Question

# If $\left( {\begin{array}{*{20}{c}} {\dfrac{1}{{25}}}&0 \\ x&{\dfrac{1}{{25}}} \end{array}} \right) = {\left( {\begin{array}{*{20}{c}} 5&0 \\ { - a}&5 \end{array}} \right)^{ - 2}}$, then the value of x is${\text{A}}{\text{. }}\dfrac{a}{{125}}$${\text{B}}{\text{. }}\dfrac{{2a}}{{125}}$${\text{C}}{\text{. }}\dfrac{{2a}}{{25}}$${\text{D}}{\text{.}}$ None of these

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Hint: If a matrix is $A = \left( {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right)$, then the inverse of the matrix is, ${A^{ - 1}} = \dfrac{1}{{\det A}}\left( {\begin{array}{*{20}{c}} d&{ - b} \\ { - c}&a \end{array}} \right)$, use this to solve.

We have been given in the question, $\left( {\begin{array}{*{20}{c}} {\dfrac{1}{{25}}}&0 \\ x&{\dfrac{1}{{25}}} \end{array}} \right) = {\left( {\begin{array}{*{20}{c}} 5&0 \\ { - a}&5 \end{array}} \right)^{ - 2}}$,

Now let $A = \left( {\begin{array}{*{20}{c}} 5&0 \\ { - a}&5 \end{array}} \right)$, so we can find the inverse of the matrix by using the formula,
${A^{ - 1}} = \dfrac{1}{{\det A}}\left( {\begin{array}{*{20}{c}} d&{ - b} \\ { - c}&a \end{array}} \right)$,
Therefore, $\det A = 25$,
Hence, the inverse is, ${A^{ - 1}} = \dfrac{1}{{25}}\left( {\begin{array}{*{20}{c}} 5&0 \\ a&5 \end{array}} \right)$,
So, the given equation becomes, $\left( {\begin{array}{*{20}{c}} {\dfrac{1}{{25}}}&0 \\ x&{\dfrac{1}{{25}}} \end{array}} \right) = {A^{ - 2}} - (1)$

Therefore, ${(A)^{ - 2}} = \dfrac{1}{{625}}{\left( {\begin{array}{*{20}{c}} 5&0 \\ a&5 \end{array}} \right)^2} = \dfrac{1}{{625}}\left( {\begin{array}{*{20}{c}} {25}&0 \\ {10a}&{25} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\dfrac{1}{{25}}}&0 \\ {\dfrac{{2a}}{{125}}}&{\dfrac{1}{{25}}} \end{array}} \right)$
Now put the value of ${A^{ - 2}}$in equation (1), we get,
$\left( {\begin{array}{*{20}{c}} {\dfrac{1}{{25}}}&0 \\ x&{\dfrac{1}{{25}}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\dfrac{1}{{25}}}&0 \\ {\dfrac{{2a}}{{125}}}&{\dfrac{1}{{25}}} \end{array}} \right) \\ \Rightarrow x = \dfrac{{2a}}{{125}} \\$
Hence, we get the value of x as 2a/125, after solving the equations.
So, the correct answer is ${\text{B}}{\text{. }}\dfrac{{2a}}{{125}}$.

Note- Whenever such types of questions appear, be careful while finding the inverse of a matrix, use the formula, ${A^{ - 1}} = \dfrac{1}{{\det A}}\left( {\begin{array}{*{20}{c}} d&{ - b} \\ { - c}&a \end{array}} \right)$, only when the matrix is a $2 \times 2$ matrix. After finding the value of ${A^{ - 2}}$, put in the given equation and solve it to find the value of x.