Q. If $\begin{bmatrix} \frac{1}{25} & 0 \\ x & \frac{1}{25} \end{bmatrix}$=$\begin{bmatrix} 5 & 0 \\ - a & 5 \end{bmatrix}^{- 2}$, then the value of $x$ is

A) $\frac{a}{125}$
B) $2\frac{a}{125}$
C) $\frac{2a}{25}$
D) None of these
Explanation

(b)
Let, $A = \begin{bmatrix}
5 & 0 \\
- a & 5
\end{bmatrix}$
$\Rightarrow$adj$(A) = \begin{bmatrix}
5 & 0 \\
a & 5
\end{bmatrix}$
$\Rightarrow A^{- 1} = \frac{1}{|A|}\begin{bmatrix}
5 & 0 \\
a & 5
\end{bmatrix} = \frac{1}{25}\begin{bmatrix}
5 & 0 \\
a & 5
\end{bmatrix}$
$\Rightarrow A^{- 2} = \left( A^{- 1} \right)^{2} = \frac{1}{25}\begin{bmatrix}
5 & 0 \\
a & 5
\end{bmatrix}\frac{1}{25}\begin{bmatrix}
5 & 0 \\
a & 5
\end{bmatrix}$
$= \frac{1}{625}\begin{bmatrix}
25 & 0 \\
10a & 25
\end{bmatrix}$
$= \begin{bmatrix}
\frac{1}{25} & 0 \\
\frac{2a}{125} & \frac{1}{25}
\end{bmatrix}$
$\begin{bmatrix}
\frac{1}{25} & 0 \\
x & \frac{1}{25}
\end{bmatrix}\begin{bmatrix}
\frac{1}{25} & 0 \\
\frac{2a}{125} & \frac{1}{25}
\end{bmatrix}$
$\Rightarrow x = \frac{2a}{125}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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