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Q)

Find the inverse of following matrix :$\begin{bmatrix} 1 & 0 & 3 \\2 & 1 & -1 \\1 & -1 & 1 \end{bmatrix}$

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A)
Toolbox:
  • Let $ A = [ a_{ij} ] $ be a square matrix x. Let $ A_{ij}$ be the cofactor of $ a_{ij}$. Then $ [ A_{ij}]$ is the matrix of cofactors and $ adjA $ ( or adjoint of the matrix A) is given by $ adjA=[A_{ij}]^T$
  • A determinant can be expanded by using the elements of any row or column.
  • The inverse of a non-singular square matrix A is given by $ A^{-1} = \frac{1}{|A|} adj\: A.$ A non-singular matrix is one whose determinant value is nonzero.
Step 1
$ A = \begin{bmatrix} 1 & 0 & 3 \\ 2 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix}$
$ |A| = \begin{vmatrix} 1 & 0 & 3 \\ 2 & 1 & -1 \\ 1 & -1 & 1 \end{vmatrix} = 1(1-1)+0+3(-2-1)$
$ = 0-9 = -9 \neq 0$
A is non-singular and so $A^{-1}$ exists.
Step 2
To find $ adj\: A$
$ [ A_{ij} ] = \begin{bmatrix} (1-1) & -(2+1) & (-2-1) \\ -(0+3) & (1-3) & -(-1-0) \\ (0-3) & -(-1-6) & (1-0) \end{bmatrix}$
$ = \begin{bmatrix} 0 & -3 & -3 \\ -3 & -2 & 1 \\ -3 & 7 & 1 \end{bmatrix}$
$ adj\: A = [ A_{ij} ]^T = \begin{bmatrix} 0 & -3 & -3 \\ -3 & -2 & 7 \\-3 & 1 & 1 \end{bmatrix}$
Step 3
$ A^{-1} = \large\frac{1}{|A|} \: adj\: A = \large\frac{1}{-9} \begin{bmatrix} 0 & -3 & -3 \\ -3 & -2 & 7 \\ -3 & 1 & 1 \end{bmatrix}$
$ = \large\frac{1}{9} \begin{bmatrix} 0 & 3 & 3 \\ 3 & 2 & -7 \\ 3 & -1 & -1 \end{bmatrix}$

 

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