Question

If A = $\begin{pmatrix}3 & 2 \\ 0 & 1\end{pmatrix}$ then $(A^{-1})^{3}$ is equal to
A. $\dfrac{1}{27}\begin{pmatrix}1 & -26 \\ 0 & 27\end{pmatrix}$
B. $\dfrac{1}{27}\begin{pmatrix}-1 & 26 \\ 0 & 27\end{pmatrix}$
C. $\dfrac{1}{27}\begin{pmatrix}1 & -26 \\ 0 & -27\end{pmatrix}$
D. $\dfrac{1}{27}\begin{pmatrix}-1 & -26 \\ 0 & -27\end{pmatrix}$

Answer 1

Hint:
We are given a matrix and need to evaluate $(A^{-1})^{3}$. For this, we need to find the inverse of A and perform matrix multiplication.

Complete Step-by-Step Answer:
Given, A = $\begin{pmatrix}3 & 2 \\ 0 & 1\end{pmatrix}$.
We need to find the inverse of A, for this first we find the adjoint matrix of A.
The adjoint matrix of A is obtained by interchanging the diagonal elements and multiplying the off-diagonal elements by a negative sign.
$\therefore$ $ adj A$ = $\begin{pmatrix}1 & -2 \\ 0& 3\end{pmatrix}$
Determinant of A, $|A| = (3\times1)-(0\times1) = 3$
Inverse of A, $A^{-1} = \dfrac{1}{|A|}adj A$
$A^{-1} = \dfrac{1}{3}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}$
Now, we need to find $(A^{-1})^{3}$. For this, we need to multiply $A^{-1}$ three times with itself.
$(A^{-1})^{3} = \dfrac{1}{3}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}\times \dfrac{1}{3}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}\times \dfrac{1}{3}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1\times1+(-2)\times0 & 1\times(-2)+(-2)\times3 \\ 0\times1 + 3\times0 & 0\times(-2)+3\times3\end{pmatrix}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1+ 0 & -2-6 \\ 0 +0 & 0 + 9\end{pmatrix}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1& -8 \\ 0 & 9\end{pmatrix}\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1\times1+(-8)\times0 & 1\times(-2)+(-8)\times3 \\ 0\times1+0\times9 & 0\times(-2)+9\times3\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1+0& -2-24 \\ 0+0 & 0+27\end{pmatrix}$
$(A^{-1})^{3} = \dfrac{1}{27}\begin{pmatrix}1& -26\\ 0 & 27\end{pmatrix}$
Therefore, the correct answer is Option A.


Note:When we have to find $(A^{-1})^{3}$, this means we have to multiply the matrix $A^{-1}$ three times and does not mean to take the cube of the elements of $A^{-1}$. Be careful while multiplying the matrices.