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

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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 $ adj\: A $ ( or adjoint of the matrix A) is given by $ adj\: A=[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} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{bmatrix} $
$ |A| = \begin{vmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{vmatrix} = 2(6-2)-2(2-1)+1(2-3)$
$ = 8-2-1=5 \neq 0$
$ \therefore A^{-1}$ exists.
Step 2
To find $ adj\: A$
$ [ A_{ij}] = \begin{bmatrix} (6-2) & -( 2-1) & (2-3) \\ -(4-2) & (4-1) & -(4-2) \\ (2-3) & -(2-1) & (6-2) \end{bmatrix} = \begin{bmatrix} 4 & -1 & -1 \\ -2 & 3 & -2 \\ -1 & -1 & 4 \end{bmatrix} $
$ adj\: A = [A_{ij} ]^T = \begin{bmatrix} 4 & -2 & -1 \\ -1 & 3 & -1 \\ -1 & 2 & -4 \end{bmatrix} $
Step 3
$ A^{-1} = \large\frac{1}{|A|} adj\: A = \large\frac{1}{5} \begin{bmatrix} 4 & -2 & -1 \\ -1 & 3 & -1 \\ -1 & 2 & -4 \end{bmatrix} $

 

answered May 23, 2013 by thanvigandhi_1
edited Jun 3, 2013 by thanvigandhi_1
 

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