Question

If $ \left[ \begin{matrix}
   \dfrac{1}{25} & 0 \\
   x & \dfrac{1}{25} \\
\end{matrix} \right]={{\left[ \begin{matrix}
   5 & 0 \\
   -a & 5 \\
\end{matrix} \right]}^{-2}} $ , then the value of x is
(a) $ \dfrac{a}{125} $
(b) $ \dfrac{2a}{125} $
(c) $ \dfrac{a}{25} $
(d) None of these

Answer 1

Hint: We start solving the problem by assuming $ A=\left[ \begin{matrix}
   \dfrac{1}{25} & 0 \\
   x & \dfrac{1}{25} \\
\end{matrix} \right] $ and $ B=\left[ \begin{matrix}
   5 & 0 \\
   -a & 5 \\
\end{matrix} \right] $ . We then multiply $ {{B}^{2}} $ on both sides of the given equation. We then find the $ {{B}^{2}} $ by using the standard rule of matrix multiplication. We then multiply the obtained $ {{B}^{2}} $ with A and equate it to the identity matrix of order 2. We then equate the corresponding elements in both matrices to get the linear equation involving x and a. We then make the necessary calculations to get the required answer.

Complete step by step answer:
According to the problem, we are given that $ \left[ \begin{matrix}
   \dfrac{1}{25} & 0 \\
   x & \dfrac{1}{25} \\
\end{matrix} \right]={{\left[ \begin{matrix}
   5 & 0 \\
   -a & 5 \\
\end{matrix} \right]}^{-2}} $ and we need to find the value of x.
Let us assume $ A=\left[ \begin{matrix}
   \dfrac{1}{25} & 0 \\
   x & \dfrac{1}{25} \\
\end{matrix} \right] $ and $ B=\left[ \begin{matrix}
   5 & 0 \\
   -a & 5 \\
\end{matrix} \right] $ .
So, we have $ A={{B}^{-2}} $ .
 $ \Rightarrow A.{{B}^{2}}={{B}^{-2}}.{{B}^{2}} $ .
 $ \Rightarrow A.{{B}^{2}}=I $ ---(1), where I is an Identity matrix of order $ 2\times 2 $ .
We know that the identity matrix is defined as a square matrix with all of its principal elements 1 and others as 0.
Now, let us find the matrix $ {{B}^{2}} $ .
So, we have $ {{B}^{2}}=\left[ \begin{matrix}
   5 & 0 \\
   -a & 5 \\
\end{matrix} \right]\left[ \begin{matrix}
   5 & 0 \\
   -a & 5 \\
\end{matrix} \right] $ .
 $ \Rightarrow {{B}^{2}}=\left[ \begin{matrix}
   \left( 5\times 5 \right)+\left( 0\times -a \right) & \left( 5\times 0 \right)+\left( 0\times 5 \right) \\
   \left( -a\times 5 \right)+\left( 5\times -a \right) & \left( -a\times 0 \right)+\left( 5\times 5 \right) \\
\end{matrix} \right] $ .
 $ \Rightarrow {{B}^{2}}=\left[ \begin{matrix}
   25+0 & 0+0 \\
   -5a-5a & 0+25 \\
\end{matrix} \right] $ .
 $ \Rightarrow {{B}^{2}}=\left[ \begin{matrix}
   25 & 0 \\
   -10a & 25 \\
\end{matrix} \right] $ . Let us substitute this in equation (1).
 $ \Rightarrow \left[ \begin{matrix}
   \dfrac{1}{25} & 0 \\
   x & \dfrac{1}{25} \\
\end{matrix} \right]\left[ \begin{matrix}
   25 & 0 \\
   -10a & 25 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right] $ .
 $ \Rightarrow \left[ \begin{matrix}
   \left( \dfrac{1}{25}\times 25 \right)+\left( 0\times -10a \right) & \left( \dfrac{1}{25}\times 0 \right)+\left( 0\times 25 \right) \\
   \left( x\times 25 \right)+\left( \dfrac{1}{25}\times -10a \right) & \left( x\times 0 \right)+\left( \dfrac{1}{25}\times 25 \right) \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right] $ .
\[\Rightarrow \left[ \begin{matrix}
   1+0 & 0+0 \\
   25x-\dfrac{2a}{5} & 0+1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]\]
\[\Rightarrow \left[ \begin{matrix}
   1 & 0 \\
   25x-\dfrac{2a}{5} & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]\].
Equating the corresponding elements on both sides, we get \[25x-\dfrac{2a}{5}=0\].
 $ \Rightarrow 25x=\dfrac{2a}{5} $ .
 $ \Rightarrow x=\dfrac{2a}{125} $ .
So, we have found the value of x as $ \dfrac{2a}{125} $ .
 $\therefore$ The correct option for the given problem is (b).

Note:
 We can also solve this solve by first finding the inverse of the matrix B i.e., $ {{B}^{-1}} $ and then performing the matrix multiplication $ {{B}^{-1}}\times {{B}^{-1}} $ to find the matrix $ {{B}^{-2}} $ . We then equate A to $ {{B}^{-2}} $ for getting the required value of x. We should not solve this problem by assuming $ A={{B}^{-2}}\Leftrightarrow {{A}^{2}}=B $ , which is one of the mistakes done by students. Similarly, we can expect problems to find the matrix $ {{A}^{-1}} $ .