Question

If we are given matrices A=$\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]$ and $I$=$\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]$. Find $k$, so that $A^2=kA-2I$.

Answer 1

Hint: We are given equations in the matrix form. So, we will first create a matrix on either side and then we will compare the elements. After doing that we will calculate the value of $k$. We need to find $A^2$ as well, which means we have to multiply the matrix $A$ by itself. The matrix multiplication is a bit of a complex process because it is not done like the real numbers. After simplifying the left hand side of the equation, we will compare the matrices element-wise and obtain the result.

Complete step-by-step solution:
To multiply the matrix $A$ by itself, we use the formula below for matrix multiplication:
If $A=[a_{ij}]$ is an $m\times n$ matrix and $B=[b_{ij}]$ is an $n\times p$ matrix,
The product AB is an $m\times p$ matrix.
$AB=[c_{ij}]$
Where$c_{ij}=a_{i1}b_{1j}+a_{i2}b_{2j}+...+a_{in}b_{nj}$
So, we have:
$A=\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]$
Using the formula we obtain:
$\begin{align}
  & {{A}^{2}}=\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]\times \left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   3\times 3+\left( -2\times 4 \right) & 3\times -2+\left( -2\times -2 \right) \\
   4\times 3+\left( -2\times 4 \right) & 4\times -2+\left( -2\times -2 \right) \\
\end{matrix} \right] \\
\end{align}$
$\Rightarrow {{A}^{2}}=\left[ \begin{matrix}
   1 & -2 \\
   4 & -4 \\
\end{matrix} \right]$
Hence, we have found $A^2$
Now, we plug these values in the equation given:
$A^2=kA-2I$
$\Rightarrow \left[ \begin{matrix}
   1 & -2 \\
   4 & -4 \\
\end{matrix} \right]=k\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]-2\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]$
$\Rightarrow \left[ \begin{matrix}
   1 & -2 \\
   4 & -4 \\
\end{matrix} \right]=\left[ \begin{matrix}
   3k & -2k \\
   4k & -2k \\
\end{matrix} \right]-\left[ \begin{matrix}
   2 & 0 \\
   0 & 2 \\
\end{matrix} \right]$
$\Rightarrow \left[ \begin{matrix}
   1 & -2 \\
   4 & -4 \\
\end{matrix} \right]=\left[ \begin{matrix}
   3k-2 & -2k \\
   4k & -2k-2 \\
\end{matrix} \right]$
Now, we compare the elements, we get:
$1=3k-2$
$\Rightarrow 3k=3$
$\Rightarrow k=1$
To cross verify, we use one more equation:
$-2k=-2$
$\Rightarrow k=1$
Since both the values match, we have found the value of $k$ correctly.
Hence, $k=1$

Note: Make sure that you add the terms before giving the resultant value in each position of the resultant matrix. Look for any calculation mistake that might occur while doing multiplication. Always check with two or three equations, if the value of $k$ occurs to be different in some cases, then there is a possibility that you have made a calculation mistake.