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Question: Let A be a 3 × 3 matrix such that A \(\left( \begin{matrix} 1 & 2 & 3 \\\ 0 & 2 & 3 \\\ ...

Let A be a 3 × 3 matrix such that A (123 023 011 )\left( \begin{matrix} 1 & 2 & 3 \\\ 0 & 2 & 3 \\\ 0 & 1 & 1 \\\ \end{matrix} \right) = (001 100 010 )\left( \begin{matrix} 0 & 0 & 1 \\\ 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ \end{matrix} \right)
Then A−1{{A}^{-1}} =

Explanation

Solution

When two matrices are multiplied to get a new matrix , we can also find one of two matrices by multiplying inverse of the other given matrix on both sides. Now you can find the inverse of the matrix by finding adjacent matrices and then multiplying the inverse matrix with the product to get that value of the unknown matrix.

Complete step by step answer:
Let us assume that the matrix (123 023 011 )\left( \begin{matrix} 1 & 2 & 3 \\\ 0 & 2 & 3 \\\ 0 & 1 & 1 \\\ \end{matrix} \right) is B.
B = (123 023 011 )\left( \begin{matrix} 1 & 2 & 3 \\\ 0 & 2 & 3 \\\ 0 & 1 & 1 \\\ \end{matrix} \right)
⇒\Rightarrow (AB) = (001 100 010 )\left( \begin{matrix} 0 & 0 & 1 \\\ 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ \end{matrix} \right)
By multiplying B−1{{B}^{-1}}on both sides on the right side of the above equation, we get
(AB)B−1{{B}^{-1}} = (001 100 010 )\left( \begin{matrix} 0 & 0 & 1 \\\ 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ \end{matrix} \right) B−1{{B}^{-1}}
By associative law of matrices we have, (AB)C = A(BC)
A(BB−1)A(B{{B}^{-1}})) = (001 100 010 )B−1\left( \begin{matrix} 0 & 0 & 1 \\\ 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ \end{matrix} \right) {{B}^{-1}}
⇒\Rightarrow A = (001 100 010 )\left( \begin{matrix} 0 & 0 & 1 \\\ 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ \end{matrix} \right) B−1{{B}^{-1}} (because BB−1{{B}^{-1}}= I and A I = A , where I is the identity matrix)
Now we need to find the value of B inverse and then multiply it with (001 100 010 )\left( \begin{matrix} 0 & 0 & 1 \\\ 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ \end{matrix} \right) to get the value of A.
We know that B−1{{B}^{-1}} = 1det⁡B(AdjB)\dfrac{1}{\det B}(AdjB)
The determinant of matrix (a11a12a13 a21a22a23 a31a32a33 )\left( \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right) will be det⁡(a11a12a13 a21a22a23 a31a32a33 )=a11(a22a33−a32a23)−a12(a21a33−a31a23)+a13(a21a32−a22a31)\det \left( \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right)={{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{32}}{{a}_{23}})-{{a}_{12}}({{a}_{21}}{{a}_{33}}-{{a}_{31}}{{a}_{23}})+{{a}_{13}}({{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}})
