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Show that AB ≠ BA in each of the following cases :

A = \(\begin{bmatrix} 1 & 2 & 3 \\[0.3em] 0& 1 & 0 \\[0.3em] 1 &1 & 0 \end{bmatrix}\) and B = \(\begin{bmatrix} -1 & 1 & 0 \\[0.3em] 0& -1 & 1 \\[0.3em] 2 &3 & 4 \end{bmatrix}\)

A = [(1,2,3)(0,1,0)(1,1,0)]

B = [(-1,1,0)(0,-1,1)(2,3,4)]

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Given : A = \(\begin{bmatrix} 1 & 2 & 3 \\[0.3em] 0 & 1 & 0\\[0.3em] 1 & 1 & 0 \end{bmatrix}\) and B = \(\begin{bmatrix} -1 & 1 & 0 \\[0.3em] 0 & -1 & 1\\[0.3em] 2 & 3 & 4 \end{bmatrix}\) 

Matrix A is of order 3 x 3, and Matrix B is of order 3 x 3

To show : matrix AB \(\neq\) BA 

The formula used :

If A is a matrix of order a x b and B is a matrix of order c x d ,

then matrix AB exists and is of order a x d ,if and only if b = c

If A is a matrix of order a x b and B is a matrix of order c x d ,

then matrix BA exists and is of order c x b, if and only if d = a

For matrix AB, a = 3,b = c = 3,d = 3 ,thus matrix AB is of order 3 x 3

Matrix AB = 

For matrix BA, a = 3,b = c = 3,d = 3 ,thus matrix AB is of order 3 x 3

Matrix BA =

Matrix AB \(\neq\) BA

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