Question

Find ${{\left[ \begin{matrix} -6 & 5 \\ -7 & 6 \\ \end{matrix} \right]}^{-1}} =$ [Karnataka CET 1994]
A. $\left[ \begin{matrix} -6 & 5 \\ -7 & 6 \\ \end{matrix} \right]$
B. $\left[ \begin{matrix} 6 & -5 \\ -7 & 6 \\ \end{matrix} \right]$
C. $\left[ \begin{matrix} 6 & 5 \\ 7 & 6 \\ \end{matrix} \right]$
D. $\left[ \begin{matrix} 6 & -5 \\ 7 & -6 \\ \end{matrix} \right]$

Answer 1

Hint:
Here you can find the inverse either by using elementary row operation or by inverse formula. Using the determinant and adjoint of the given matrix you can determine the inverse. Divide each component of an adjoint matrix obtained by the determinant of that matrix.

Formula Used:
Determinant of $2 \times 2$ matrix A (say) is given by $|A|$ or det $A=(a_{11} a_{22}-a_{12} a_{21})$
Adjoint of $2 \times 2$ matrix A(say) is given by $adjA=\left[\begin{matrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \\ \end{matrix}\right]$
Inverse matrix formula $A^{-1}=\dfrac{adjA}{|A|}$

Complete step-by-step solution:
Let $A= \left[ \begin{matrix} -6 & 5 \\ -7 & 6 \\ \end{matrix} \right]$
Determinant;
$|A|=(-6\times6)-(-7\times5)\\
|A|=-36+35\\
|A|=-1$
Alternating the principal diagonal components now produces the Adjoint of A. Flip the components' signs in the other diagonal.
Thus, Adjoint of $A=Adj A=\left[ \begin{matrix} 6 & -5 \\ 7 & -6 \\ \end{matrix} \right]$

Therefore,
$A^{-1}=-1.\left[ \begin{matrix} 6 & -5 \\ 7 & -6 \\ \end{matrix} \right]\\
A^{-1}=\left[ \begin{matrix} -6 & 5 \\ -7 & 6 \\ \end{matrix} \right]$

So, option is A correct.

Additional Information:
Inverse using Row Operations for $2\times 2$ Matrix
To determine the inverse of a $2\times 2$ matrix, $A$, we can use the fundamental row operations.
1. As the identity matrix of order $2\times 2$, $A$ and $I$ are first written as an augmented matrix with a line separating them so that $A$ is on the left and $I$ is on the right.
2. Apply row operations so that identity matrix $I$ is created from the left side matrix.
Following that, $A^{-1}$ is the right side matrix.

Note:
The alternate method involves applying elementary operations to find the inverse of the $2 \times 2$ matrix. Write $A = IA$ and do a number of row operations on $A = IA$ until we get $I = BA$ if $A$ is a matrix such that $A-1$ exists. This will use straightforward row operations to compute the inverse of A, or $A-1$. Matrix $B$ will be the inverse of matrix $A$. In a similar manner, if finding $A^{-1}$ via column operations is what you're after, then write $A = AI$ and use a series of column operations on it until we get $AB = I$.