Question

select the correct answer.
if $a = \

$$\begin{bmatrix} 5 & -8 \\\\ 3 & 0 \\end{bmatrix}$$

$ and $b = \

$$\begin{bmatrix} 2 & -9 \\\\ 1 & 0 \\end{bmatrix}$$

$, which matrix is the result of $2a - 3b$?
\bigcirc a. $\

$$\begin{bmatrix} 4 & 11 \\\\ 3 & 0 \\end{bmatrix}$$

$
\bigcirc b. $\

$$\begin{bmatrix} 4 & 38 \\\\ 9 & 0 \\end{bmatrix}$$

$
\bigcirc c. $\

$$\begin{bmatrix} 4 & 11 \\\\ 9 & 0 \\end{bmatrix}$$

$
\bigcirc d. $\

$$\begin{bmatrix} 1 & -18 \\\\ 1 & 0 \\end{bmatrix}$$

$

Explanation:

Step1: Calculate 2A

To find \( 2A \), we multiply each element of matrix \( A \) by 2.
Given \( A = \begin{bmatrix} 5 & -8 \\ 3 & 0 \end{bmatrix} \), then:
\( 2A = 2\times\begin{bmatrix} 5 & -8 \\ 3 & 0 \end{bmatrix} = \begin{bmatrix} 2\times5 & 2\times(-8) \\ 2\times3 & 2\times0 \end{bmatrix} = \begin{bmatrix} 10 & -16 \\ 6 & 0 \end{bmatrix} \)

Step2: Calculate 3B

To find \( 3B \), we multiply each element of matrix \( B \) by 3.
Given \( B = \begin{bmatrix} 2 & -9 \\ 1 & 0 \end{bmatrix} \), then:
\( 3B = 3\times\begin{bmatrix} 2 & -9 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 3\times2 & 3\times(-9) \\ 3\times1 & 3\times0 \end{bmatrix} = \begin{bmatrix} 6 & -27 \\ 3 & 0 \end{bmatrix} \)

Step3: Calculate \( 2A - 3B \)

Subtract the corresponding elements of \( 3B \) from \( 2A \).
\( 2A - 3B = \begin{bmatrix} 10 - 6 & -16 - (-27) \\ 6 - 3 & 0 - 0 \end{bmatrix} = \begin{bmatrix} 4 & 11 \\ 3 & 0 \end{bmatrix} \)? Wait, no, wait:
Wait, \( -16 - (-27) = -16 + 27 = 11 \)
\( 6 - 3 = 3 \)? Wait, no, wait my mistake:
Wait \( 2A \) is \( \begin{bmatrix} 10 & -16 \\ 6 & 0 \end{bmatrix} \)
\( 3B \) is \( \begin{bmatrix} 6 & -27 \\ 3 & 0 \end{bmatrix} \)
So subtracting:
First row, first column: \( 10 - 6 = 4 \)
First row, second column: \( -16 - (-27) = -16 + 27 = 11 \)
Second row, first column: \( 6 - 3 = 3 \)
Second row, second column: \( 0 - 0 = 0 \)
Wait, but option A is \( \begin{bmatrix} 4 & 11 \\ 3 & 0 \end{bmatrix} \), option C is \( \begin{bmatrix} 4 & 11 \\ 9 & 0 \end{bmatrix} \). Wait, did I make a mistake in calculating \( 2A \) or \( 3B \)?

Wait, no, \( A = \begin{bmatrix} 5 & -8 \\ 3 & 0 \end{bmatrix} \), so \( 2A \) is \( \begin{bmatrix} 10 & -16 \\ 6 & 0 \end{bmatrix} \)
\( B = \begin{bmatrix} 2 & -9 \\ 1 & 0 \end{bmatrix} \), so \( 3B \) is \( \begin{bmatrix} 6 & -27 \\ 3 & 0 \end{bmatrix} \)
Then \( 2A - 3B \) is:
First element: \( 10 - 6 = 4 \)
Second element (first row, second column): \( -16 - (-27) = -16 + 27 = 11 \)
Third element (second row, first column): \( 6 - 3 = 3 \)
Fourth element (second row, second column): \( 0 - 0 = 0 \)
Wait, but option A is \( \begin{bmatrix} 4 & 11 \\ 3 & 0 \end{bmatrix} \), which matches. Wait, but let me check again.

