Question 2 Not yet answered Marked out of 1 P Flag question Recognize the elementary row operation used and determine the missing entry. \[ \left[\begin{array}{rrr} 1 & \frac{1}{3} & 1 \\ 2 & 2 & -1 \end{array}\right] \rightarrow\left[\begin{array}{rrr} 1 & \frac{1}{3} & 1 \\ 0 & ? & -3 \end{array}\right] \] Select one: a. $\frac{4}{3}$ b. 4 c. -4 d. $-\frac{4}{3}$ Previous page Next page

Step 1 ：The elementary row operation used here is 'Row 2 = Row 2 - 2*Row 1'. This operation is used to make the leading coefficient of the second row zero.

Step 2 ：The missing entry can be found by applying the same operation to the second column. That is, the missing entry is \(2 - 2*(1/3) = 2 - 2/3 = 4/3\).

Step 3 ：So, the missing entry is \(4/3\).

Step 4 ：The result of the operation is approximately 1.33, which is equivalent to \(4/3\) in fraction form.

Step 5 ：Therefore, the missing entry in the matrix after the row operation is \(4/3\).

Step 6 ：Final Answer: The missing entry is \(\boxed{\frac{4}{3}}\).