Browse Questions

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

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 $(AB)^T=B^TA^T$