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.