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

Answer 1

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.

Complete step-by-step answer:

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.