D.4-5-2]-6 -2B-1[-89.3Е.[4-58B-19.L22-3] Determine theinverse of matrix B.-3-21B =| 1-4 -2-341A.[-5B-15-6-2[-89.3В.-421B-1= |-5 6 289 3.С.-81|-5 -6-4B-1%3|I-2-23

Given matrix is B=0−3−21−4−2−341 We have to find the inverse of the matrix B The inverse of B is…

Answered: Determine the inverse of matrix B. -3…

Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN:9781305658004

Author:Ron Larson

Publisher:Ron Larson

Chapter6: Linear Transformations

Section6.4: Transistion Matrices And Similarity

Problem 29E

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Question

D.
4
-5
-2]
-6 -2
B-1
[-8
9.
3
Е.
[4
-5
8
B-1
9.
L2
2
-3]

Determine the
inverse of matrix B.
-3
-21
B =| 1
-4 -2
-3
4
1
A.
[-5
B-1
5
-6
-2
[-8
9.
3
В.
-4
21
B-1
= |-5 6 2
8
9 3.
С.
-81
|-5 -6
-4
B-1
%3|
I-2
-2
3

Expert Solution

Step 1

Given matrix is $B = [\begin{matrix} 0 & - 3 & - 2 \\ 1 & - 4 & - 2 \\ - 3 & 4 & 1 \end{matrix}]$

We have to find the inverse of the matrix $B$

The inverse of $B$ is defined as $B^{- 1} = \frac{a d j (B)}{\det (B)}$

First of all we will find the determinant of $B$

$\begin{matrix} \det (B) = 0 (- 4 + 8) - (- 3) (1 - 6) - 2 (4 - 12) \\ = 0 + 3 (- 5) - 2 (- 8) \\ = 0 + 3 (- 5) - 2 (- 8) \\ = 0 - 15 + 16 \\ = 1 \end{matrix}$

Therefore, $\det (B) = 1$

Step 2

Now, find the cofactors of $B$

$b_{11} = |\begin{matrix} - 4 & - 2 \\ 4 & 1 \end{matrix}| = (- 4 + 8) = 4$

$b_{12} = - |\begin{matrix} 1 & - 2 \\ - 3 & 1 \end{matrix}| = - (1 - 6) = 5$

$b_{13} = |\begin{matrix} 1 & - 4 \\ - 3 & 4 \end{matrix}| = (4 - 12) = - 8$

$b_{21} = - |\begin{matrix} - 3 & - 2 \\ 4 & 1 \end{matrix}| = - (- 3 + 8) = - 5$

$b_{22} = |\begin{matrix} 0 & - 2 \\ - 3 & 1 \end{matrix}| = (0 - 6) = - 6$

$b_{23} = - |\begin{matrix} 0 & - 3 \\ - 3 & 4 \end{matrix}| = - (0 - 9) = 9$

$b_{31} = |\begin{matrix} - 3 & - 2 \\ - 4 & - 2 \end{matrix}| = (6 - 8) = - 2$

$b_{32} = - |\begin{matrix} 0 & - 2 \\ 1 & - 2 \end{matrix}| = - (0 + 2) = - 2$

$b_{33} = |\begin{matrix} 0 & - 3 \\ 1 & - 4 \end{matrix}| = (0 + 3) = 3$

Now, write the cofactor matrix is

$\begin{array}{l} \Rightarrow c o f (B) = [\begin{matrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{matrix}] \\ \Rightarrow c o f (B) = [\begin{matrix} 4 & 5 & - 8 \\ - 5 & - 6 & 9 \\ - 2 & - 2 & 3 \end{matrix}] \end{array}$

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