Question

Find the product, if possible of : $\left[\begin{array}{ccc}3&4&9\\0&-1&5\\2&6&12\end{array}\right] \left[\begin{array}{ccc}13&-2&0\\0&4&1\end{array}\right]$

Answer 1

Answer:

AB is not possible,but B.A exists as

$B.A= \left[\begin{array}{ccc}39&54&107\\2&2&32\end{array}\right]$

Step-by-step explanation:

If  

$A= \left[\begin{array}{ccc}3&4&9\\0&-1&5\\2&6&12\end{array}\right] _{3\times3}\\\\\\B=\left[\begin{array}{ccc}13&-2&0\\0&4&1\end{array}\right] _{2\times3}$

As we know that matrix multiplication is possible only if number of columns of first matrix is equal to the number of rows of second matrix.

Thus AB is not exist.

But B.A is possible because number of column in B are 3 and so as number of rows in A.

$BA=\left[\begin{array}{ccc}13&-2&0\\0&4&1\end{array}\right]\times \left[\begin{array}{ccc}3&4&9\\0&-1&5\\2&6&12\end{array}\right]\\\\\\= \left[\begin{array}{ccc}13(3)+-2(0)+0(2)&13(4)+-2(-1)+0(6)&13(9)+-2(5)+0(12)\\0(3)+4(0)+1(2)&0(4)+4(-1)+1(6)&0(9)+5(4)+1(12)\end{array}\right] \\\\\\BA= \left[\begin{array}{ccc}39&54&107\\2&2&32\end{array}\right]$