Find ‘k’ if $A=\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]$, if \[{{A}^{2}}=kA-2I\]

Hint: To solve this question firstly, we will find the value of \[{{A}^{2}}\] and 2I. then by re – arranging the equation \[{{A}^{2}}=kA-2I\] as \[{{A}^{2}}+2I=kA\]. And then we will find the inverse of matrix A and will multiply the inverse A on both side of equation, \[{{A}^{2}}+2I=kA\], we will get matrix k, which is equals to\[k={{A}^{-1}}\left( {{A}^{2}}+2I \right)\].

Complete step by step answer:
Now, let us first find the value of \[{{A}^{2}}\].
\[{{A}^{2}}\] means A.A or,
\[{{A}^{2}}=A.A=\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]\]
To multiply two matrix, what we do is, we multiply first row of first matrix with elements of first and second column of second matrix and values are considered as new row of new matrix formd and same goes with second raw of first matrix to second column of second matrix.
So, \[{{A}^{2}}=\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]\]
\[=\left[ \begin{matrix}
   3\times 3+4\times -2 & 3\times -2+(-2)\times (-2) \\
   4\times 3+4\times -2 & 4\times -2+(-2)\times (-2) \\
\end{matrix} \right]\]
On simplification, we get
\[=\left[ \begin{matrix}
   9+(-8) & -6+4 \\
   12-8 & -8+4 \\
\end{matrix} \right]\]
\[=\left[ \begin{matrix}
   1 & -2 \\
   4 & -4 \\
\end{matrix} \right]\]
Now, let us find ${{A}^{-1}}$ of matrix $A=\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]$
We know that, ${{A}^{-1}}=\dfrac{1}{ad-bc}\left[ \begin{matrix}
   d & -c \\
   -b & a \\
\end{matrix} \right]$
Here, a = 3, b = 4, c = -2 and d = -2.
So, ${{A}^{-1}}=\dfrac{1}{3\times -2-(-2)\times 4}\left[ \begin{matrix}
   -2 & 2 \\
   -4 & 3 \\
\end{matrix} \right]$
On simplification, we get
${{A}^{-1}}=\dfrac{1}{-6+8}\left[ \begin{matrix}
   -2 & 2 \\
   -4 & 3 \\
\end{matrix} \right]$
${{A}^{-1}}=\dfrac{1}{2}\left[ \begin{matrix}
   -2 & 2 \\
   -4 & 3 \\
\end{matrix} \right]$
And,$2I=2\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]$
$=\left[ \begin{matrix}
   2 & 0 \\
   0 & 2 \\
\end{matrix} \right]$
Now, we have in question that, \[{{A}^{2}}=kA-2I\]
So, putting all the values, we get
\[{{A}^{2}}=kA-2I\]
\[\left[ \begin{matrix}
   1 & -2 \\
   4 & -4 \\
\end{matrix} \right]=k\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]-\left[ \begin{matrix}
   2 & 0 \\
   0 & 2 \\
\end{matrix} \right]\]
Adding matrix $\left[ \begin{matrix}
   2 & 0 \\
   0 & 2 \\
\end{matrix} \right]$on both sides, we get
\[\left[ \begin{matrix}
   1 & -2 \\
   4 & -4 \\
\end{matrix} \right]+\left[ \begin{matrix}
   2 & 0 \\
   0 & 2 \\
\end{matrix} \right]=k\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]-\left[ \begin{matrix}
   2 & 0 \\
   0 & 2 \\
\end{matrix} \right]+\left[ \begin{matrix}
   2 & 0 \\
   0 & 2 \\
\end{matrix} \right]\]
\[\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]=k\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]\]
On multiplying, with matrix ${{A}^{-1}}=\dfrac{1}{2}\left[ \begin{matrix}
   -2 & 2 \\
   -4 & 3 \\
\end{matrix} \right]$, we get
\[\dfrac{1}{2}\left[ \begin{matrix}
   -2 & 2 \\
   -4 & 3 \\
\end{matrix} \right]\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]=k\dfrac{1}{2}\left[ \begin{matrix}
   -2 & 2 \\
   -4 & 3 \\
\end{matrix} \right]\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]\]
On solving using matrix multiplication as discussed above, we get
\[\dfrac{1}{2}\left[ \begin{matrix}
   -2\times 3+2\times 4 & -2\times -2+2\times -2 \\
   -4\times 3+3\times 4 & -4\times -2+3\times -2 \\
\end{matrix} \right]=k\dfrac{1}{2}\left[ \begin{matrix}
   -2\times 3+2\times 4 & -2\times -2+2\times -2 \\
   -4\times 3+3\times 4 & -4\times -2+3\times -2 \\
\end{matrix} \right]\]
On simplifying, we get
\[\dfrac{1}{2}\left[ \begin{matrix}
   2 & 0 \\
   0 & 2 \\
\end{matrix} \right]=k\dfrac{1}{2}\left[ \begin{matrix}
   2 & 0 \\
   0 & 2 \\
\end{matrix} \right]\]
On simplifying, we get
\[\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]=k\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]\]
Now, on both sides we can see that we have an identity matrix and for this condition, k must be unit number.
So, k = 1

Note: Always remember that if we have matrix A of order $n\times n$ , then matrix ${{A}^{n}}$ and inverse of matrix A,${{A}^{-1}}$ will also be of order $n\times n$. Remember that if we have $A=\left[ \begin{matrix}
   a & c \\
   b & d \\
\end{matrix} \right]$, then inverse of A, ${{A}^{-1}}$ will be equal to ${{A}^{-1}}=\dfrac{1}{ad-bc}\left[ \begin{matrix}
   d & -c \\
   -b & a \\
\end{matrix} \right]$. I matrix is called identity matrix and is denoted by $I=\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]$. Try not to make any calculation errors.