Question

select the correct answer.
for which pair of matrices is $ab \
eq ba$
\\(\boldsymbol{\text{a.}}\\) \\(a = \

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

\\) \\(b = \

$$\begin{bmatrix}7 & 0 \\\\ 3 & 4\\end{bmatrix}$$

\\)
\\(\boldsymbol{\text{b.}}\\) \\(a = \

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

\\) \\(b = \

$$\begin{bmatrix}8 & 0 \\\\ 11 & -3\\end{bmatrix}$$

\\)
\\(\boldsymbol{\text{c.}}\\) \\(a = \

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

\\) \\(b = \

$$\begin{bmatrix}6 & 0 \\\\ 9 & -4\\end{bmatrix}$$

\\)
\\(\boldsymbol{\text{d.}}\\) \\(a = \

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

\\) \\(b = \

$$\begin{bmatrix}9 & 0 \\\\ 3 & 6\\end{bmatrix}$$

\\)

Explanation:

Step1: Recall matrix multiplication rule

For $2\times2$ matrices $X=

$$\begin{bmatrix}x_{11}&x_{12}\\x_{21}&x_{22}\end{bmatrix}$$

$ and $Y=

$$\begin{bmatrix}y_{11}&y_{12}\\y_{21}&y_{22}\end{bmatrix}$$

$, $XY=

$$\begin{bmatrix}x_{11}y_{11}+x_{12}y_{21}&x_{11}y_{12}+x_{12}y_{22}\\x_{21}y_{11}+x_{22}y_{21}&x_{21}y_{12}+x_{22}y_{22}\end{bmatrix}$$

$

Step2: Test Option A

Calculate $AB$:

$$\begin{bmatrix}1&0\\3&-2\end{bmatrix}\begin{bmatrix}7&0\\3&4\end{bmatrix}=\begin{bmatrix}1*7+0*3&1*0+0*4\\3*7+(-2)*3&3*0+(-2)*4\end{bmatrix}=\begin{bmatrix}7&0\\15&-8\end{bmatrix}$$

Calculate $BA$:

$$\begin{bmatrix}7&0\\3&4\end{bmatrix}\begin{bmatrix}1&0\\3&-2\end{bmatrix}=\begin{bmatrix}7*1+0*3&7*0+0*(-2)\\3*1+4*3&3*0+4*(-2)\end{bmatrix}=\begin{bmatrix}7&0\\15&-8\end{bmatrix}$$

$AB=BA$, so A is not the answer.

Step3: Test Option B

Calculate $AB$:

$$\begin{bmatrix}1&0\\3&-2\end{bmatrix}\begin{bmatrix}8&0\\11&-3\end{bmatrix}=\begin{bmatrix}1*8+0*11&1*0+0*(-3)\\3*8+(-2)*11&3*0+(-2)*(-3)\end{bmatrix}=\begin{bmatrix}8&0\\2&6\end{bmatrix}$$

Calculate $BA$:

$$\begin{bmatrix}8&0\\11&-3\end{bmatrix}\begin{bmatrix}1&0\\3&-2\end{bmatrix}=\begin{bmatrix}8*1+0*3&8*0+0*(-2)\\11*1+(-3)*3&11*0+(-3)*(-2)\end{bmatrix}=\begin{bmatrix}8&0\\2&6\end{bmatrix}$$

$AB=BA$, so B is not the answer.

Step4: Test Option C

Calculate $AB$:

$$\begin{bmatrix}1&0\\3&-2\end{bmatrix}\begin{bmatrix}6&0\\9&-4\end{bmatrix}=\begin{bmatrix}1*6+0*9&1*0+0*(-4)\\3*6+(-2)*9&3*0+(-2)*(-4)\end{bmatrix}=\begin{bmatrix}6&0\\0&8\end{bmatrix}$$

Calculate $BA$:

$$\begin{bmatrix}6&0\\9&-4\end{bmatrix}\begin{bmatrix}1&0\\3&-2\end{bmatrix}=\begin{bmatrix}6*1+0*3&6*0+0*(-2)\\9*1+(-4)*3&9*0+(-4)*(-2)\end{bmatrix}=\begin{bmatrix}6&0\\-3&8\end{bmatrix}$$

$AB
eq BA$, so C is a candidate.

Step5: Verify Option D (optional)

Calculate $AB$:

$$\begin{bmatrix}1&0\\-1&2\end{bmatrix}\begin{bmatrix}9&0\\3&6\end{bmatrix}=\begin{bmatrix}1*9+0*3&1*0+0*6\\-1*9+2*3&-1*0+2*6\end{bmatrix}=\begin{bmatrix}9&0\\-3&12\end{bmatrix}$$

Calculate $BA$:

$$\begin{bmatrix}9&0\\3&6\end{bmatrix}\begin{bmatrix}1&0\\-1&2\end{bmatrix}=\begin{bmatrix}9*1+0*(-1)&9*0+0*2\\3*1+6*(-1)&3*0+6*2\end{bmatrix}=\begin{bmatrix}9&0\\-3&12\end{bmatrix}$$

$AB=BA$, so D is not the answer.

Answer:

C. $A =

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

\ B =

$$\begin{bmatrix}6&0\\9&-4\end{bmatrix}$$

$