Question

If A = $\left[\begin{array}{ccc}2&0&1\\-1&1&5\end{array}\right]$ and B = $\left[\begin{array}{ccc}-1&1&0\\0&1&-2\end{array}\right]$ then find AB'

Answer 1

Answer:

$AB^{'}=\left[\begin{array}{ccc}-2&-2\\2&-9\end{array}\right]$

Step-by-step explanation:

As

$A=\left[\begin{array}{ccc}2&0&1\\-1&1&5\end{array}\right]_{2\times 3}$

$B=\left[\begin{array}{ccc}-1&1&0\\0&1&-2\end{array}\right]$

$B^{'}=\left[\begin{array}{ccc}-1&0\\1&1\\0&-2\end{array}\right]_{3\times 2}$

now from A and B' we can check that both are compitable for multiplication.

$AB^{'}=\left[\begin{array}{ccc}2&0&1\\-1&1&5\end{array}\right]\times\left[\begin{array}{ccc}-1&0\\1&1\\0&-2\end{array}\right]\\\\\\=\left[\begin{array}{ccc}2(-1)+0(1)+1(0)&2(0)+0(1)+1(-2)\\-1(-1)+1(1)+0(5)&-1(0)+1(1)+5(-2)\end{array}\right]\\\\\\AB^{'}=\left[\begin{array}{ccc}-2&-2\\2&-9\end{array}\right]$