Question

Find inverse of matrix using adjoint method
( 0 1 2 )
( 1 2 3 )
( 3 1 1 )​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

Step-by-step explanation:

Given matrix = $\left[\begin{array}{ccc}0&1&2\\1&2&3\\3&1&1\end{array}\right]$

We need to find its inverse

We know that for a matrix A, A⁻¹ = $\frac{adj A}{|A|}$

Therefore, here

|A| = 0(2*1 - 3*1) -1(1*1 - 3*3) + 2(1*1 - 3*2)

    = 0  - 1( 1-9) + 2(1-6)

    = 0 -1(-8) + 2(-5)

    = 0 + 8 - 10

    = -2

We know that adjoint of a matrix is the transpose of cofactor matrix.

Therefore, we first find the cofactor matrix of A.

=> A₁₁ = $\left|\begin{array}{cc}2&3\\1&1\end{array}\right|$ = 2*1 - 3*1 = 2 - 3 = -1

    A₁₂ = $- \left|\begin{array}{cc}1&3\\3&1\end{array}\right|$ = -(1*1 - 3*3) = -(1-9) = -(-8) = 8

    A₁₃ = $\left|\begin{array}{cc}1&2\\3&1\end{array}\right|$ = 1*1 - 3*2 = 1 - 6 = -5

    A₂₁ =  $- \left|\begin{array}{cc}1&2\\1&1\end{array}\right|$ = -(1*1 - 1*2) = -(1-2) = -(-1) = 1

    A₂₂ = $\left|\begin{array}{cc}0&2\\3&1\end{array}\right|$ = 0*3 - 3*2 = 0 - 6 = -6

    A₂₃ = $-\left|\begin{array}{cc}0&1\\3&1\end{array}\right|$ = -(0*1 - 3*1) = -(0 - 3) = -(-3) = 3

    A₃₁ = $\left|\begin{array}{cc}1&2\\2&3\end{array}\right|$ = 1*3 - 2*2 = 3 - 4 = -1

    A₃₂ = $-\left|\begin{array}{cc}0&2\\1&3\end{array}\right|$ = -(0*3 - 1*2) = -(0 - 2) = -(-2) = 2

    A₃₃ = $\left|\begin{array}{cc}0&1\\1&2\end{array}\right|$ = 0*2 - 1*1 = 0 - 1 = -1

Therefore, the cofactor matrix = $\left[\begin{array}{ccc}-1&8&-5\\1&-6&3\\-1&2&-1\end{array}\right]$

Transpose of the cofactor matrix = $\left[\begin{array}{ccc}-1&1&-1\\8&-6&2\\-5&3&-1\end{array}\right]$

Therefore, as A⁻¹ = $\frac{adj A}{|A|}$

=> A⁻¹ = $\frac{-1}{2}$* $\left[\begin{array}{ccc}-1&1&-1\\8&-6&2\\-5&3&-1\end{array}\right]$

          =  $\left[\begin{array}{ccc}\frac{1}{2} &\frac{-1}{2}&\frac{1}{2}\\-4&3&-1\\\frac{5}{2}&\frac{-3}{2}&\frac{1}{2}\end{array}\right]$

Therefore,  A⁻¹ =  $\left[\begin{array}{ccc}\frac{1}{2} &\frac{-1}{2}&\frac{1}{2}\\-4&3&-1\\\frac{5}{2}&\frac{-3}{2}&\frac{1}{2}\end{array}\right]$