1
Question
For the matrix A=⎡⎢⎣11112−32−13⎤⎥⎦ show that A3−6A2+5A+11I=0. Hence find A−1.
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Solution
Given: A=⎡⎢⎣11112−32−13⎤⎥⎦
⇒A2=⎡⎢⎣11112−32−13⎤⎥⎦⎡⎢⎣11112−32−13⎤⎥⎦
⇒A2=⎡⎢⎣1+1+21+2−11−3+31+2−61+4+31−6−92−1+62−2−32+3+9⎤⎥⎦
⇒A2=⎡⎢⎣421−38−147−314⎤⎥⎦
A3=A2×A=⎡⎢⎣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⎤⎥⎦
∴A3=⎡⎢⎣871−2327−6932−1358⎤⎥⎦
Now, proving A3−6A2+5A+11I=0
Solving L.H.S.
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⎤⎥⎦
⇒L.H.S.=⎡⎢⎣24−2412−126−6−23+23−48+4884−84−42+4218−1884−84⎤⎥⎦
⇒L.H.S.=⎡⎢⎣000000000⎤⎥⎦=0
∴L.H.S.=R.H.S.
Hence proved.
Now,
A3−6A2+5A+11I=0
Multiplying A−1 both sides
⇒(A3−6A2+5A+11I)A−1=0A−1
⇒A3⋅A−1−6A2⋅A−1+5AA−1+11IA−1=0
⇒A2⋅AA−1−6A⋅AA−1+5AA−1+11IA−1=0
⇒A2I−6AI+5I+11IA−1=0 (AA−1=1)
⇒A2−6A+5I+11A−1=0 (AI=A)
⇒11A−1=−A2+6A−5I
⇒A−1=111(−A2+6A−5I)
⇒A−1=111⎛⎜⎝⎡⎢⎣−4−2−13−814−73−14⎤⎥⎦+⎡⎢⎣666612−1812−618⎤⎥⎦−⎡⎢⎣500050005⎤⎥⎦⎞⎟⎠
⇒A−1=111⎡⎢⎣−3459−1−45−3−1⎤⎥⎦
⇒A2=⎡⎢⎣11112−32−13⎤⎥⎦⎡⎢⎣11112−32−13⎤⎥⎦
⇒A2=⎡⎢⎣1+1+21+2−11−3+31+2−61+4+31−6−92−1+62−2−32+3+9⎤⎥⎦
⇒A2=⎡⎢⎣421−38−147−314⎤⎥⎦
A3=A2×A=⎡⎢⎣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⎤⎥⎦
∴A3=⎡⎢⎣871−2327−6932−1358⎤⎥⎦
Now, proving A3−6A2+5A+11I=0
Solving L.H.S.
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⎤⎥⎦
⇒L.H.S.=⎡⎢⎣24−2412−126−6−23+23−48+4884−84−42+4218−1884−84⎤⎥⎦
⇒L.H.S.=⎡⎢⎣000000000⎤⎥⎦=0
∴L.H.S.=R.H.S.
Hence proved.
Now,
A3−6A2+5A+11I=0
Multiplying A−1 both sides
⇒(A3−6A2+5A+11I)A−1=0A−1
⇒A3⋅A−1−6A2⋅A−1+5AA−1+11IA−1=0
⇒A2⋅AA−1−6A⋅AA−1+5AA−1+11IA−1=0
⇒A2I−6AI+5I+11IA−1=0 (AA−1=1)
⇒A2−6A+5I+11A−1=0 (AI=A)
⇒11A−1=−A2+6A−5I
⇒A−1=111(−A2+6A−5I)
⇒A−1=111⎛⎜⎝⎡⎢⎣−4−2−13−814−73−14⎤⎥⎦+⎡⎢⎣666612−1812−618⎤⎥⎦−⎡⎢⎣500050005⎤⎥⎦⎞⎟⎠
⇒A−1=111⎡⎢⎣−3459−1−45−3−1⎤⎥⎦
Suggest Corrections
42
show
that A3 − 6A2 + 5A +
11 I = O. Hence, find A−1.