Consider the matrix, A=⎣⎡11212−11−33⎦⎤ then show that A3−6A2+5A+11I3=0. Hence, find A−1.

Question

Consider the matrix, A=⎣⎡​112​12−1​1−33​⎦⎤​ then show that A3−6A2+5A+11I3​=0. Hence, find A−1.

Filo · Accepted Answer

We have, A=⎣⎡​112​1 2−1​1−33​⎦⎤​∴A2=A A=⎣⎡​112​1 2−1​1−33​⎦⎤​⎣⎡​112​1 2−1​1−33​⎦⎤​=⎣⎡​4−37​2 8−3​1−1414​⎦⎤​and, A3=A2 A=⎣⎡​4−37​2 8−3​1−1414​⎦⎤​⎣⎡​112​1 2−1​1−33​⎦⎤​⎣⎡​8−2332​7 27−13​1−6958​⎦⎤​∴A3−6A2+5A+11I3​=⎣⎡​8−2332​7 27−13​1−6958​⎦⎤​−6⎣⎡​4−37​2 8−3​1−1414​⎦⎤​+5⎣⎡​112​1 2−1​1−33​⎦⎤​+11⎣⎡​100​0 10​001​⎦⎤​⎣⎡​8−24+5+11−23+18+5+032−42+10+0​7−12+5+0 27−48+10+11−13+18−5+0​1−6+5+0−69+84−15+058−84+15+1​⎦⎤​=⎣⎡​000​0 00​000​⎦⎤​=ONow, A3−6A2+5A+11I3​=O⇒A−1(A3−6A2+5A+11I3​)=A−1O [Multiplying both sides by A−1]⇒A2−6A+5I+11A−1=O⇒11A−1=−A2+6A−5I⇒11A−1 =−⎣⎡​4−37​2 8−3​1−1414​⎦⎤​+6⎣⎡​112​1 2−1​1−33​⎦⎤​−5⎣⎡​100​0 10​001​⎦⎤​⇒11A−1 =⎣⎡​−4+6−53+6+0−7+12+0​−2+6+0 −8+12−53−6+0​−1+6+014−18+0−14+18−5​⎦⎤​=⎣⎡​−395​4 −1−3​5−4−1​⎦⎤​A−1=111​ ⎣⎡​−395​4 −1−3​5−4−1​⎦⎤​

Question Text	Consider the matrix, $A = ⎣ ⎡ 112 12 - 1 1 - 3 3 ⎦ ⎤$ then show that $A^{3} - 6 A^{2} + 5 A + 11 I_{3} = 0.$ Hence, find $A^{- 1} .$
Answer Type	Text solution:1
Upvotes	150

Consider the matrix, $A = ⎣ ⎡ 112 12 - 1 1 - 3 3 ⎦ ⎤$ then show that $A^{3} - 6 A^{2} + 5 A + 11 I_{3} = 0.$ Hence, find $A^{- 1} .$

Text solutionVerified

Practice questions from similar books

Consider the matrix, A=⎣⎡​112​12−1​1−33​⎦⎤​ then show that A3−6A2+5A+11I3​=0. Hence, find A−1.

Text solutionVerified

Practice questions from similar books

Consider the matrix, $A = ⎣ ⎡ 112 12 - 1 1 - 3 3 ⎦ ⎤$ then show that $A^{3} - 6 A^{2} + 5 A + 11 I_{3} = 0.$ Hence, find $A^{- 1} .$