Question

If the matrix is given as $X=\left[ \begin{matrix} 4 & 1 \\\ -1 & 2 \\\ \end{matrix} \right]$, then show that $6X-{{X}^{2}}-9I=0$?

Answer 1

We start solving the problem by finding the matrices ${{X}^{2}}$, $-6X$. We then recall the definition of identity matrix $I$ and find the matrix $9I$. We substitute all these matrices in the equation $6X-{{X}^{2}}-9I$ and make the necessary calculations to get a matrix. We now recall the definition of the null matrix and compare the null matrix with the resultant matrix to complete the required proof.

Complete step-by-step solution:
According to the problem, we are given a matrix $X=\left[ \begin{matrix} 4 & 1 \\\ -1 & 2 \\\ \end{matrix} \right]$ and we need to prove that $6X-{{X}^{2}}-9I=0$ ---(1).
Let us find the matrix ${{X}^{2}}$.
So, we have ${{X}^{2}}=\left[ \begin{matrix} 4 & 1 \\\ -1 & 2 \\\ \end{matrix} \right]\times \left[ \begin{matrix} 4 & 1 \\\ -1 & 2 \\\ \end{matrix} \right]$.
$\Rightarrow {{X}^{2}}=\left[ \begin{matrix} \left( 4\times 4 \right)+\left( 1\times -1 \right) & \left( 4\times 1 \right)+\left( 1\times 2 \right) \\\ \left( -1\times 4 \right)+\left( 2\times -1 \right) & \left( -1\times 1 \right)+\left( 2\times 2 \right) \\\ \end{matrix} \right]$.
$\Rightarrow {{X}^{2}}=\left[ \begin{matrix} 16-1 & 4+2 \\\ -4-2 & -1+4 \\\ \end{matrix} \right]$.
$\Rightarrow {{X}^{2}}=\left[ \begin{matrix} 15 & 6 \\\ -6 & 3 \\\ \end{matrix} \right]$ ---(2).
Now, let us find $6X$.
We have $6X=6\times \left[ \begin{matrix} 4 & 1 \\\ -1 & 2 \\\ \end{matrix} \right]$.
$\Rightarrow 6X=\left[ \begin{matrix} 6\times 4 & 6\times 1 \\\ 6\times \left( -1 \right) & 6\times 2 \\\ \end{matrix} \right]$.
$\Rightarrow 6X=\left[ \begin{matrix} 24 & 6 \\\ -6 & 12 \\\ \end{matrix} \right]$ ---(3).
We know that $I$ is defined as the identity matrix which contains all elements of principal diagonal as 1 and all the other elements as 0. Since the given matrices are of order $2\times 2$, we also take the order of the identity matrix as $2\times 2$.
So, using the definition of identity matrix we get $I=\left[ \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right]$.
Now, let us find the matrix $9I$.
We have $9I=9\times \left[ \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right]$.
$\Rightarrow 9I=\left[ \begin{matrix} 9\times 1 & 9\times 0 \\\ 9\times 0 & 9\times 1 \\\ \end{matrix} \right]$.
$\Rightarrow 9I=\left[ \begin{matrix} 9 & 0 \\\ 0 & 9 \\\ \end{matrix} \right]$ ---(4).
Let us substitute the results obtained from equations (2), (3) and (4) in equation (1) in order to prove the result.
Let us consider $6X-{{X}^{2}}-9I$.
$\Rightarrow 6X-{{X}^{2}}-9I=\left[ \begin{matrix} 24 & 6 \\\ -6 & 12 \\\ \end{matrix} \right]-\left[ \begin{matrix} 15 & 6 \\\ -6 & 3 \\\ \end{matrix} \right]-\left[ \begin{matrix} 9 & 0 \\\ 0 & 9 \\\ \end{matrix} \right]$.
$\Rightarrow 6X-{{X}^{2}}-9I=\left[ \begin{matrix} 24-15-9 & 6-6-0 \\\ -6-\left( -6 \right)-0 & 12-3-9 \\\ \end{matrix} \right]$.
$\Rightarrow 6X-{{X}^{2}}-9I=\left[ \begin{matrix} 0 & 0 \\\ 0 & 0 \\\ \end{matrix} \right]$ ---(5).
We know that the matrix $0$ is known as null or zero matrix which has all the elements in it as zero. Since the given matrices are of order $2\times 2$, we also take the order of the null matrix as $2\times 2$.
So, using the definition of null matrix we get $0=\left[ \begin{matrix} 0 & 0 \\\ 0 & 0 \\\ \end{matrix} \right]$. Using this in equation (5) we get $6X-{{X}^{2}}-9I=0$.
$\therefore$ We have proved the result $6X-{{X}^{2}}-9I=0$.

Note: We can also prove this result by first finding the quadratic equation that gives us the eigenvalues for the given matrix. We then use the fact that the given matrix will satisfy its quadratic equation whose roots are eigenvalues. We can also find the eigenvalues and eigenvectors using this quadratic equation. We should not confuse with the definitions and formulas of matrices. We should not make calculation mistakes while solving this problem.