Question

which matrix is equivalent to the given matrix by performing two row operations?\

$$\begin{bmatrix}0&-2&8\\\\6&-5&3\\\\3&1&4\\end{bmatrix}$$

\
\
\bigcirc a. \

$$\begin{bmatrix}0&-2&8\\\\4&1&3\\\\3&1&4\\end{bmatrix}$$

\
\bigcirc b. \

$$\begin{bmatrix}3&1&4\\\\0&-2&8\\\\6&-5&0\\end{bmatrix}$$

\
\bigcirc c. \

$$\begin{bmatrix}3&1&4\\\\9&-4&7\\\\0&-2&8\\end{bmatrix}$$

\
\bigcirc d. \

$$\begin{bmatrix}0&-2&8\\\\6&-5&3\\\\3&1&0\\end{bmatrix}$$

Explanation:

Step1: Analyze Option A

Check row operations: Row 2 of original is

$$\begin{bmatrix}6&-5&3\end{bmatrix}$$

, Option A row 2 is

$$\begin{bmatrix}4&1&3\end{bmatrix}$$

. No valid row operations (swap, scalar multiply, row addition) can turn the original row 2 to this in two steps. Reject A.

Step2: Analyze Option B

First, swap Row 1 and Row 3:

$$\begin{bmatrix}3&1&4\\6&-5&3\\0&-2&8\end{bmatrix}$$

Second, subtract 2×Row 1 from Row 2:
Row 2 new values: $6-2(3)=0$, $-5-2(1)=-7$, $3-2(4)=-5$. This does not match Option B's row 2

$$\begin{bmatrix}6&-5&0\end{bmatrix}$$

. Reject B.

Step3: Analyze Option C

First, swap Row 1 and Row 3:

$$\begin{bmatrix}3&1&4\\6&-5&3\\0&-2&8\end{bmatrix}$$

Second, add Row 1 and Row 2:
Row 2 new values: $6+3=9$, $-5+1=-4$, $3+4=7$. Result is:

$$\begin{bmatrix}3&1&4\\9&-4&7\\0&-2&8\end{bmatrix}$$

This matches Option C.

Step4: Analyze Option D

Only Row 3's last element changes from 4 to 0. This would require one row operation (subtract 4 from Row 3, which is not a valid row operation; or scalar multiply, which doesn't fit). Reject D.

Answer:

C.

$$\begin{bmatrix}3&1&4\\9&-4&7\\0&-2&8\end{bmatrix}$$