If A=\left[\begin{array}{rr}3 & -2 \\ 4 & -2\end{array}\right], find k so that A^{2}=k A-21.
If A=\left[\begin{array}{rr}3 & -2 \\ 4 & -2\end{array}\right], find k so that A^{2}=k A-21.

Solution:

We have \boldsymbol{A}=\left(\begin{array}{ll}\mathbf{3} & -\mathbf{2} \\ \mathbf{4} & -\mathbf{2}\end{array}\right).
Now addition/subtraction of two matrices is possible if order of both the matrices are same and multiplication of two matrices is possible if number of columns in left matrix is equals to the number of rows in right matrix
Next discuss the order of the matrices which are given. The order of matrix \boldsymbol{A} is \mathbf{2} \times \mathbf{2}.
Therefore the multiplications we can proceed as
(i) The multiplication A^{2} is possible.
A^{2}=A A=\left(\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right)\left(\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right)=\left(\begin{array}{ll} 1 & -2 \\ 4 & -4 \end{array}\right)
(ii) We have the equation A^{2}=\boldsymbol{k A}-\mathbf{2 I}.
\begin{array}{l} A^{2}=k A-2 I \\ \Rightarrow\left(\begin{array}{ll} 1 & -2 \\ 4 & -4 \end{array}\right)=k\left(\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right)-2\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) \\ \Rightarrow\left(\begin{array}{ll} 1 & -2 \\ 4 & -4 \end{array}\right)=\left(\begin{array}{cc} 3 k-2 & -2 k \\ 4 k & -2 k-2 \end{array}\right) \end{array}
Therefore we have
3 k-2=1 \Rightarrow k=1