Question

If A=\(\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\) and I=\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\),find k so that A²=kA-2I

Answer 1

A²=A.A

\(\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\)\(\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\) 

=\(\begin{bmatrix}3(3)+(-2)(4)&3(-2)+(-2)(-2)\\4(3)+(-2)(4)&4(-2)+(-2)(-2)\end{bmatrix}\)=\(\begin{bmatrix}1&-2\\4&-4\end{bmatrix}\)

Now A²=kA-2I
\(\Rightarrow \begin{bmatrix}1&-2\\4&-4\end{bmatrix}\)=\(\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\)-2\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)

\(\Rightarrow\) \(\begin{bmatrix}1&-2\\4&-4\end{bmatrix}\)=\(\begin{bmatrix}3k&-2k\\4k&-2k\end{bmatrix}\)-2\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)

\(\Rightarrow\) \(\begin{bmatrix}1&-2\\4&-4\end{bmatrix}\)=\(\begin{bmatrix}3k-2&-2k\\4k-2k&-2\end{bmatrix}\)
Comparing the corresponding elements, we have:
3k-2=1
3k=2
\(\Rightarrow\) k=1

Thus, the value of k is 1.