Given the two matrices, A=begin{bmatrix}1 & 2&3 1 & 1&20&1&2 end{bmatrix} text{ and } B=begin{bmatrix}1 & 1&1 2 & 1&23&1&2 end{bmatrix} (a) Find det A

Cabiolab 2021-02-15 Answered

Given the two matrices,
\(A=\begin{bmatrix}1 & 2&3 \\1 & 1&2\\0&1&2 \end{bmatrix} \text{ and } B=\begin{bmatrix}1 & 1&1 \\2 & 1&2\\3&1&2 \end{bmatrix}\)
(a) Find det A, det B , det(AB) , det(BA) , det(5A) , \(det A^T\) and \(det(B^6)\)
(b) Find adj A and adj B
(c) Find \(A^{-1}\) and \(B^{-1}\) using the adjoint matrices you found in part (b)

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Expert Answer

odgovoreh
Answered 2021-02-16 Author has 14992 answers

Step 1
We have given the matrices
\(A=\begin{bmatrix}1 & 2&3 \\1 & 1&2\\0&1&2 \end{bmatrix} \text{ and } B=\begin{bmatrix}1 & 1&1 \\2 & 1&2\\3&1&2 \end{bmatrix}\)
Step 2
Part(a)
Find det A:
\(det A=det\begin{bmatrix}1 & 2&3 \\1 & 1&2\\0&1&2 \end{bmatrix} =1 \cdot det\begin{bmatrix}1 & 2 \\1 & 2 \end{bmatrix}-2 \cdot det \begin{bmatrix}1 & 2 \\0 & 2 \end{bmatrix}+3 \cdot det \begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}\)
\(=1 \cdot 0 - 2 \cdot 2 +3 \cdot 1\)
=-1 Find det B:
\(det B=det\begin{bmatrix}1 & 1&1 \\2 & 1&2\\3&1&2 \end{bmatrix}\) \(=1 \cdot det \begin{bmatrix}1 & 2 \\1 & 2 \end{bmatrix}- 1 \cdot det \begin{bmatrix}2 & 2 \\3 & 2 \end{bmatrix}+1 \cdot det \begin{bmatrix}2 & 1 \\3 & 1 \end{bmatrix}\)
\(=1 \cdot 0 - 1 \cdot (-2) +1 \cdot (-1)\)
=1
Step 3
Find det(AB) and det(BA)
According to determinant properties,
\(det(AB)=det A \times det B\)
\(=-1 \times 1\)
=-1
\(det(BA)=det B \times det A\)
\(=1 \times -1\)
=-1
Step 4
Find det(5A)
\(det(5A)=5^3 \times det A\)
\(=125 \times -1\)
=-125
Find \(det A^T:\)
\(det A^T = det A\)
=-1
Find \(det(B^6):\)
\(det(B^6)=(det B)^6\)
\(=1^6\)
=1
Step 5
Part (b)
Find adj A and adj B
\(A=\begin{bmatrix}1 & 2&3 \\1 & 1&2\\0&1&2 \end{bmatrix}\)
The cofactors matrix is
\(C=\begin{bmatrix}+det\begin{bmatrix}1 & 2 \\1 & 2 \end{bmatrix} & -det\begin{bmatrix}1 & 2 \\0 & 2 \end{bmatrix}& +det\begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}\\-det\begin{bmatrix}2 & 3 \\1 & 2 \end{bmatrix} & +det\begin{bmatrix}1 & 3 \\0 & 2 \end{bmatrix}&-det\begin{bmatrix}1 & 2 \\0 & 1 \end{bmatrix}\\+det\begin{bmatrix}2 & 3 \\1 & 2 \end{bmatrix}&-det\begin{bmatrix}1 & 3 \\1 & 2 \end{bmatrix}&+det\begin{bmatrix}1 & 2 \\1 & 1 \end{bmatrix} \end{bmatrix}\)
\(C=\begin{bmatrix}+(2-2) & -(2-0)&+(1-0) \\-(4-3) & +(2-0)&-(1-0)\\+(4-3)&-(2-3)&+(1-2) \end{bmatrix}\)
\(C=\begin{bmatrix}0 & -2&1 \\-1 & 2&-1\\1&1&-1 \end{bmatrix}\)
\(adj A=C^T=\begin{bmatrix}0 & -1&1 \\-2 & 2&1\\1&-1&-1 \end{bmatrix}\)
Step 6
\(B=\begin{bmatrix}1 & 1&1 \\2 & 1&2\\3&1&2 \end{bmatrix}\)
The cofactors matrix is
\(C=\begin{bmatrix}+det\begin{bmatrix}1 & 2 \\1 & 2 \end{bmatrix} & -det\begin{bmatrix}2 & 2 \\3 & 2 \end{bmatrix}& +det\begin{bmatrix}2 & 1 \\3 & 1 \end{bmatrix}\\-det\begin{bmatrix}1 & 1 \\1 & 2 \end{bmatrix} & +det\begin{bmatrix}1 & 1 \\3 & 2 \end{bmatrix}&-det\begin{bmatrix}1 & 1 \\3 & 1 \end{bmatrix}\\+det\begin{bmatrix}1 & 1 \\1 & 2 \end{bmatrix}&-det\begin{bmatrix}1 & 1 \\2 & 2 \end{bmatrix}&+det\begin{bmatrix}1 & 1 \\2 & 1 \end{bmatrix} \end{bmatrix}\)
\(C=\begin{bmatrix}+(2-2) & -(4-6)&+(2-3) \\-(2-1) & +(2-3)&-(1-3)\\+(2-1)&-(2-2)&+(1-2) \end{bmatrix}\)
\(C=\begin{bmatrix}0 & 2&-1 \\-1 & -1&2\\1&0&-1 \end{bmatrix}\)
\(adj B=C^T=\begin{bmatrix}0 & -1&1 \\2 & -1&0\\-1&2&-1 \end{bmatrix}\)
Step 7
Part (c)
Find \(A^{-1} and B^{-1}\)
\(A^{-1}=\frac{1}{det A}adj A\)
\(=\frac{1}{-1}\begin{bmatrix}0 & -1&1 \\-2 & 2&1\\1&-1&-1 \end{bmatrix}\)
\(=\begin{bmatrix}0 & 1&-1 \\2 & -2&-1\\-1&1&1 \end{bmatrix}\)
\(B^{-1}=\frac{1}{det B}adj B\)
\(=\frac{1}{1}\begin{bmatrix}0 & -1&1 \\2 & -1&0\\-1&2&-1 \end{bmatrix}\)
\(=\begin{bmatrix}0 & -1&1 \\2 & -1&0\\-1&2&-1 \end{bmatrix}\)

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