Question

The inverse of $\begin{bmatrix}2 & -3 \\-4 & 2 \end{bmatrix}$ is
A. $\dfrac{-1}{8}\begin{bmatrix}2 & 3 \\ 4 & 2 \end{bmatrix}$
B. $\dfrac{-1}{8}\begin{bmatrix}3 & 2 \\ 2 & 4 \end{bmatrix}$
C. $\dfrac{1}{8}\begin{bmatrix}2 & 3 \\ 4 & 2 \end{bmatrix}$
D. $\dfrac{1}{8}\begin{bmatrix}3 & 2 \\ 2 & 4 \end{bmatrix}$

Answer 1

Hint: Inverse of the matrix can be determined by first calculating the value of the determinant of the given matrix. Afterward, determine the co-factor matrix in order to find the adjoint of the given matrix. Finally, substitute both these values in the required formula to determine the inverse of the matrix.

Formula used: $A^{-1}=\dfrac{adj\lgroup~A\rgroup}{\det\lgroup~A\rgroup}$
$C_{ij}=\lgroup-1\rgroup^{i+j}\det\lgroup~M_{ij}\rgroup$

Complete step by step solution: The given matrix, in accordance with the question, is,
$\begin{bmatrix}2 & -3 \\-4 & 2 \end{bmatrix}$
Consider the given matrix to be A.
Let us now determine the determinant of the given matrix A.,
$\begin{vmatrix} 2 & -3 \\ -4 & 2 \end{vmatrix}$
Now,
$|A|=2\times2-\lgroup-3\rgroup\times\lgroup-4\rgroup$
$\Rightarrow|A|=4-12=-8$
We must now find the adjoint of matrix A. To determine the adjoint of matrix A, first, find the cofactor matrix of a given matrix A and afterward take the transpose of the cofactor matrix.
The cofactor of the matrix can be obtained as:
$C_{ij}=\lgroup-1\rgroup^{i+j}\det\lgroup~M_{ij}\rgroup$
Here, $M_{ij}$ denotes the $\lgroup~i,j\rgroup^{th}$ minor matrix after eliminating the ith row and jth column.
Now, the Cofactor of matrix A can be written as:
$\begin{bmatrix}c_{11} & c_{12} \\c_{21} & c_{22} \end{bmatrix}=\begin{bmatrix}2 & -\lgroup-4\rgroup \\-\lgroup-3\rgroup & 2 \end{bmatrix}$
This implies,
$\begin{bmatrix}c_{11} & c_{12} \\c_{21} & c_{22} \end{bmatrix}=\begin{bmatrix}2 & 4 \\3 & 2 \end{bmatrix}$
Thus, the adjoint of the matrix A is,
$adj\lgroup~A\rgroup=\begin{bmatrix}2 & 3 \\4 & 2 \end{bmatrix}$
The inverse of a matrix can be determined by the following formula,
$A^{-1}=\dfrac{adj\lgroup~A\rgroup}{\det\lgroup~A\rgroup}$
The value of adjoint A and the determinant A are already determined. After substituting these values in the above formula we get,
$A^{-1}=\dfrac{-1}{8}\begin{bmatrix}2 & 3 \\4 & 2 \end{bmatrix}$

Thus, Option (A) is correct.

Note: It should be emphasized that in order to find the inverse matrix, the square matrix must be non-singular which is to have a determinant value that is not zero. The new matrix obtained by interchanging the rows and columns of the previous matrix is described as the matrix transpose.