Question

Show that the matrix $ A = \left[ {\begin{array}{*{20}{c}}
2&3\\
1&2
\end{array}} \right] $ satisfies the equation $ {A^2} - 4A + I = 0 $ , where $ I $ is $ 2 \times 2 $ identity matrix and $ 0 $ is $ 2\, \times 2 $ zero matrix. Using the equation find $ {A^{ - 1}} $ .

Answer 1

Hint: In this problem, initially to prove and satisfy the equation take left side and substitute the value of $ A $ . To determine the value of $ {A^{ - 1}} $ multiply with $ {A^{ - 1}} $ in the equation then accordingly subtract and add them. The matrix addition and subtraction is point wise.

Complete step-by-step answer:
The matrix $ A $ is $ \left[ {\begin{array}{*{20}{c}}
2&3\\
1&2
\end{array}} \right] $ .
The quadratic equation is $ {A^2} - 4A + I = 0 $ .
Since, we know that the matrix $ A $ is $ \left[ {\begin{array}{*{20}{c}}
2&3\\
1&2
\end{array}} \right] $ .
Now, we will substitute the value of A in the quadratic equation $ {A^2} - 4A + I = 0 $ .
To show that for $ A $ satisfies the equation we obtain,
To get $ {A^2} $ we do a multiplication of matrices that is $ A \times A $ .
Then $ {A^2} = \left[ {\begin{array}{*{20}{c}}
2&3\\
1&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
2&3\\
1&2
\end{array}} \right] $
To multiply two matrices, first, multiply the first row with the first column and substitute in the first. Now, multiply the first row with the second column and substitute the in the second. Multiply the second row with the first column and substitute in three and multiply the second row with the second column to substitute the obtained value in the fourth.

 $ \begin{array}{c}
{A^2} = \left[ {\begin{array}{*{20}{c}}
{4 + 3}&{6 + 6}\\
{2 + 2}&{3 + 4}
\end{array}} \right]\\
 = \left[ {\begin{array}{*{20}{c}}
7&{12}\\
4&7
\end{array}} \right]
\end{array} $
Substitute the values of $ {A^2} $ , $ A $ and $ I $ in the given equation $ {A^2} - 4A + I $ .
 $ \begin{array}{c}
{A^2} - 4A + I = \left[ {\begin{array}{*{20}{c}}
7&{12}\\
4&7
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
8&12\\
4&8
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right]\\
 = \left[ {\begin{array}{*{20}{c}}
0&0\\
0&0
\end{array}} \right]
\end{array} $
Hence this proves that the equation is equal to zero that $ {A^2} - 4A + I = 0 $ .
To get $ {A^{ - 1}} $ we obtain,
Multiply $ {A^{ - 1}} $ with the equation $ {A^2} - 4A + I = 0 $ , then we obtain
 $ \begin{array}{c}
{A^{ - 1}}\left( {{A^2} - 4A + I} \right) = 0\\
A - 4I + {A^{ - 1}} = 0\\
{A^{ - 1}} = 4I - A
\end{array} $
Substitute the value of $ A $ and $ I $ in the above equation we obtain,
 $ \begin{array}{c}
{A^{ - 1}} = 4\left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
2&3\\
1&2
\end{array}} \right]\\
 = \left[ {\begin{array}{*{20}{c}}
2&{ - 3}\\
{ - 1}&{ - 1}
\end{array}} \right]
\end{array} $
Therefore, the value for the matrix $ {A^{ - 1}} $ is $ \left[ {\begin{array}{*{20}{c}}
2&{ - 3}\\
{ - 1}&{ - 1}
\end{array}} \right] $ .

Note: Be careful with multiplication for matrices, do not multiply with each column wise and substitute each multiplication value column wise in the matrix. The matrix addition for the matrix is point wise. $ {A^{ - 1}} $ can be found in different ways, that is to find the detriment of the matrix and reverse the right diagonal terms and take minus for left diagonal terms.