Question

We have ( begin{array}{r}A^{2}=left[begin{array}{rr}3 & -5 -4 & 2end{array}right]left[begin{array}{rr}3 & -5 -4 & 2end{array}right] =left[begin{array}{cc}3 cdot 3+(-5)(-4) & 3 cdot(-5)+(-5) cdot 2 -4 cdot 3+2 cdot(-4) & -4 cdot(-5)+2 cdot 2end{array}right]=left[begin{array}{rr}29 & -25 -20 & 24end{array}right] -5 A=(-5)left[begin{array}{rr}3 & -5 -4 & 2end{array}right]=left[begin{array}{rr}-15 & 25 20 & -10end{array}right] ; -14 I=(-14)left[begin{array}{rr}1 & 0 0 & 1end{array}right]=left[begin{array}{rr}-14 & 0 0 & -14end{array}right] therefore quad A^{2}-5 A-14 I=A^{2}+(-5) A+(-14 I) = & {left[begin{array}{rr}29 & -25 -20 & 24end{array}right]+left[begin{array}{rr}-15 & 25 20 & -10end{array}right]+left[begin{array}{rr}-14 & 0 0 & -14end{array}right]} & =left[begin{array}{cc}29 & -15 -20 & -20+0 & 24+(-10)+(-14)end{array}right] & =left[begin{array}{cc}0 & 0 0 & 0end{array}right]end{array} )

# Сл EXAMPLE 7 s 3 If A = 1-4 -5 7 show that A2 - 5A - 141 = 0. 21

Solution