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Q)

Find the inverse of following matrix: $\begin{bmatrix} 1 & 3 & 7 \\4 & 2 & 3 \\1 & 2 & 1 \end{bmatrix}$

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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 $ 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} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{bmatrix}$
$ |A| = \begin{vmatrix} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{vmatrix} = 1(2-6)-3(4-3)+7(8-2) $
$ = -4-3+42 = 35 \neq 0$
A is non-singular. $A^{-1}$ exists.
Step 2
To find $ adj\: A$
$ [A_{ij} ] = \begin{bmatrix} (2-6) & -(4-3) & (8-2) \\ -(3-14) & (1-7) & -(2-3) \\ (9-14) & -(3-28) & (2-12) \end{bmatrix} = \begin{bmatrix} -4 & -1 & 6 \\ 11 & -6 & 1 \\ -5 & 25 & -10 \end{bmatrix} $
$ adj\: A = [ A_{ij}]^T = \begin{bmatrix} -4 & 11 & -5 \\ -1 & -6 & 25 \\ 6 & 1 & -10 \end{bmatrix} $
Step 3
$ A^{-1} = \frac{1}{|A|} adj\: A$
$ \frac{1}{35} \begin{bmatrix} -4 & 11 & -5 \\ -1 & -6 & 25 \\ 6 & 1 & -10 \end{bmatrix} $

 

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