Comparing (a11a12a13 a21a22a23 a31a32a33 )\left( \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right) with (123 023 011 )\left( \begin{matrix} 1 & 2 & 3 \\\ 0 & 2 & 3 \\\ 0 & 1 & 1 \\\ \end{matrix} \right)
(a11=1a12=2a13=3 a21=0a22=2a23=3 a31=0a32=1a33=1 )\left( \begin{matrix} {{a}_{11}}=1 & {{a}_{12}}=2 & {{a}_{13}}=3 \\\ {{a}_{21}}=0 & {{a}_{22}}=2 & {{a}_{23}}=3 \\\ {{a}_{31}}=0 & {{a}_{32}}=1 & {{a}_{33}}=1 \\\ \end{matrix} \right)
detB = 1(2−3)2(0−0)+3(0−0)=−1detB\text{ }=\text{ }1\left( 2-3 \right)2\left( 0-0 \right)+3\left( 0-0 \right)=-1
Now we need the value of Adj B
For finding the adjoint matrix we will first find the minors of each element and then writing the minors matrix by taking determinant values of each minor and then convert it into cofactor matrix and then finally taking the transpose of cofactor matrix.
The minors of the matrix (a11=1a12=2a13=3 a21=0a22=2a23=3 a31=0a32=1a33=1 )\left( \begin{matrix} {{a}_{11}}=1 & {{a}_{12}}=2 & {{a}_{13}}=3 \\\ {{a}_{21}}=0 & {{a}_{22}}=2 & {{a}_{23}}=3 \\\ {{a}_{31}}=0 & {{a}_{32}}=1 & {{a}_{33}}=1 \\\ \end{matrix} \right) will be
M11=minor of a11=det⁡(23 11 )=2(1)−3(1)=−1 M12=minor of a12=det⁡(03 01 )=0−0=0 M13=minor of a13=det⁡(02 01 )=0−0=0 M21=minor of a21=det⁡(23 11 )=2−3=−1 M22=minor of a22=det⁡(13 01 )=1−0=1 M23=minor of a23=det⁡(12 01 )=1−0=1 M31=minor of a31=det⁡(23 23 )=6−6=0 M32=minor of a32=det⁡(13 03 )=3−0=3 M33=minor of a33=det⁡(12 02 )=2−0=2 \begin{aligned} & {{M}_{11}}=\text{minor of }{{\text{a}}_{11}}=\det \left( \begin{matrix} 2 & 3 \\\ 1 & 1 \\\ \end{matrix} \right)=2(1)-3(1)=-1 \\\ & {{M}_{12}}=\text{minor of }{{\text{a}}_{12}}=\det \left( \begin{matrix} 0 & 3 \\\ 0 & 1 \\\ \end{matrix} \right)=0-0=0 \\\ & {{M}_{13}}=\text{minor of }{{\text{a}}_{13}}=\det \left( \begin{matrix} 0 & 2 \\\ 0 & 1 \\\ \end{matrix} \right)=0-0=0 \\\ & {{M}_{21}}=\text{minor of }{{\text{a}}_{21}}=\det \left( \begin{matrix} 2 & 3 \\\ 1 & 1 \\\ \end{matrix} \right)=2-3=-1 \\\ & {{M}_{22}}=\text{minor of }{{\text{a}}_{22}}=\det \left( \begin{matrix} 1 & 3 \\\ 0 & 1 \\\ \end{matrix} \right)=1-0=1 \\\ & {{M}_{23}}=\text{minor of }{{\text{a}}_{23}}=\det \left( \begin{matrix} 1 & 2 \\\ 0 & 1 \\\ \end{matrix} \right)=1-0=1 \\\ & {{M}_{31}}=\text{minor of }{{\text{a}}_{31}}=\det \left( \begin{matrix} 2 & 3 \\\ 2 & 3 \\\ \end{matrix} \right)=6-6=0 \\\ & {{M}_{32}}=\text{minor of }{{\text{a}}_{32}}=\det \left( \begin{matrix} 1 & 3 \\\ 0 & 3 \\\ \end{matrix} \right)=3-0=3 \\\ & {{M}_{33}}=\text{minor of }{{\text{a}}_{33}}=\det \left( \begin{matrix} 1 & 2 \\\ 0 & 2 \\\ \end{matrix} \right)=2-0=2 \\\ \end{aligned}
The minor matrix will be
