Question

Show that \[A = \left( {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  3&4
\end{array}} \right)\] satisfies the equation \[{x^2} - 6x + 17 = 0\]. Hence, find \[{A^{ - 1}}\].

Answer 1

Hint: To solve this question first we put the value of the matrix and check whether this equation satisfies this equation or not. To check this first we find the square of the matrix and all other required values by multiplying by a constant. And then we multiply that by the inverse of the matrix and again put the value of the matrix and find the inverse of that.

Complete step by step answer:
We have given a matrix. \[A = \left( {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  3&4
\end{array}} \right)\]
To check whether the matrix satisfies the equation or not we have to put the value of \[x\] in terms of matrix \[A\]. So the given equation is \[{x^2} - 6x + 17 = 0\] of putting the value of \[A\].
\[{A^2} - 6A + 17 = 0\]

Now solving the value of \[{A^2}\].
\[{A^2} = A.A\]
\[\Rightarrow {A^2} = \left( {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  3&4
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  3&4
\end{array}} \right)\]
On multiplying both the matrices.
\[{A^2} = \left( {\begin{array}{*{20}{c}}
  {4 - 9}&{ - 6 - 12} \\
  {6 + 12}&{ - 9 + 16}
\end{array}} \right)\]
On further calculating
\[{A^2} = \left( {\begin{array}{*{20}{c}}
  { - 5}&{ - 18} \\
  {18}&7
\end{array}} \right)\]

Now on calculating the value of \[6A\].
\[6A = 6\left( {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  3&4
\end{array}} \right)\]
On multiplying this constant.
\[6A = \left( {\begin{array}{*{20}{c}}
  {12}&{ - 18} \\
  {18}&7
\end{array}} \right)\]
On solving the value of \[17I\].
\[17I = 17\left( {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right)\]
On multiplying this constant.
\[17I = \left( {\begin{array}{*{20}{c}}
  {17}&0 \\
  0&{17}
\end{array}} \right)\]

On putting all these values in the given \[{A^2} - 6A + 17 = 0\] equation.
\[{A^2} - 6A + 17 = \left( {\begin{array}{*{20}{c}}
  { - 5}&{ - 18} \\
  {18}&7
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
  {12}&{ - 18} \\
  {18}&7
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  {17}&0 \\
  0&{17}
\end{array}} \right)\]
On further solving this equation.
\[{A^2} - 6A + 17 = \left( {\begin{array}{*{20}{c}}
  { - 5 - 12 + 17}&{ - 18 + 18 + 0} \\
  {18 - 18 + 0}&{7 - 24 + 17}
\end{array}} \right)\]
\[\Rightarrow {A^2} - 6A + 17 = \left( {\begin{array}{*{20}{c}}
  0&0 \\
  0&0
\end{array}} \right)\]
Hence, matrix \[A\] satisfies the equation, \[{x^2} - 6x + 17 = 0\].
\[{A^2} - 6A + 17 = 0\]

On rearranging this equation.
\[{A^2} - 6A = - 17\]
Multiplying this equation by \[{A^{ - 1}}\] both sides.
\[A - 6{I_2} = - 17{A^{ - 1}}\]
On rearranging this equation.
\[{A^{ - 1}} = \dfrac{1}{{17}}\left( {6{I_2} - A} \right)\]
On putting the value of A and I.
\[{A^{ - 1}} = \dfrac{1}{{17}}\left( {\left( {\begin{array}{*{20}{c}}
  6&0 \\
  0&6
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  3&4
\end{array}} \right)} \right)\]
\[\therefore {A^{ - 1}} = \dfrac{1}{{17}}\left( {\begin{array}{*{20}{c}}
  4&3 \\
  { - 3}&2
\end{array}} \right)\]

Hence, the given matrix satisfies the given quadratic equation and the value of \[{A^{ - 1}}\] is \[{A^{ - 1}} = \dfrac{1}{{17}}\left( {\begin{array}{*{20}{c}}
  4&3 \\
  { - 3}&2
\end{array}} \right)\].

Note: