a_{32}}{a_{24}}} \right) + {a_{14}}\left( {{a_{22}}{a_{33}} - {a_{23}}{a_{32}}} \right)$
Substitute all the values in equation (1). And calculate the value of the determinant.

Note:
To solve this type of question, first of all, make sure that we are dealing with a square matrix i.e. the number of rows and the number of columns is the same. If it is a square matrix then we can solve the question by applying the general formula for calculating determinants as below.
$\Delta = \sum\limits_{i,j = 1}^4 {{{( - 1)}^{i + j}}{a_{i1}}{M_{i1}}}$