For the matrix A=⎡⎢⎣11112−32−13⎤⎥⎦. Show that A3−6A2+5A+11I=0. Hence, find A−1
For the matrix A=⎡⎢⎣11112−32−13⎤⎥⎦. Show that A3−6A2+5A+11I=0. Hence, find A−1
Given, A=⎡⎢⎣11112−32−13⎤⎥⎦
∴ A2=AA=⎡⎢⎣11112−32−13⎤⎥⎦⎡⎢⎣11112−32−13⎤⎥⎦
=⎡⎢⎣1+1+21+2−11−3+31+2−61+4+31−6−92−1+62−2−32+3+9⎤⎥⎦=⎡⎢⎣421−38−147−314⎤⎥⎦
and A3=A2A=⎡⎢⎣421−38−147−314⎤⎥⎦⎡⎢⎣11112−32−13⎤⎥⎦
=⎡⎢⎣4+2+24+4−14−6+3−3+8−28−3+16+14−3−24−427−3+287−6−147+9+42⎤⎥⎦=⎡⎢⎣871−2327−6932−1358⎤⎥⎦
∴ A3−6A2+5A+11I
⎡⎢⎣871−2327−6932−1358⎤⎥⎦−6⎡⎢⎣421−38−147−314⎤⎥⎦+5⎡⎢⎣11112−32−13⎤⎥⎦+11⎡⎢⎣100010001⎤⎥⎦
=⎡⎢⎣871−2327−6932−1358⎤⎥⎦−⎡⎢⎣24126−1848−8442−1884⎤⎥⎦+⎡⎢⎣555510−1510−515⎤⎥⎦+⎡⎢⎣110001100011⎤⎥⎦
=⎡⎢⎣8−24+5+117−12+5+01−6+5+0−23+18+5+027−48+10+11−69+84−15+032−42+10+0−13+18−5+058−84+15+11⎤⎥⎦=⎡⎢⎣000000000⎤⎥⎦
Now, A3−6A2+5A+11I=0
⇒(AAA)A−1−6(AA)A−1+5AA−1+11IA−1=0
(Pre-multiplying by A−1 as |A|≠0)
⎡⎢⎣∵|A|=∣∣
∣∣11112−32−13∣∣
∣∣=1(6−3)−1(3+6)+1(−1−4)=3−9−5=−11≠0⎤⎥⎦
⇒AA(AA−1)−6A(AA−1)+5(AA−1)+11(IA−1)=0
⇒AAI−6AI+5I+11A−1=0
⇒(using AA−1=I and IA−1=A−1)
⇒A2−6A+5I=−11A−1 (Using AAI=A2 and AI=A)
⇒A−1=−111(A2−6A+5I)⇒A−1=111(−A2+6A−5I)
=111⎧⎪⎨⎪⎩−⎡⎢⎣421−38−147−314⎤⎥⎦+6⎡⎢⎣11112−32−13⎤⎥⎦−5⎡⎢⎣100010001⎤⎥⎦⎫⎪⎬⎪⎭
=111⎧⎪⎨⎪⎩⎡⎢⎣−4−2−13−814−73−14⎤⎥⎦+⎡⎢⎣666612−1812−618⎤⎥⎦−⎡⎢⎣500050005⎤⎥⎦⎫⎪⎬⎪⎭
=111⎡⎢⎣−4+6−5−2+6−0−1+6−03+6−0−8+12−514−18−0−7+12−03−6+0−14+18−5⎤⎥⎦=111⎡⎢⎣−3459−1−45−3−1⎤⎥⎦
Given, A=⎡⎢⎣11112−32−13⎤⎥⎦
∴ A2=AA=⎡⎢⎣11112−32−13⎤⎥⎦⎡⎢⎣11112−32−13⎤⎥⎦
=⎡⎢⎣1+1+21+2−11−3+31+2−61+4+31−6−92−1+62−2−32+3+9⎤⎥⎦=⎡⎢⎣421−38−147−314⎤⎥⎦
and A3=A2A=⎡⎢⎣421−38−147−314⎤⎥⎦⎡⎢⎣11112−32−13⎤⎥⎦
=⎡⎢⎣4+2+24+4−14−6+3−3+8−28−3+16+14−3−24−427−3+287−6−147+9+42⎤⎥⎦=⎡⎢⎣871−2327−6932−1358⎤⎥⎦
∴ A3−6A2+5A+11I
⎡⎢⎣871−2327−6932−1358⎤⎥⎦−6⎡⎢⎣421−38−147−314⎤⎥⎦+5⎡⎢⎣11112−32−13⎤⎥⎦+11⎡⎢⎣100010001⎤⎥⎦
=⎡⎢⎣871−2327−6932−1358⎤⎥⎦−⎡⎢⎣24126−1848−8442−1884⎤⎥⎦+⎡⎢⎣555510−1510−515⎤⎥⎦+⎡⎢⎣110001100011⎤⎥⎦
=⎡⎢⎣8−24+5+117−12+5+01−6+5+0−23+18+5+027−48+10+11−69+84−15+032−42+10+0−13+18−5+058−84+15+11⎤⎥⎦=⎡⎢⎣000000000⎤⎥⎦
Now, A3−6A2+5A+11I=0
⇒(AAA)A−1−6(AA)A−1+5AA−1+11IA−1=0
(Pre-multiplying by A−1 as |A|≠0)
⎡⎢⎣∵|A|=∣∣
∣∣11112−32−13∣∣
∣∣=1(6−3)−1(3+6)+1(−1−4)=3−9−5=−11≠0⎤⎥⎦
⇒AA(AA−1)−6A(AA−1)+5(AA−1)+11(IA−1)=0
⇒AAI−6AI+5I+11A−1=0
⇒(using AA−1=I and IA−1=A−1)
⇒A2−6A+5I=−11A−1 (Using AAI=A2 and AI=A)
⇒A−1=−111(A2−6A+5I)⇒A−1=111(−A2+6A−5I)
=111⎧⎪⎨⎪⎩−⎡⎢⎣421−38−147−314⎤⎥⎦+6⎡⎢⎣11112−32−13⎤⎥⎦−5⎡⎢⎣100010001⎤⎥⎦⎫⎪⎬⎪⎭
=111⎧⎪⎨⎪⎩⎡⎢⎣−4−2−13−814−73−14⎤⎥⎦+⎡⎢⎣666612−1812−618⎤⎥⎦−⎡⎢⎣500050005⎤⎥⎦⎫⎪⎬⎪⎭
=111⎡⎢⎣−4+6−5−2+6−0−1+6−03+6−0−8+12−514−18−0−7+12−03−6+0−14+18−5⎤⎥⎦=111⎡⎢⎣−3459−1−45−3−1⎤⎥⎦
show
that A3 − 6A2 + 5A +
11 I = O. Hence, find A−1.