Consider the matrices


Show that multiplication of matrices is not commutative be determining the product matrices ST and TS. Enter your answer by filling in the boxes.

Answer:

Answer attached

Step-by-step explanation:

ST

3*-2 + 4*-4 = -22                3*0 + 4*5 = 20

-1*-2 + -2*-4 = 10                -1*0 + -2*5 = -10

try TS by yourself

the product of two given matrices is [tex]ST=\left[\begin{array}{cc}-22&20\\10&-10\end{array}\right][/tex] and [tex]TS=\left[\begin{array}{cc}-6&-8\\-17&-26\end{array}\right][/tex].

Given that,

[tex]S=\left[\begin{array}{cc}3&4\\-1&-2\end{array}\right][/tex] and [tex]T=\left[\begin{array}{cc}-2&0\\-4&5\end{array}\right][/tex]

The product of two matrices A and B is defined if the number of columns of A is equal to the number of rows of B. If both A and B are square matrices of the same order, then both AB and BA are defined. If AB and BA are both defined, it is not necessary that AB = BA.

Now, [tex]ST=\left[\begin{array}{cc}3&4\\-1&-2\end{array}\right] \times \left[\begin{array}{cc}-2&0\\-4&5\end{array}\right][/tex]

[tex]ST=\left[\begin{array}{cc}(-6-16)&(0+20)\\(2+8)&(0-10)\end{array}\right][/tex]

[tex]ST=\left[\begin{array}{cc}-22&20\\10&-10\end{array}\right][/tex]

[tex]TS= \left[\begin{array}{cc}-2&0\\-4&5\end{array}\right]\times \left[\begin{array}{cc}3&4\\-1&-2\end{array}\right][/tex]

[tex]TS=\left[\begin{array}{cc}(-6-0)&(-8-0)\\(-12-5)&(-16-10)\end{array}\right][/tex]

[tex]TS=\left[\begin{array}{cc}-6&-8\\-17&-26\end{array}\right][/tex]

Therefore, the product of two given matrices is [tex]ST=\left[\begin{array}{cc}-22&20\\10&-10\end{array}\right][/tex] and [tex]TS=\left[\begin{array}{cc}-6&-8\\-17&-26\end{array}\right][/tex].

To learn more about the product of matrices visit:

https://brainly.com/question/24810141.

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