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Question: If A and B are square matrices of same order then...

If A and B are square matrices of same order then

A

(AB)′=A′B′(AB)^{'} = A^{'}B^{'}

B

(AB)′=B′A′(AB)^{'} = B^{'}A^{'}

C

AB=0,if∣A∣=0or∣B∣=0AB = 0,\text{if}|A| = 0or|B| = 0

D

AB=0,if∣A∣=IorB=IAB = 0,\text{if}|A| = IorB = I

Answer

(AB)′=B′A′(AB)^{'} = B^{'}A^{'}

Explanation

Solution

A=[aij]n×n,B=[bjk]n×nA = \lbrack a_{ij}\rbrack_{n \times n},B = \lbrack b_{jk}\rbrack_{n \times n}, AB=[aij]n×n[bjk]n×n=[cik]n×nAB = \lbrack a_{ij}\rbrack_{n \times n}\lbrack b_{jk}\rbrack_{n \times n} = \lbrack c_{ik}\rbrack_{n \times n},

where cik=aijbjkc_{ik} = a_{ij}b_{jk}

(AB)′=[cik]n×n′=[cki]n×n=[bkj]n×n[aji]n×n(AB)^{'} = \lbrack c_{ik}\rbrack_{n \times n}^{'} = \lbrack c_{ki}\rbrack_{n \times n} = \lbrack b_{kj}\rbrack_{n \times n}\lbrack a_{ji}\rbrack_{n \times n}=B′A′B^{'}A^{'}

Alternatively, Let A=[1234]2×2,B=[1304]2×2A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}_{2 \times 2},B = \begin{bmatrix} 1 & 3 \\ 0 & 4 \end{bmatrix}_{2 \times 2};

1 & 11 \\ 3 & 25 \end{bmatrix}$$ $(AB)^{'} = \begin{bmatrix} 1 & 3 \\ 11 & 25 \end{bmatrix}$ …..(i) and $B^{'}A^{'} = \begin{bmatrix} 1 & 0 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 11 & 25 \end{bmatrix}$ …..(ii) From (i) and (ii), $(AB) = B^{'}A^{'}$