Question

If $A=\left[ \begin{matrix} 4 & 8 \\\ -2 & -4 \\\ \end{matrix} \right]$ then prove that ${{A}^{2}}=0$ .

Answer 1

At first, we need to find the square of the matrix $A=\left[ \begin{matrix} 4 & 8 \\\ -2 & -4 \\\ \end{matrix} \right]$ . The multiplication of two matrices of order $2\times 2$ can be shown as,
$\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]\left[ \begin{matrix} d & e \\\ f & g \\\ \end{matrix} \right]=\left[ \begin{matrix} ad+bf & ae+bg \\\ cd+df & ce+dg \\\ \end{matrix} \right]$
Similarly, the square of matrix A can also be found out. If the square turns out to be a zero matrix, then the required is proved.

Complete step by step answer:
The matrix that we are given in this problem is,
$A=\left[ \begin{matrix} 4 & 8 \\\ -2 & -4 \\\ \end{matrix} \right]$
We have to prove that the square of the matrix is a zero matrix, or that ${{A}^{2}}=0$ . Now, in order to prove this, we need to find the value of the square of the matrix A. For that, we need to multiply the matrix A with itself. The matrix multiplication of two matrices of order $2\times 2$ can be shown as,
$\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]\left[ \begin{matrix} d & e \\\ f & g \\\ \end{matrix} \right]=\left[ \begin{matrix} ad+bf & ae+bg \\\ cd+df & ce+dg \\\ \end{matrix} \right]$
In the above way of multiplication, we can find out the multiplication of A with itself. The multiplication goes as,
$\Rightarrow A\times A=\left[ \begin{matrix} 4 & 8 \\\ -2 & -4 \\\ \end{matrix} \right]\left[ \begin{matrix} 4 & 8 \\\ -2 & -4 \\\ \end{matrix} \right]=\left[ \begin{matrix} 4\times 4+8\left( -2 \right) & 4\times 8+8\left( -4 \right) \\\ \left( -2 \right)4+\left( -4 \right)\left( -2 \right) & \left( -2 \right)8+\left( -4 \right)\left( -4 \right) \\\ \end{matrix} \right]$
Simplifying the above matric by necessary addition and subtractions in the element spaces, we can write it as,
$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 0 & 0 \\\ 0 & 0 \\\ \end{matrix} \right]$
Now, we know that the zero matrix $\left[ \begin{matrix} 0 & 0 \\\ 0 & 0 \\\ \end{matrix} \right]$ can also be shown as $0$ for a representation.
Thus, we can conclude that ${{A}^{2}}=0$ .

Note: There are various types of matrices. Out of them, the matrix whose square is a zero matrix is called a nilpotent matrix. For a nilpotent matrix,
$\begin{aligned} & {{A}^{2}}=0 \\\ & \Rightarrow {{A}^{-1}}.{{A}^{2}}={{A}^{-1}}.0 \\\ & \therefore A=0 \\\ \end{aligned}$
But this is not true for this case. This means multiplication of ${{A}^{-1}}$ is not feasible or that it is undefined. This only means that $\left| A \right|=0$ . $\left| A \right|=4\left( -4 \right)-8\left( -2 \right)=0$ which is true. So, we can also prove this way.