(M11M12M13 M21M22M23 M31M32M33 )=\left( \begin{matrix} {{M}_{11}} & {{M}_{12}} & {{M}_{13}} \\\ {{M}_{21}} & {{M}_{22}} & {{M}_{23}} \\\ {{M}_{31}} & {{M}_{32}} & {{M}_{33}} \\\ \end{matrix} \right)= (−100 −111 032 )\left( \begin{matrix} -1 & 0 & 0 \\\ -1 & 1 & 1 \\\ 0 & 3 & 2 \\\ \end{matrix} \right)
now the co-factor matrix will be(−1)i+jaij{{\left( -1 \right)}^{i+j}}{{a}_{ij}}
co-factor matrix= (−1−(0)0 −(−1)1−(1) 0−(3)2 )=\left( \begin{matrix} -1 & -(0) & 0 \\\ -(-1) & 1 & -(1) \\\ 0 & -(3) & 2 \\\ \end{matrix} \right)= (−100 11−1 0−32 )\left( \begin{matrix} -1 & 0 & 0 \\\ 1 & 1 & -1 \\\ 0 & -3 & 2 \\\ \end{matrix} \right)
the transpose of cofactor matrix that is AdjB will be
(co-factor matrix)T=(−100 11−1 0−32 )T={{\left( \text{co-factor matrix} \right)}^{T}}={{\left( \begin{matrix} -1 & 0 & 0 \\\ 1 & 1 & -1 \\\ 0 & -3 & 2 \\\ \end{matrix} \right)}^{T}}= (−110 01−3 0−12 )\left( \begin{matrix} -1 & 1 & 0 \\\ 0 & 1 & -3 \\\ 0 & -1 & 2 \\\ \end{matrix} \right)
now B−1{{B}^{-1}} will be 1det⁡B(AdjB)\dfrac{1}{\det B}(AdjB)
that implies
B−1{{B}^{-1}}= 1(−1)(−110 01−3 0−12 )\dfrac{1}{(-1)}\left( \begin{matrix} -1 & 1 & 0 \\\ 0 & 1 & -3 \\\ 0 & -1 & 2 \\\ \end{matrix} \right)
B−1{{B}^{-1}}=(1−10 0−13 01−2 )\left( \begin{matrix} 1 & -1 & 0 \\\ 0 & -1 & 3 \\\ 0 & 1 & -2 \\\ \end{matrix} \right)
Now by multiplying (001 100 010 )\left( \begin{matrix} 0 & 0 & 1 \\\ 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ \end{matrix} \right) with B−1{{B}^{-1}} =(1−10 0−13 01−2 )\left( \begin{matrix} 1 & -1 & 0 \\\ 0 & -1 & 3 \\\ 0 & 1 & -2 \\\ \end{matrix} \right)we get A
A=(001 100 010 )×(1−10 0−13 01−2 )=(01−2 1−10 0−13 )\left( \begin{matrix} 0 & 0 & 1 \\\ 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ \end{matrix} \right)\times \left( \begin{matrix} 1 & -1 & 0 \\\ 0 & -1 & 3 \\\ 0 & 1 & -2 \\\ \end{matrix} \right)=\left( \begin{matrix} 0 & 1 & -2 \\\ 1 & -1 & 0 \\\ 0 & -1 & 3 \\\ \end{matrix} \right)
A = (01−2 1−10 0−13 )\left( \begin{matrix} 0 & 1 & -2 \\\ 1 & -1 & 0 \\\ 0 & -1 & 3 \\\ \end{matrix} \right)
In the similar let us find A−1{{A}^{-1}},
A−1{{A}^{-1}}=1det⁡A(adjA)\dfrac{1}{\det A}(adjA)
det⁡(01−2 1−10 0−13 )=0(−3)−1(3)−2(−1)=−1 det⁡A=−1 \begin{aligned} & \det \left( \begin{matrix} 0 & 1 & -2 \\\ 1 & -1 & 0 \\\ 0 & -1 & 3 \\\ \end{matrix} \right)=0(-3)-1(3)-2(-1)=-1 \\\ & \det A=-1 \\\ \end{aligned}
First let us find the minor matrix of A,
Comparing (01−2 1−10 0−13 )\left( \begin{matrix} 0 & 1 & -2 \\\ 1 & -1 & 0 \\\ 0 & -1 & 3 \\\ \end{matrix} \right),(a11a12a13 a21a22a23 a31a32a33 )\left( \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right)we have (a11=0a12=1a13=−2 a21=1a22=−1a23=0 a31=0a32=−1a33=3 )\left( \begin{matrix} {{a}_{11}}=0 & {{a}_{12}}=1 & {{a}_{13}}=-2 \\\ {{a}_{21}}=1 & {{a}_{22}}=-1 & {{a}_{23}}=0 \\\ {{a}_{31}}=0 & {{a}_{32}}=-1 & {{a}_{33}}=3 \\\ \end{matrix} \right)

