Question

The multiplicative inverse of matrix \[\left[ {\begin{array}{*{20}{c}}
  2&1 \\
  7&4
\end{array}} \right]\] is
A. \[\left[ {\begin{array}{*{20}{c}}
   4&{ - 1} \\
   7&2
 \end{array}} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}}
   4&{ - 1} \\
   { - 7}&2
 \end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}}
  4&{ - 1} \\
   { - 7}&{ - 2}
 \end{array}} \right]\]
D. \[\left[ {\begin{array}{*{20}{c}}
   { - 4}&{ - 1} \\
   7&{ - 2}
 \end{array}} \right]\]

Answer 1

Hint: First, we will use the formula of the inverse of the matrix \[A = \left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right]\] by, \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}\left[ {\begin{array}{*{20}{c}}
  d&{ - b} \\
  { - c}&a
\end{array}} \right]\], where \[\left| A \right|\] is the determinant\[A\]. Then we will find the value of \[a\], \[b\], \[c\] and \[d\] from the given matrix \[A\] and then substitute them in the formula of inverse of matrix to find the required value.

Complete step by step answer:

We are given that the matrix is \[\left[ {\begin{array}{*{20}{c}}
  2&1 \\
  7&4
\end{array}} \right]\].
We know that the inverse of the matrix \[A = \left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right]\] by using the formula, \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}\left[ {\begin{array}{*{20}{c}}
  d&{ - b} \\
  { - c}&a
\end{array}} \right]\], where \[\left| A \right|\] is the determinant of \[A\].

Finding the value of \[a\], \[b\], \[c\] and \[d\] from the given matrix \[A\], we get
\[ \Rightarrow a = 2\]
\[ \Rightarrow b = 1\]
\[ \Rightarrow c = 7\]
\[ \Rightarrow d = 4\]

Then we will compute the value of determinant of \[A\] using the above values, we get

\[
   \Rightarrow \left| A \right| = \left| {\begin{array}{*{20}{c}}
  2&1 \\
  7&4
\end{array}} \right| \\
   \Rightarrow \left| A \right| = 8 - 7 \\
   \Rightarrow \left| A \right| = 1 \\
 \]

Substituting the above values of the determinant of A, \[a\], \[b\], \[c\] and \[d\] in the formula of inverse of matrix, we get

\[
   \Rightarrow {A^{ - 1}} = \dfrac{1}{1}\left[ {\begin{array}{*{20}{c}}
  4&{ - 1} \\
  { - 7}&2
\end{array}} \right] \\
   \Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  4&{ - 1} \\
  { - 7}&2
\end{array}} \right] \\
 \]

Hence, option B is correct.

Note: In these types of questions, the key concept is to find the inverse by putting the values in the formula of inverse. Students should know that the matrix \[\left[ {\begin{array}{*{20}{c}}
  d&{ - b} \\
  { - c}&a
\end{array}} \right]\] is the adjoint matrix of \[A\]. We can remember this matrix to save some time in the \[2 \times 2\] matrix but we have to compute the value of \[adjA\] in the matrix more than 2 rows and 2 columns. When a student knows the formula of inverse, the solution is very simple and easy.