Wait, maybe I miscalculated \( 2A \) or \( 3B \). Wait, \( A \) is \( \begin{bmatrix}5 & -8 \\ 3 & 0\end{bmatrix} \), so \( 2A \) is \( 2*5=10 \), \( 2*(-8)=-16 \), \( 2*3=6 \), \( 2*0=0 \). Correct.

\( B \) is \( \begin{bmatrix}2 & -9 \\ 1 & 0\end{bmatrix} \), so \( 3B \) is \( 3*2=6 \), \( 3*(-9)=-27 \), \( 3*1=3 \), \( 3*0=0 \). Correct.

Then \( 2A - 3B \):

First row, first column: \( 10 - 6 = 4 \)

First row, second column: \( -16 - (-27) = 11 \)

Second row, first column: \( 6 - 3 = 3 \)

Second row, second column: \( 0 - 0 = 0 \)

So the matrix is \( \begin{bmatrix}4 & 11 \\ 3 & 0\end{bmatrix} \), which is option A. Wait, but earlier I thought option C had 9, but no, my mistake. So the correct answer is option A.

Wait, but let me check the options again:

Option A: \( \begin{bmatrix}4 & 11 \\ 3 & 0\end{bmatrix} \)

Option C: \( \begin{bmatrix}4 & 11 \\ 9 & 0\end{bmatrix} \)

Ah, I see my mistake! Oh no, \( 2A \) is \( \begin{bmatrix}10 & -16 \\ 6 & 0\end{bmatrix} \), \( 3B \) is \( \begin{bmatrix}6 & -27 \\ 3 & 0\end{bmatrix} \)

Wait, \( 6 - 3 = 3 \)? No, wait \( 2A \) second row first element is 6, \( 3B \) second row first element is 3, so 6 - 3 = 3. But option A has 3, option C has 9. Wait, maybe I made a mistake in \( 2A \) or \( 3B \). Wait, no, \( A \) is \( \begin{bmatrix}5 & -8 \\ 3 & 0\end{bmatrix} \), so \( 2A \) is \( \begin{bmatrix}10 & -16 \\ 6 & 0\end{bmatrix} \). Correct. \( B \) is \( \begin{bmatrix}2 & -9 \\ 1 & 0\end{bmatrix} \), so \( 3B \) is \( \begin{bmatrix}6 & -27 \\ 3 & 0\end{bmatrix} \). Correct. Then \( 6 - 3 = 3 \). So the matrix is \( \begin{bmatrix}4 & 11 \\ 3 & 0\end{bmatrix} \), which is option A.

Wait, but let me check the options again. The options are:

A. \( \begin{bmatrix}4 & 11 \\ 3 & 0\end{bmatrix} \)

B. \( \begin{bmatrix}4 & 38 \\ 9 & 0\end{bmatrix} \)

C. \( \begin{bmatrix}4 & 11 \\ 9 & 0\end{bmatrix} \)

D. \( \begin{bmatrix}1 & -18 \\ 1 & 0\end{bmatrix} \)

So according to my calculation, the correct answer is A.

Wait, but maybe I made a mistake in the subtraction. Let me recheck:

\( 2A = \begin{bmatrix} 5*2 & -8*2 \\ 3*2 & 0*2 \end{bmatrix} = \begin{bmatrix}10 & -16 \\ 6 & 0\end{bmatrix} \)

\( 3B = \begin{bmatrix}2*3 & -9*3 \\ 1*3 & 0*3\end{bmatrix} = \begin{bmatrix}6 & -27 \\ 3 & 0\end{bmatrix} \)

Now subtract \( 3B \) from \( 2A \):

First element: 10 - 6 = 4

Second element: -16 - (-27) = -16 + 27 = 11

Third element: 6 - 3 = 3

Fourth element: 0 - 0 = 0

So the result is \( \begin{bmatrix}4 & 11 \\ 3 & 0\end{bmatrix} \), which is option A.

Answer:

A. \(

$$\begin{bmatrix}4 & 11 \\ 3 & 0\end{bmatrix}$$

\)