M11=minor of a11=det⁡(−10 −13 )=−1(3)−0(−1)=−3 M12=minor of a12=det⁡(10 03 )=3−0=3 M13=minor of a13=det⁡(1−1 0−1 )=−1−0=−1 M21=minor of a21=det⁡(1−2 −13 )=3−2=1 M22=minor of a22=det⁡(0−2 03 )=0−0=0 M23=minor of a23=det⁡(01 0−1 )=0−0=0 M31=minor of a31=det⁡(1−2 −10 )=0−2=−2 M32=minor of a32=det⁡(0−2 10 )=0+2=2 M33=minor of a33=det⁡(01 1−1 )=0−1=−1 \begin{aligned} & {{M}_{11}}=\text{minor of }{{\text{a}}_{11}}=\det \left( \begin{matrix} -1 & 0 \\\ -1 & 3 \\\ \end{matrix} \right)=-1(3)-0(-1)=-3 \\\ & {{M}_{12}}=\text{minor of }{{\text{a}}_{12}}=\det \left( \begin{matrix} 1 & 0 \\\ 0 & 3 \\\ \end{matrix} \right)=3-0=3 \\\ & {{M}_{13}}=\text{minor of }{{\text{a}}_{13}}=\det \left( \begin{matrix} 1 & -1 \\\ 0 & -1 \\\ \end{matrix} \right)=-1-0=-1 \\\ & {{M}_{21}}=\text{minor of }{{\text{a}}_{21}}=\det \left( \begin{matrix} 1 & -2 \\\ -1 & 3 \\\ \end{matrix} \right)=3-2=1 \\\ & {{M}_{22}}=\text{minor of }{{\text{a}}_{22}}=\det \left( \begin{matrix} 0 & -2 \\\ 0 & 3 \\\ \end{matrix} \right)=0-0=0 \\\ & {{M}_{23}}=\text{minor of }{{\text{a}}_{23}}=\det \left( \begin{matrix} 0 & 1 \\\ 0 & -1 \\\ \end{matrix} \right)=0-0=0 \\\ & {{M}_{31}}=\text{minor of }{{\text{a}}_{31}}=\det \left( \begin{matrix} 1 & -2 \\\ -1 & 0 \\\ \end{matrix} \right)=0-2=-2 \\\ & {{M}_{32}}=\text{minor of }{{\text{a}}_{32}}=\det \left( \begin{matrix} 0 & -2 \\\ 1 & 0 \\\ \end{matrix} \right)=0+2=2 \\\ & {{M}_{33}}=\text{minor of }{{\text{a}}_{33}}=\det \left( \begin{matrix} 0 & 1 \\\ 1 & -1 \\\ \end{matrix} \right)=0-1=-1 \\\ \end{aligned}
Minor matrix would be ,
(M11M12M13 M21M22M23 M31M32M33 )=\left( \begin{matrix} {{M}_{11}} & {{M}_{12}} & {{M}_{13}} \\\ {{M}_{21}} & {{M}_{22}} & {{M}_{23}} \\\ {{M}_{31}} & {{M}_{32}} & {{M}_{33}} \\\ \end{matrix} \right)= (−33−1 100 −22−1 )\left( \begin{matrix} -3 & 3 & -1 \\\ 1 & 0 & 0 \\\ -2 & 2 & -1 \\\ \end{matrix} \right)
now the co-factor matrix will be(−1)i+jaij{{\left( -1 \right)}^{i+j}}{{a}_{ij}}
co-factor matrix= (−3−(3)−1 −(1)0−(0) −2−(2)−1 )=(−3−3−1 −100 −2−2−1 )\left( \begin{matrix} -3 & -(3) & -1 \\\ -(1) & 0 & -(0) \\\ -2 & -(2) & -1 \\\ \end{matrix} \right)=\left( \begin{matrix} -3 & -3 & -1 \\\ -1 & 0 & 0 \\\ -2 & -2 & -1 \\\ \end{matrix} \right)
the transpose of cofactor matrix that is AdjA will be
(co-factor matrix)T=(−3−3−1 −100 −2−2−1 )T=(−3−1−2 −30−2 −10−1 )=adjA{{\left( \text{co-factor matrix} \right)}^{T}}={{\left( \begin{matrix} -3 & -3 & -1 \\\ -1 & 0 & 0 \\\ -2 & -2 & -1 \\\ \end{matrix} \right)}^{T}}=\left( \begin{matrix} -3 & -1 & -2 \\\ -3 & 0 & -2 \\\ -1 & 0 & -1 \\\ \end{matrix} \right)=adjA
A−1{{A}^{-1}}=1det⁡A(adjA)\dfrac{1}{\det A}(adjA)

A−1{{A}^{-1}}=1(−1)(−3−1−2 −30−2 −10−1 )=(312 302 101 )\dfrac{1}{(-1)}\left( \begin{matrix} -3 & -1 & -2 \\\ -3 & 0 & -2 \\\ -1 & 0 & -1 \\\ \end{matrix} \right)=\left( \begin{matrix} 3 & 1 & 2 \\\ 3 & 0 & 2 \\\ 1 & 0 & 1 \\\ \end{matrix} \right)
A−1{{A}^{-1}} = (312 302 101 )\left( \begin{matrix} 3 & 1 & 2 \\\ 3 & 0 & 2 \\\ 1 & 0 & 1 \\\ \end{matrix} \right)

Note: Make sure that the determinant of the matrix for which you are finding the inverse is not 0. Because if it is zero , then the matrix will not be invertible , hence we cannot find the inverse of that particular matrix. Make sure there are no calculation mistakes as there are many calculation processes.