Answer:
its the second and third option
Step-by-step explanation:
I took the test
In order to multiply a matrix by another matrix, we multiply the rows in the first matrix by the columns in the other matrix (How this is done is shown below)
To determine the pairs of matrices that AB=BA, we will determine AB and BA for each of the options below.
For the first option
[tex]A= \left[\begin{array}{cc}1&0&-2&1&\end{array}\right]; B= \left[\begin{array}{cc}5&0&3&2&\end{array}\right] \\[/tex]
[tex]AB= \left[\begin{array}{cc}(1\times5)+(0\times3)&(1\times0)+(0\times 2)&(-2\times5)+(1\times3)&(-2\times0)+(1\times2)&\end{array}\right]\\AB= \left[\begin{array}{cc}5+0&0+0&-10+3&0+2&\end{array}\right]\\AB = \left[\begin{array}{cc}5&0&-7&2&\end{array}\right] \\[/tex]
and
[tex]BA= \left[\begin{array}{cc}(5\times1)+(0\times-2)&(5\times0)+(0\times 1)&(3\times1)+(2\times-2)&(3\times0)+(1\times2)&\end{array}\right]\\BA= \left[\begin{array}{cc}5+0&0+0&3+-4&0+2&\end{array}\right]\\BA = \left[\begin{array}{cc}5&0&-1&2&\end{array}\right] \\[/tex]
∴ AB≠BA
For the second option
[tex]A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right]; B= \left[\begin{array}{cc}3&0&6&-3&\end{array}\right] \\[/tex]
[tex]AB= \left[\begin{array}{cc}(1\times3)+(0\times6)&(1\times0)+(0\times -3)&(-1\times3)+(2\times6)&(-1\times0)+(2\times-3)&\end{array}\right]\\AB= \left[\begin{array}{cc}3+0&0+0&-3+12&0+-6&\end{array}\right]\\AB = \left[\begin{array}{cc}3&0&9&-6&\end{array}\right] \\[/tex]
and
[tex]BA= \left[\begin{array}{cc}(3\times1)+(0\times-1)&(3\times0)+(0\times 2)&(6\times1)+(-3\times-1)&(6\times0)+(-3\times2)&\end{array}\right]\\BA= \left[\begin{array}{cc}3+0&0+0&6+3&0+-6&\end{array}\right]\\BA = \left[\begin{array}{cc}3&0&9&-6&\end{array}\right] \\[/tex]
Here AB = BA
For the third option
[tex]A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right]; B= \left[\begin{array}{cc}5&0&3&2&\end{array}\right] \\[/tex]
[tex]AB= \left[\begin{array}{cc}(1\times5)+(0\times3)&(1\times0)+(0\times 2)&(-1\times5)+(2\times3)&(-1\times0)+(2\times2)&\end{array}\right]\\AB= \left[\begin{array}{cc}5+0&0+0&-5+6&0+4&\end{array}\right]\\AB = \left[\begin{array}{cc}5&0&1&4&\end{array}\right] \\[/tex]
and
[tex]BA= \left[\begin{array}{cc}(5\times1)+(0\times-1)&(5\times0)+(0\times 2)&(3\times1)+(2\times-1)&(3\times0)+(2\times2)&\end{array}\right]\\BA= \left[\begin{array}{cc}5+0&0+0&3+-2&0+4&\end{array}\right]\\BA = \left[\begin{array}{cc}5&0&1&4&\end{array}\right] \\[/tex]
Here also, AB=BA
For the fourth option
[tex]A= \left[\begin{array}{cc}1&0&-2&1&\end{array}\right]; B= \left[\begin{array}{cc}3&0&6&-3&\end{array}\right] \\[/tex]
[tex]AB= \left[\begin{array}{cc}(1\times3)+(0\times6)&(1\times0)+(0\times -3)&(-2\times3)+(1\times6)&(-2\times0)+(1\times-3)&\end{array}\right]\\AB= \left[\begin{array}{cc}3+0&0+0&-6+6&0+-3&\end{array}\right]\\AB = \left[\begin{array}{cc}3&0&0&-3&\end{array}\right] \\[/tex]
and
[tex]BA= \left[\begin{array}{cc}(3\times1)+(0\times-2)&(3\times0)+(0\times 1)&(6\times1)+(-3\times-2)&(6\times0)+(-3\times1)&\end{array}\right]\\BA= \left[\begin{array}{cc}3+0&0+0&6+6&0+-3&\end{array}\right]\\BA = \left[\begin{array}{cc}3&0&12&-3&\end{array}\right] \\[/tex]
Here, AB≠BA
Hence, it is only in the second and third options that AB = BA
[tex]A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right] B= \left[\begin{array}{cc}3&0&6&-3&\end{array}\right] \\[/tex] and [tex]A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right]B= \left[\begin{array}{cc}5&0&3&2&\end{array}\right] \\[/tex]
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