Find the inverse of each of the following matrices. [101−111−1−2−3]

Step 1 Given matrix is A=[101−111−1−2−3] firstly, we will find all cofactors by using the formula Aij=(−1)i+jMij where Mij are minors. A11=(−1)1+1(−3+2)=−1 A12=(−1)1+2(3+1)=−4 A13=(−1)1+3(2+1)=3 A21=(−1)2+1(0+2)=−2 A22=(−1)2+2(−3+1)=−2 A23=(−1)2+3(−2+0)=2 A31=(−1)3+1(0−1)=−1 A32=(−1)3+2(1+1)=−2 A33=(−1)3+3(1+0)=1 Now, Adjoint A matrix is transpose of cofactors matrix. AdjA=[−1−43−2−22−1−21]T=[−1−2−1−4−2−2321] Step 2 Now, we ill evaluate the determinant of matrix A |A|=a11A11+a12A12+a13A13 =1(−1)+0(−4)+1(3)=2 Since, determinant of A is non-zero, therefore, inverse matrix of A exists and is obtained as A−1=1|A|AdjA A−1=12[−1−2−1−4−2−2321] Step 3 Ans: A−1=12[−1−2−1−4−2−2321]

Solved: "Find the inverse of each of the following..."

postillan4

Answered question

2020-12-05

Find the inverse of each of the following matrices.

[\begin{matrix} 1 & 0 & 1 \\ - 1 & 1 & 1 \\ - 1 & - 2 & - 3 \end{matrix}]

Answer & Explanation

Nathalie Redfern

Skilled2020-12-06Added 99 answers

Step 1
Given matrix is

A = [\begin{matrix} 1 & 0 & 1 \\ - 1 & 1 & 1 \\ - 1 & - 2 & - 3 \end{matrix}]

firstly, we will find all cofactors by using the formula

A_{i j} = (- 1)^{i + j} M_{i j}

where

M_{i j}

are minors.

A_{11} = (- 1)^{1 + 1} (- 3 + 2) = - 1

A_{12} = (- 1)^{1 + 2} (3 + 1) = - 4

A_{13} = (- 1)^{1 + 3} (2 + 1) = 3

A_{21} = (- 1)^{2 + 1} (0 + 2) = - 2

A_{22} = (- 1)^{2 + 2} (- 3 + 1) = - 2

A_{23} = (- 1)^{2 + 3} (- 2 + 0) = 2

A_{31} = (- 1)^{3 + 1} (0 - 1) = - 1

A_{32} = (- 1)^{3 + 2} (1 + 1) = - 2

A_{33} = (- 1)^{3 + 3} (1 + 0) = 1

Now, Adjoint A matrix is transpose of cofactors matrix.

A d j A = {[\begin{matrix} - 1 & - 4 & 3 \\ - 2 & - 2 & 2 \\ - 1 & - 2 & 1 \end{matrix}]}^{T} = [\begin{matrix} - 1 & - 2 & - 1 \\ - 4 & - 2 & - 2 \\ 3 & 2 & 1 \end{matrix}]

Step 2
Now, we ill evaluate the determinant of matrix A

| A | = a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13}

= 1 (- 1) + 0 (- 4) + 1 (3) = 2

Since, determinant of A is non-zero, therefore, inverse matrix of A exists and is obtained as

A^{- 1} = \frac{1}{| A |} A d j A

A^{- 1} = \frac{1}{2} [\begin{matrix} - 1 & - 2 & - 1 \\ - 4 & - 2 & - 2 \\ 3 & 2 & 1 \end{matrix}]

Step 3
Ans:

A^{- 1} = \frac{1}{2} [\begin{matrix} - 1 & - 2 & - 1 \\ - 4 & - 2 & - 2 \\ 3 & 2 & 1 \end{matrix}]

Jeffrey Jordon

Expert2022-01-27Added 2605 answers

Answer is given below (on video)

Jeffrey Jordon

Expert2022-08-23Added 2605 answers

Answer is given below (on video)

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