Question

If A=$\left[ \begin{matrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \\ \end{matrix} \right] $then$ A^{-1}$ [DCE 1999]
a)$A$
b) $A^2$
c) $A^3$
d) $A^4$

Answer 1

Hint: We have to find $A^{-1}$ of this $3\times3$ matrix, but in the options we have $A^2$, $A^3$, and $A^4$. So for that firstly we will determine the determinant of the matrix $|A|$ then, make $adjA$ and put the values in the inverse matrix formula. After getting the inverse, evaluate the options given and compare results.



Formula used: Inverse Matrix formula: $A^{-1}=\dfrac{AdjA}{|A|}$



Complete step by step solution: Let $A=\left[ \begin{matrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\0 & -1 & 1 \\ \end{matrix} \right]$
Finding determinant of $A$;
$|A|=3[(-3)(1)-4(-1)]+3[2(1)-4(0)]+4[2(-1)-0(4)]\\
|A|=3(-3+4)+3(2)+4(-2)\\
|A|=3+6-8\\
|A|=1$

Since its determinant is not zero, it's an invertible matrix.
Now, find the adjoint of matrix $A$ by taking the transpose of the cofactor matrix.
Cofactors for each element of matrix A are given by;
$C_{11}=\begin{vmatrix}-3 & 4 \\ -1 & 1 \\ \end{vmatrix}\\
\Rightarrow C_{11}=-3+4\\
\Rightarrow C_{11}=1
C_{12}=-\begin{vmatrix}2 & 4 \\ 0 & 1 \\ \end{vmatrix}\\
\Rightarrow C_{12}=-2-0\\
\Rightarrow C_{12}=-2
C_{13}=\begin{vmatrix}2 & -3 \\ 0 & -1 \\ \end{vmatrix}\\
\Rightarrow C_{13}=-2-0\\
\Rightarrow C_{13}=-2$

$C_{21}=\begin{vmatrix}-3 & 4 \\ -1 & 1 \\ \end{vmatrix}\\
\Rightarrow C_{21}=3
C_{22}=-\begin{vmatrix}3 & -3 \\ 0 & -1 \\ \end{vmatrix}\\
\Rightarrow C_{22}=-(-3)\\
\Rightarrow C_{22}=3
C_{23}=\begin{vmatrix}-3 & 4 \\ -3 & 4 \\ \end{vmatrix}\\
\Rightarrow C_{23}=0$

$C_{31}=\begin{vmatrix}-3 & 4 \\ -3 & 4 \\ \end{vmatrix}\\
\Rightarrow C_{31}=0
C_{32}=-\begin{vmatrix}3 & 4 \\ 2 & 4 \\ \end{vmatrix}\\
\Rightarrow C_{32}=-4
C_{33}=\begin{vmatrix} 3 & -3 \\ 2 & -3 \\ \end{vmatrix}\\
\Rightarrow C_{33}=-3$

Therefore, the adjoint matrix is
$Adj A=\left[ \begin{matrix} 1 & -2 & -2 \\ -1 & 3 & 3 \\0 & -4 & -3 \\ \end{matrix} \right]^{-1}\\
Adj A=\left[ \begin{matrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\-2 & 3 & -3 \\ \end{matrix} \right]$

Substitute the adjoint matrix and determine in the inverse matrix formula we get;
$A^{-1}=\left[ \begin{matrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\-2 & 3 & -3 \\ \end{matrix} \right]$

Let’s evaluate $A^2$ i.e.$ A^2=A.A$
$A^2=\left[ \begin{matrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\0 & -1 & 1 \\ \end{matrix} \right].\left[ \begin{matrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\0 & -1 & 1 \\ \end{matrix} \right]\\
=\left[ \begin{matrix} 9-6 & -9+9-4 & 12-12+4 \\ 6-6 & -6+9-4 & 8-12+4 \\-2 & 3-1 & -4+1 \\ \end{matrix} \right]\\
=\left[ \begin{matrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\-2 & 2 & -3 \\ \end{matrix} \right]$
Since it is not equal to $A^{-1}$.
$A^3=A^2.A$
$A^3$
=$\left[ \begin{matrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\-2 & 2 & -3 \\ \end{matrix} \right].\left[ \begin{matrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\0 & -1 & 1 \\ \end{matrix} \right]\\$
=$\left[ \begin{matrix} 9-8 & -9+12-4 & 12-16+4 \\ -2 & 3 & -4 \\-6+4 & 6-6+3 & -8+8-3 \\ \end{matrix} \right]\\$
=$\left[ \begin{matrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\-2 & 3 & -3 \\ \end{matrix} \right]$

Hence,$ A^{-1}=A^3$



Thus, Option (C) is correct.

Note: Remember ${{A}^{-1}}$ exists only when $|A|\ne 0$. We have to remember the formula for ${{A}^{-1}}$. Sometimes students make mistakes while solving $adjA$ and $|A|$ for a $3\times3$ matrix. If the inverse of a matrix exists, we can find the adjoint of the given matrix and divide it by the determinant of the matrix.