Question

For the matrix \[A = \left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&{\dfrac{{1 + bc}}{a}}
\end{array}} \right]\], show that \[a{A^{ - 1}} = \left( {{a^2} + bc + 1} \right)I - aA\].

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 and multiply it with \[a\]for left sides of the equation. Then we will substitute the value of A and I in the right hand side of the equation to show the required result.

Complete step-by-step answer:
We are given that the matrix is \[A = \left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&{\dfrac{{1 + bc}}{a}}
\end{array}} \right]\].
We know that the inverse of the matrix \[A = \left[ {\begin{array}{*{20}{c}}
  h&i \\
  j&k
\end{array}} \right]\] by using the formula, \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}\left[ {\begin{array}{*{20}{c}}
  k&{ - i} \\
  { - j}&h
\end{array}} \right]\], where \[\left| A \right|\] is the determinant of \[A\].
Finding the value of \[h\], \[i\], \[k\] and \[l\] from the given matrix \[A\], we get
\[ \Rightarrow h = a\]
\[ \Rightarrow i = b\]
\[ \Rightarrow j = c\]
\[ \Rightarrow l = \dfrac{{1 + bc}}{a}\]
Then we will compute the value of determinant of \[A\] using the above values, we get
\[
   \Rightarrow \left| A \right| = \left| {\begin{array}{*{20}{c}}
  a&b \\
  c&{\dfrac{{1 + bc}}{a}}
\end{array}} \right| \\
   \Rightarrow \left| A \right| = a\left( {\dfrac{{1 + bc}}{a}} \right) - bc \\
   \Rightarrow \left| A \right| = 1 + bc - bc \\
   \Rightarrow \left| A \right| = 1 \\
 \]
Substituting the above values of the determinant of A, \[h\], \[i\], \[j\] and \[k\] in the formula of inverse of matrix, we get
\[
   \Rightarrow {A^{ - 1}} = \dfrac{1}{1}\left[ {\begin{array}{*{20}{c}}
  {\dfrac{{1 + bc}}{a}}&{ - b} \\
  { - c}&a
\end{array}} \right] \\
   \Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{{1 + bc}}{a}}&{ - b} \\
  { - c}&a
\end{array}} \right] \\
 \]
Multiplying the above equation by \[a\] on both sides, we get
\[ \Rightarrow a{A^{ - 1}} = a\left[ {\begin{array}{*{20}{c}}
  {\dfrac{{1 + bc}}{a}}&b \\
  c&a
\end{array}} \right]\]
Using the property in the above equation of matrix if a matrix multiplies a constant, then every element of the matrix will multiply it, we get
\[
   \Rightarrow a{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  {a\left( {\dfrac{{1 + bc}}{a}} \right)}&{ab} \\
  {ac}&{{a^2}}
\end{array}} \right] \\
   \Rightarrow a{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  {1 + bc}&{ab} \\
  {ac}&{{a^2}}
\end{array}} \right]{\text{ ......eq.(1)}} \\
 \]
Now we will substitute the value of matrix A and identity matrix in the equation\[\left( {{a^2} + bc + 1} \right)I - aA\], we get
\[
   \Rightarrow \left( {{a^2} + bc + 1} \right)\left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] - a\left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&{\dfrac{{1 + bc}}{a}}
\end{array}} \right] \\
   \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {\left( {{a^2} + bc + 1} \right)}&0 \\
  0&{\left( {{a^2} + bc + 1} \right)}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  {{a^2}}&{ab} \\
  {ac}&{a\left( {\dfrac{{1 + bc}}{a}} \right)}
\end{array}} \right] \\
   \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {\left( {{a^2} + bc + 1} \right)}&0 \\
  0&{\left( {{a^2} + bc + 1} \right)}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  {{a^2}}&{ab} \\
  {ac}&{1 + bc}
\end{array}} \right] \\
   \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {{a^2} + bc + 1 - {a^2}}&{0 - ab} \\
  {0 - ac}&{{a^2} + bc + 1 - bc - 1}
\end{array}} \right] \\
   \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {bc + 1}&{ - ab} \\
  { - ac}&{{a^2}}
\end{array}} \right]{\text{ ......eq.(2)}} \\
 \]
From equation (1) and equation (2), we get
\[a{A^{ - 1}} = \left( {{a^2} + bc + 1} \right)I - aA\]
Hence, proved.

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 with more than 2 rows and 2 columns. When students know the formula of inverse, the solution is very simple and easy. We need to know that an identity matrix I is a \[2 \times 2\] matrix \[\left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]\].