Question

If matrix \[A=\left[ \begin{matrix}
   2\ \ \ -2 \\
   -2\ \ \ 2 \\
\end{matrix} \right]\]and${{A}^{2}}=pA$, then write the value of $p$.

Answer 1

Hint: Recall that ${{A}^{2}}=A\cdot A$ and use this relation to find the matrix ${{A}^{2}}$ . Then find the matrix $pA$ (where $p$ is a scalar) and use these matrices to determine the value of $p$ by using the condition of equality of matrices. Remember that two matrices are equal if and only if all elements of the two matrices at the same index are equal.

Complete step by step answer:
First, we find the matrix ${{A}^{2}}$ by using the following relation:
$\begin{align}
  & {{A}^{2}}=A\cdot A \\
 & \ \ \ \ \ =\left[ \begin{matrix}
   2\ \ \ -2 \\
   -2\ \ \ 2 \\
\end{matrix} \right]\left[ \begin{matrix}
   2\ \ \ -2 \\
   -2\ \ \ 2 \\
\end{matrix} \right] \\
\end{align}$
Now, we’ll multiply the two matrices. To get the first element of the matrix ${{A}^{2}}$ , we’ll multiply the elements in the first row of the first matrix with the elements in the first column of the second matrix and then add them. So, the first element of the matrix ${{A}^{2}}$ will be $2(2)+(-2)(-2)$ . Similarly, the other elements can be computed as shown below:
$\begin{align}
  & {{A}^{2}}=\left[ \begin{matrix}
   2(2)+(-2)(-2)\ \ \ \ \ 2(-2)+(-2)2 \\
   (-2)2+2(-2)\ \ \ \ \ (-2)(-2)+2(2) \\
\end{matrix} \right] \\
 & \ \ \ \ \ =\left[ \begin{matrix}
   4+4\ \ \ -4-4 \\
   -4-4\ \ \ 4+4 \\
\end{matrix} \right] \\
 & \ \ \ \ \ =\left[ \begin{matrix}
   8\ \ \ -8 \\
   -8\ \ \ 8 \\
\end{matrix} \right] \\
\end{align}$
Now, we'll find the matrix $pA$ . Since $p$ is a constant, we’ll multiply it with all the elements of $A$ to obtain
$pA=\left[ \begin{matrix}
   2p\ \ \ -2p \\
   -2p\ \ \ 2p \\
\end{matrix} \right]$
Now, we use the important fact quoted in question ${{A}^{2}}=pA$ and the condition of equality of matrices (two matrices are equal if and only if each element of the matrices at the same index is equal).
${{A}^{2}}=pA$
$\left[ \begin{matrix}
   8\ \ \ -8 \\
   -8\ \ \ 8 \\
\end{matrix} \right]=\left[ \begin{matrix}
   2p\ \ \ -2p \\
   -2p\ \ \ 2p \\
\end{matrix} \right]$
We see that $2p=8,\ -2p=-8$ and so on.
We conclude that $2p=8\Rightarrow p=4$ and the equality of all the elements in the matrices at the same index holds for this value of $p$ .

Hence, the value of $p$ is $4$ .

Note: It is important to ensure that the equality of all elements of the two matrices at the same index holds with the value of $p$ found. If the equality of elements at a particular index does not hold, the matrices would not be equal and there would be no value of $p$ for which ${{A}^{2}}=pA$ . Also, observe that ${{A}^{2}}$ is the product of the matrices $A$ and $A$ . A common mistake is to consider ${{A}^{2}}$ as the matrix of the squares of all the elements of $A$ . This should be avoided.