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Q : 9         Find the inverse of each of the matrices (if it exists).

                 \small \begin{bmatrix} 2 &1 &3 \\ 4 &-1 &0 \\ -7 &2 &1 \end{bmatrix}

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Given the matrix :  \small \begin{bmatrix} 2 &1 &3 \\ 4 &-1 &0 \\ -7 &2 &1 \end{bmatrix} =A

To find the inverse we have to first find adjA then as we know the relation:

A^{-1} = \frac{1}{|A|}adjA

So, calculating |A| :

|A| = 2(-1-0)-1(4-0)+3(8-7) =-2-4+3 = -3

Now, calculating the cofactors terms and then adjA.

A_{11} = (-1)^{1+1} (-1-0) = -1                  A_{12} = (-1)^{1+2} (4-0) = -4

A_{13} = (-1)^{1+3} (8-7) =1                      A_{21} = (-1)^{2+1} (1-6) = 5

A_{22} = (-1)^{2+2} (2+21) = 23             A_{23} = (-1)^{2+1} (4+7) = -11

A_{31} = (-1)^{3+1} (0+3) = 3             A_{32} = (-1)^{3+2} (0-12) =12

A_{33} = (-1)^{3+3} (-2-4) = -6               

So, we have adjA = \begin{bmatrix} -1 &5 &3 \\ -4& 23 &12 \\ 1& -11 &-6 \end{bmatrix}

Therefore inverse of A will be:

A^{-1} = \frac{1}{|A|}adjA

A^{-1} = \frac{1}{-3} \begin{bmatrix} -1 &5 &3 \\ -4& 23 &12 \\ 1& -11 &-6 \end{bmatrix}

 

Posted by

Divya Prakash Singh

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