Question

Let Upper A equals left bracket Start 2 By 2 Matrix 1st Row 1st Column negative 2 2nd Column 4 2nd Row 1st Column 1 2nd Column 3 EndMatrix right bracketA=−2 41 3​, and Upper B equals left bracket Start 2 By 2 Matrix 1st Row 1st Column negative 2 2nd Column 1 2nd Row 1st Column 3 2nd Column 7 EndMatrix right bracketB=−2 13 7.a. FindABAB​,if possible. b. FindBABA​,if possible.c. Are the answers in parts a and b the ​same?d. In​ general, for matrices A and B such that AB and BA both​ exist, does AB always equal​ BA?a. FindABAB​,if possible. Select the correct choice below​ and, if​necessary, fill in the answer boxes to complete your choice.

Answer 1

Answer:

See explanation

Step-by-step explanation:

Given:

$A=\left[\begin{array}{cc}-2&4\\1&3\end{array}\right]$

$B=\left[\begin{array}{cc}-2&1\\3&7\end{array}\right]$

A. Find AB:

$AB=\left[\begin{array}{cc}-2&4\\1&3\end{array}\right]\cdot \left[\begin{array}{cc}-2&1\\3&7\end{array}\right]=\left[\begin{array}{cc}-2\cdot (-2)+4\cdot 3&-2\cdot 1+4\cdot 7\\1\cdot (-2)+3\cdot 3&1\cdot 1+3\cdot 7\end{array}\right]=\left[\begin{array}{cc}16&26\\7&22\end{array}\right]$

B. Find BA:

$BA=\left[\begin{array}{cc}-2&1\\3&7\end{array}\right]\cdot \left[\begin{array}{cc}-2&4\\1&3\end{array}\right]=\left[\begin{array}{cc}-2\cdot (-2)+1\cdot 1&-2\cdot 4+1\cdot 3\\3\cdot (-2)+7\cdot 1&3\cdot 4+7\cdot 3\end{array}\right]=\left[\begin{array}{cc}5&-5\\1&33\end{array}\right]$

C. Answers are not the same

D. Matrices multiplication is not commutastive in general, so

$AB\neq BA$