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 4
1 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 1
3 7
.a. Find

ABAB​,

if possible. b. Find

BABA​,

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. Find

ABAB​,

Answer 1

Answer: a) AB = $\left[\begin{array}{ccc}-14&26\\-11&22\end{array}\right]$ ; b) $\left[\begin{array}{ccc}5&-5\\13&9\end{array}\right]$ ; c) No, they are different; d) No, they are never the same;

Step-by-step explanation:

a) A = $\left[\begin{array}{ccc}-2&4\\1&3\end{array}\right]$ , B = $\left[\begin{array}{ccc}-2&1\\-3&7\end{array}\right]$. This multiplication is possible, because matrix A has 2 columns and matrix B has 2 rows. When those value are equal, the multiplication is possible.

AB = $\left[\begin{array}{ccc}-2&4\\1&3\end{array}\right]$ . $\left[\begin{array}{ccc}-2&1\\-3&7\end{array}\right]$ = $\left[\begin{array}{ccc}-14&26\\-11&22\end{array}\right]$

b) BA = $\left[\begin{array}{ccc}-2&1\\-3&7\end{array}\right]$ . $\left[\begin{array}{ccc}-2&4\\1&3\end{array}\right]$ = $\left[\begin{array}{ccc}5&-5\\13&9\end{array}\right]$

c) As, we can see, parts a and b are differents.

d) With matrices, multiplication is not commutative, which means AB≠BA

Answer 2

Answer:

(a)$=\left(\begin{array}{cc}14 & 26\\7 & 22\end{array}\right)$

(b)$\left(\begin{array}{cc}5 & -5\\1 & 33\end{array}\right)$

(c)No

(d)AB≠BA

Step-by-step explanation:

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

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

(a)

$AB=\left(\begin{array}{cc}-2 & 4\\1 & 3\end{array}\right) \left(\begin{array}{cc}-2 & 1\\3 & 7end{array}\right)$

$=\left(\begin{array}{cc}-2*-2+4*3 &-2*1+4*7\\1*-2+3*3 & 1*1+3*7\end{array}\right)$

$=\left(\begin{array}{cc}14 & 26\\7 & 22\end{array}\right)$

(b) $BA=\left(\begin{array}{cc}-2 & 1\\3&7\end{array}\right)\left(\begin{array}{cc}-2 & 4\\1 & 3\end{array}\right)$

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

(c)No

(d)For Matrices A and B, AB≠BA.

Matrix Multiplication is not Commutative.