Question

If $A=\left[ \begin{matrix}
   1 & 0 \\
   -1 & 7 \\
\end{matrix} \right]$ and ${{A}^{2}}=8A+K{{I}_{2}}$ , then $K$ is equal to:
$A)-1$
 $B)1$
$C)-7$
$D)7$

Answer 1

Hint : To solve this question we need to have the knowledge of matrix and its operation. Firstly we will find the value for ${{A}^{2}}$. The matrix $I$denotes the identity matrix of four elements. We will do the operation of the matrix as per given in the question. Then we will equate the elements in the Right Hand Side of the matrix to that in the Left Hand Side of the matrix.

Complete step-by-step solution:
The question ask us to find value of $K$, if the condition given is ${{A}^{2}}=8A+K{{I}_{2}}$ and the matrix A is $A=\left[ \begin{matrix}
   1 & 0 \\
   -1 & 7 \\
\end{matrix} \right]$. The first step is to calculate ${{A}^{2}}$ . On multiplying the two matrices we get:
$\Rightarrow {{A}^{2}}=\left[ \begin{matrix}
   1 & 0 \\
   -1 & 7 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 0 \\
   -1 & 7 \\
\end{matrix} \right]$
$\Rightarrow {{A}^{2}}=\left[ \begin{matrix}
   1\times 1+0\times 1 & -1\times 0+7\times 0 \\
   -1\times 1+7\times -1 & 0\times -1+7\times 7 \\
\end{matrix} \right]$
On solving the matrix we get:
$\Rightarrow {{A}^{2}}=\left[ \begin{matrix}
   1 & 0 \\
   -8 & 49 \\
\end{matrix} \right]$
Now we will solve for the L.H.S of the expression ${{A}^{2}}=8A+K{{I}_{2}}$. On solving we get:
$\Rightarrow 8A+K{{I}_{2}}=8\left[ \begin{matrix}
   1 & 0 \\
   -1 & 7 \\
\end{matrix} \right]+K\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]$
$\Rightarrow 8A+K{{I}_{2}}=\left[ \begin{matrix}
   8 & 0 \\
   -8 & 56 \\
\end{matrix} \right]+\left[ \begin{matrix}
   K & 0 \\
   0 & K \\
\end{matrix} \right]$
Since the above two matrix of same length are being added so each element will be added giving the matrix as:
$\Rightarrow 8A+K{{I}_{2}}=\left[ \begin{matrix}
   8+K & 0 \\
   -8 & 56+K \\
\end{matrix} \right]$
Now the R.H.S. of the expression is $\left[ \begin{matrix}
   8+K & 0 \\
   -8 & 56+K \\
\end{matrix} \right]$. We will equate to $\left[ \begin{matrix}
   1 & 0 \\
   -8 & 49 \\
\end{matrix} \right]$.On equating we will get:
$\Rightarrow \left[ \begin{matrix}
   1 & 0 \\
   -8 & 49 \\
\end{matrix} \right]=\left[ \begin{matrix}
   8+K & 0 \\
   -8 & 56+K \\
\end{matrix} \right]$
On equating we will get:
$\Rightarrow 8+K=1$
$\Rightarrow K=1-8$
$\Rightarrow K=-7$
$\therefore $ The value for $K$ is equal to $C)-7$.

Note: Do remember that all the matrices cannot be multiplied. For two matrices to be multiplied when the number of rows of the second matrix is the same as the number of columns of the first matrix. The above answer can be checked by substituting the value of $K$in the matrix $\left[ \begin{matrix}
   8+K & 0 \\
   -8 & 56+K \\
\end{matrix} \right]$. ON putting $K=-7$in the matrix, we get
$\Rightarrow \left[ \begin{matrix}
   8-7 & 0 \\
   -8 & 56-7 \\
\end{matrix} \right]$
$\Rightarrow \left[ \begin{matrix}
   1 & 0 \\
   -8 & 49 \\
\end{matrix} \right]$
Since the above matrix is same as ${{A}^{2}}$, so the answer we got is correct.