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Find the inverse of the matrix $A=\begin{bmatrix} 3 & -3 & 4 \\2 & -3 & 4 \\0 & -1 & 1 \end{bmatrix}$ and verify that $A^3=A^{-1}$

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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$
  • Multiplication of two matrices A and B is possible when the number of columns in the first matrix equals the number of rows in the second matrix B. If A is of type $ m $ x $ p $ and B is of type $ p $ x $ x $ then AB is of type $ m $ x $ n $ $AB = [ c_{ij}]$ where $ c_{ij} = \sum^p_{\substack{k=1}}a_{ik}b_{kj}$
  • 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
To find $ A^{-1} = \frac{1}{|A|}adj\: A$
$ |A| = \begin{vmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{vmatrix} = 3(-3+4)-2(-3+4)$
$ = 3-2=1 \neq 0 \Rightarrow A^{-1}$ exists.
To find $ adj\: A$
$ [ A_{ij}] = \begin{bmatrix} (-3+4) & -(2-0) & (-2-0) \\ -(-3+4) & (3-0) & -(-3-0) \\ (-12+12) & -(12-8) & (-9+6) \end{bmatrix}$
$ = \begin{bmatrix} 1 & -2 & -2 \\ -1 & 3 & 3 \\ 0 & -4 & -3 \end{bmatrix}$
$ adj\: A = [ A_{ij}]^T = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$
$ A^{-1} = \frac{1}{|A|} adj\: A = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$
Step 2
To verify $ A^3 = A^{-1}$
$ A^2 = AA = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$
$ = \begin{bmatrix} (3)(3)+(3)(2)+0 & (3)(-3)+(-3)+(4)(-1) & (3)(1)+(-3)(1)+(4)(1) \\ (2)(3)+(-3)(2)+0 & (2)(-3)+(-3)(-3)+(4)(-1) & (2)(4)+(-3)(4)+(4)(1) \\ 0+(-1)(3)+0 & 0+(-1)(-3)+(1)(-1) & 0+(-1)(4)+(1)(1) \end{bmatrix}$
$ \begin{bmatrix} 9-6 & -9+9-4 & 12-12+4 \\ 6-6+0 & -6+9-4 & 8-12+4 \\ 0-2+0 & 0+3-1 & 0-4+1 \end{bmatrix} = \begin{bmatrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix}$
$ A^3 = AA^2 = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix}$
$= \begin{bmatrix} (3)(3)+(-3)(0)+(4)(-2) & (3)(-4)+(-3)(-1)+(4)(2) & (3)(4)+0+(4)(-3) \\ (2)(3)+0+(4)(-2) & (2)(-4)+(-3)(-1)+(4)(2) & (2)(+4)+0+(4)(-3) \\ 0+0+(1)(-2) & 0+(-1)(-1)+(1)(2) & 0+0+(1)(-3) \end{bmatrix}$
$ = \begin{bmatrix} 9+0-8 & -12+3+8 & 12-12 \\ 6-8 & -8+3+8 & 8-12 \\ -2 & 1+2 & -3 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$
From step 1 $ A^{-1} = A^3$
answered May 23, 2013 by thanvigandhi_1
 

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