Question

if $a = \

$$\begin{bmatrix} 3 & -1 & 2 \\\\ -2 & 4 & 0 \\end{bmatrix}$$

$ and $b = \

$$\begin{bmatrix} 2 & 0 \\\\ -1 & 4 \\\\ -3 & 2 \\end{bmatrix}$$

$, then what is $ba$?
if $a = \

$$\begin{bmatrix} 1 & x \\\\ 2 & 3 \\end{bmatrix}$$

$ and $b = \

$$\begin{bmatrix} 1 & 1 \\\\ 1 & 2 \\end{bmatrix}$$

$ and $ab = ba$ then what is the value of $x$?
if $a = \

$$\begin{bmatrix} 2 & 0 & -1 \\\\ 3 & 5 & 2 \\\\ -4 & 1 & 4 \\end{bmatrix}$$

$ and $b = \

$$\begin{bmatrix} 5 & 1 & -2 \\\\ -1 & 0 & 4 \\\\ 2 & -3 & 3 \\end{bmatrix}$$

$, then what is $ab$?

Explanation:

First Problem: Calculate $BA$

Step1: Recall matrix multiplication rule

For $B_{3 \times 2}$ and $A_{2 \times 3}$, $BA$ is $3 \times 3$. Element $(i,j)$: $\sum_{k=1}^2 B_{ik}A_{kj}$

Step2: Compute first row of $BA$

Row1: $2*3 + 0*(-2) = 6$, $2*(-1) + 0*4 = -2$, $2*2 + 0*0 = 4$

Step3: Compute second row of $BA$

Row2: $-1*3 + 4*(-2) = -11$, $-1*(-1) + 4*4 = 17$, $-1*2 + 4*0 = -2$

Step4: Compute third row of $BA$

Row3: $-3*3 + 2*(-2) = -13$, $-3*(-1) + 2*4 = 11$, $-3*2 + 2*0 = -6$

Step1: Calculate $AB$

$AB =

$$\begin{bmatrix}1*1 + x*1 & 1*1 + x*2 \\2*1 + 3*1 & 2*1 + 3*2\end{bmatrix}$$

=

$$\begin{bmatrix}1+x & 1+2x \\5 & 8\end{bmatrix}$$

$

Step2: Calculate $BA$

$BA =

$$\begin{bmatrix}1*1 + 1*2 & 1*x + 1*3 \\1*1 + 2*2 & 1*x + 2*3\end{bmatrix}$$

=

$$\begin{bmatrix}3 & x+3 \\5 & x+6\end{bmatrix}$$

$

Step3: Equate corresponding elements

Set $1+x=3$ and $1+2x=x+3$

Step4: Solve for $x$

From $1+x=3$, $x=3-1=2$

Step1: Recall matrix multiplication rule

For $A_{3 \times 3}$ and $B_{3 \times 3}$, $AB$ is $3 \times 3$. Element $(i,j)$: $\sum_{k=1}^3 A_{ik}B_{kj}$

Step2: Compute first row of $AB$

Row1: $2*5+0*(-1)+(-1)*2=8$, $2*1+0*0+(-1)*(-3)=5$, $2*(-2)+0*4+(-1)*3=-7$

Step3: Compute second row of $AB$

Row2: $3*5+5*(-1)+2*2=16$, $3*1+5*0+2*(-3)=-3$, $3*(-2)+5*4+2*3=20$

Step4: Compute third row of $AB$

Row3: $-4*5+1*(-1)+4*2=-13$, $-4*1+1*0+4*(-3)=-16$, $-4*(-2)+1*4+4*3=24$

Answer:

$$\begin{bmatrix} 6 & -2 & 4 \\ -11 & 17 & -2 \\ -13 & 11 & -6 \end{bmatrix}$$

---

Second Problem: Find $x$ such that $AB=BA$