Question

Let Aequals [Start 2 By 2 Matrix 1st Row 1st Column 3 2nd Column 2 2nd Row 1st Column negative 1 2nd Column 2 EndMatrix ]and Bequals [Start 2 By 2 Matrix 1st Row 1st Column 2 2nd Column 6 2nd Row 1st Column negative 3 2nd Column k EndMatrix ]. What​ value(s) of​ k, if​ any, will make ABequals ​BA?

Answer 1

Answer:

No value of k will make AB=BA

Step-by-step explanation:

$A=\left(\begin{array}{ccc}3&2\\-1&2\end{array}\right), B=\left(\begin{array}{ccc}2&6\\-3&k\end{array}\right) \\\\\\AB=\left(\begin{array}{ccc}3&2\\-1&2\end{array}\right)\left(\begin{array}{ccc}2&6\\-3&k\end{array}\right)=\left(\begin{array}{ccc}3*2+2*-3&3*6+2*k\\-1*2+2*-3&-1*6+2k\end{array}\right)=\left(\begin{array}{ccc}0&18+2k\\-8&-6+2k\end{array}\right)$

$BA=\left(\begin{array}{ccc}2&6\\-3&k\end{array}\right)\left(\begin{array}{ccc}3&2\\-1&2\end{array}\right)=\left(\begin{array}{ccc}0&16\\-6&-6+2k\end{array}\right)$

We can see that $AB \neq BA$. Therefore, there is no value of k that will make it equal. In general, matrix multiplication is not commutative.