Question

Let $A=\left[ \begin{matrix} 2 & 3 \\\ -1 & 2 \\\ \end{matrix} \right]$ and $f\left( x \right)={{x}^{2}}-4x+7$ show that $f\left( A \right)=0$. Use this to find ${{A}^{5}}$.

Answer 1

To solve this question we should know the concept of matrix. The question has more than one part. Firstly find the value for the function $f\left( A \right)$in which we need to multiply the matrix given in the question and then substitute values in the question given. For multiplication of matrices the number of columns in the first matrix be equal to the number of columns in the second matrix.

Complete step by step answer:
The question ask us to find the value of ${{A}^{5}}$ where is $A$ matrix given which is equal to $A=\left[ \begin{matrix} 2 & 3 \\\ -1 & 2 \\\ \end{matrix} \right]$ and the $f\left( A \right)$ satisfies the function ${{x}^{2}}-4x+7$.
The function in matrix form could be written as $f\left( A \right)={{A}^{2}}-4A+7I$ , Here $I$ is an identity matrix of size same as $2\times 2$. So identity matrix of size $2$ can be written as $\left[ \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right]$ .
We will find the value of the function for $A$ individually. First we will find the value of ${{A}^{2}}$ which means
$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 2 & 3 \\\ -1 & 2 \\\ \end{matrix} \right]\left[ \begin{matrix} 2 & 3 \\\ -1 & 2 \\\ \end{matrix} \right]$
On multiplying the matrix together we need to follow certain rules which are the elements in the row of the first matrix should be multiplied to the elements in the column of the second matrix.
$= \left[ \begin{matrix} 2\times 2+3\times \left( -1 \right) & 2\times 3+3\times 2 \\\ -1\times 2+\left( -1 \right)\times 2 & \left( -1 \right)\times 3+2\times 2 \\\ \end{matrix} \right]$
On multiplying and adding the numbers in the matrix we get:
$= \left[ \begin{matrix} 4-3 & 6+6 \\\ -2-2 & -3+4 \\\ \end{matrix} \right]$
$= \left[ \begin{matrix} 1 & 12 \\\ -4 & 1 \\\ \end{matrix} \right]$
Now we will find the value of $4A$ which is
$= 4\left[ \begin{matrix} 2 & 3 \\\ -1 & 2 \\\ \end{matrix} \right]$
On multiplying $4$ to all the elements given in the matrix we get
$= \left[ \begin{matrix} 8 & 12 \\\ -4 & 8 \\\ \end{matrix} \right]$
Now we will calculate the $7\text{I}$ where $\text{I}$ is the identity matrix
$= 7\left[ \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right]$
$= \left[ \begin{matrix} 7 & 0 \\\ 0 & 7 \\\ \end{matrix} \right]$
Putting these value in the equation$f\left( A \right)={{A}^{2}}-4A+7I$we get
$= \left[ \begin{matrix} 1 & 12 \\\ -4 & 1 \\\ \end{matrix} \right]-\left[ \begin{matrix} 8 & 12 \\\ -4 & 8 \\\ \end{matrix} \right]+\left[ \begin{matrix} 7 & 0 \\\ 0 & 7 \\\ \end{matrix} \right]$
On calculating the matrix we get:
$= \left[ \begin{matrix} 1-8+7 & 12-12 \\\ -4+4 & 1-8+7 \\\ \end{matrix} \right]$
$= \left[ \begin{matrix} 0 & 0 \\\ 0 & 0 \\\ \end{matrix} \right]$
Hence proved that $f\left( A \right)=0$
Now the second part of the question ask us to find the value of ${{A}^{5}}$ . Now ${{A}^{5}}$could be written as the product of ${{A}^{2}},{{A}^{2}}$ and $A$ which mathematically is written as:
${{A}^{5}}={{A}^{2}}\times {{A}^{2}}\times A$
Now substituting the values of each of the matrix we get:
$= {{A}^{5}}=\left( \left[ \begin{matrix} 1 & 12 \\\ -4 & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 & 12 \\\ -4 & 1 \\\ \end{matrix} \right] \right)\left[ \begin{matrix} 2 & 3 \\\ -1 & 2 \\\ \end{matrix} \right]$
We will multiply the first two matrices first and then will multiply the resultant matrix to the third matrix.
$= \left[ \begin{matrix} 1\times 1+12\times \left( -4 \right) & 1\times 12+1\times 12 \\\ -4\times 1+1\times -4 & \left( -4 \right)\times 12+1\times 1 \\\ \end{matrix} \right]$
$= \left[ \begin{matrix} 1-48 & 12+12 \\\ -4-4 & -48+1 \\\ \end{matrix} \right]$
$= \left[ \begin{matrix} -47 & 24 \\\ -8 & -47 \\\ \end{matrix} \right]$
Multiplying this term to the matrix $A$ we get:
$= \left[ \begin{matrix} -47 & 24 \\\ -8 & -47 \\\ \end{matrix} \right]\left[ \begin{matrix} 2 & 3 \\\ -1 & 2 \\\ \end{matrix} \right]$
On multiplying the matrix we get
$= \left[ \begin{matrix} -94-24 & -141+48 \\\ -16+47 & -24+\left( -94 \right) \\\ \end{matrix} \right]$
$= \left[ \begin{matrix} -118 & -93 \\\ +31 & -118 \\\ \end{matrix} \right]$
$\therefore$ The value of ${{A}^{5}}$ is $\left[ \begin{matrix} -118 & -93 \\\ +31 & -118 \\\ \end{matrix} \right]$.

Note: Always keep in mind that to multiply the two matrices these conditions should be followed: The number of columns present in the first matrix should be equal to the number of rows present in the second matrix. The matrix address $\left[ 0,0 \right]$ means the element which is common in the first row and the first column.