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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\((AB)^{T}=B^{T}A^{T}\)

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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 $ adjA $ ( or adjoint of the matrix A) is given by $ adjA=[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 )
  • 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} 8 & -3 \\ 1 & -4 \end{bmatrix}$
$(AB)^T = \begin{bmatrix} 8 & 11 \\ -3 & -4 \end{bmatrix}$
$ A^T = \begin{bmatrix} 5 & 7 \\ 2 & 3 \end{bmatrix} \: \: B^T = \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}$
Step 2
$ B^TA^T = \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix}5 & 7 \\ 2 & 3 \end{bmatrix}$
$ \begin{bmatrix} (2)(5)+(-1)(2) & (2)(7)+(-1)(3) \\ (-1)(5)+(1)(2) & (-1)(7)+(1)(3) \end{bmatrix}$
$ = \begin{bmatrix} 10-2 & 14-3 \\ -5+2 & -7+3 \end{bmatrix} = \begin{bmatrix} 8 & 11 \\ -3 & -4 \end{bmatrix}$
From step 1 $ (AB)^T=B^TA^T$
answered Jun 3, 2013 by thanvigandhi_1
 

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