Question

If $A=\left[\begin{array}{cc}3& -5\\ -4& 2\end{array}\right]$, then find A² − 5A − 14I. Hence, obtain A³.

Answer 1

Given: $A=\left[\begin{array}{cc}3& -5\\ -4& 2\end{array}\right]$

$$\begin{gathered} A^2=\left[\begin{array}{cc}3& -5\\ -4& 2\end{array}\right]\left[\begin{array}{cc}3& -5\\ -4& 2\end{array}\right] \\ =\left[\begin{array}{cc}9+20& -15-10\\ -12-8& 20+4\end{array}\right] \\ =\left[\begin{array}{cc}29& -25\\ -20& 24\end{array}\right] \end{gathered}$$

$$\begin{gathered} A^2-5A-14I=\left[\begin{array}{cc}29& -25\\ -20& 24\end{array}\right]-5\left[\begin{array}{cc}3& -5\\ -4& 2\end{array}\right]-14\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ =\left[\begin{array}{cc}29-15-14& -25+25-0\\ -20+20-0& 24-10-14\end{array}\right] \\ =\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \end{gathered}$$

Therefore, A² − 5A − 14I = 0 ...(1)

Premultiplying the (1) by A, we get

A(A² − 5A − 14I) = A.0
⇒ A³ − 5A² − 14A = 0
⇒ A³ = 5A² + 14A
$$\begin{gathered} \therefore A^3=5\left[\begin{array}{cc}29& -25\\ -20& 24\end{array}\right]+14\left[\begin{array}{cc}3& -5\\ -4& 2\end{array}\right] \\ =\left[\begin{array}{cc}145+42& -125-70\\ -100-56& 120+28\end{array}\right] \\ =\left[\begin{array}{cc}187& -195\\ -156& 148\end{array}\right] \end{gathered}$$