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If$A=\begin{bmatrix} 5 & 2 \\7 & 3 \end{bmatrix}$ and$B=\begin{bmatrix} 2 & -1 \\-1 & 1 \end{bmatrix}$ verify that \((AB)^{-1}=B^{-1}A^{-1}\)

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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$
  • Let A be a square matrix of order 2. $ A = \begin{bmatrix} a  &  b  \\  c  &  d  \end{bmatrix}$. Then $ adj\: A = \begin{bmatrix} d  &  -b  \\  -c  &  a  \end{bmatrix}$ ( shortcut for finding the adjoint in multiple choice questions / objective questions )
  • 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
AB = $ \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}$
$ = \begin{bmatrix} (5)(2)+(2)(-1) & (5)(-1)+(2)(1) \\ (7)(2)+(3)(-1) & (7)(-1)+(3)(1) \end{bmatrix}$
$ = \begin{bmatrix} 10-2 & 5+2 \\ 14-3 & -7+3 \end{bmatrix} = \begin{bmatrix} 8 & -3 \\ 11 & -4 \end{bmatrix}$
$ (AB)^T = \begin{bmatrix} 8 & 11 \\ -3 & -4 \end{bmatrix}$
Step 2
To find the inverses
$ |A| = \begin{vmatrix} 5 & 2 \\ 7 & 3 \end{vmatrix} = 15-14=1$
$ adj\: A = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$
$ A^{-1} = \frac{1}{|A|} adj\: A = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$
$ |B| = \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix} = 2-1 = 1$
$ adj\: B = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$
$ B^{-1} = \frac{1}{|B|} adj\: B = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$
$ |AB| = \begin{vmatrix} 8 & -3 \\ 11 & -4 \end{vmatrix} = -32+33=1$
$ adj\: AB = \begin{bmatrix} -4 & 3 \\ -11 & 8 \end{bmatrix}$
$ (AB)^{-1} = \begin{bmatrix} -4 & 3 \\ -11 & 8 \end{bmatrix}$
Step 3
$ B^{-1}\: A^{-1} = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$
$ = \begin{bmatrix} (1)(3)+(1)(-7) & (1)(-2)+(1)(5) \\ (1)(3)+(2)(-7) & (1)(-2)+(2)(5) \end{bmatrix}$
$ = \begin{bmatrix} 3-7 & -2+5 \\ 3-14 & -2+10 \end{bmatrix} = \begin{bmatrix} -4 & 3 \\ -11 & 8 \end{bmatrix}$
From step 2 $(AB)^{-1} = B^{-1}A^{-1}$

 

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

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