Matrices!

Clearly Matrix BA is of order $3 \times 3$ . Now,

Rank of Matrix A $\leq 2$

Rank of Matrix B $\leq 2$

Hence, Rank of Matrix BA $\leq 2$

Therefore, $|BA|=0$

Details and Properties used

• A whole number r is said to be rank of matrix A if A has atleast one non-singular sub-matrix of order r and all submatrices of order more than r are singular.

• Rank of matrix A of order m $\times$ n $\leq$ min(m,n)

• Rank of matrix AB $\leq$ min(Rank of A, Rank of B).

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Consider a $3\times3$ matrix $B^{'}$ whose first two columns are identical to $B$ and all elements of third column are zero. Similarly, consider another $3\times3$ matrix $A^{'}$ whose first two rows are identical to $A$ and all elements of third row are zero.

So, we have $\lvert A^{'}\rvert=\lvert B^{'}\rvert=0$ .

Also clearly, $B^{'}A^{'}=BA$ .

Therefore, we have

$\lvert BA\rvert =\lvert B^{'}A^{'}\rvert=\lvert A^{'}\rvert\lvert B^{'}\rvert=0$