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Recognize the elementary row operation used and determine the missing entry.
$[\begin{array}{rrr} 1 & \frac{1}{3} & 1 \\ 2 & 2 & - 1 \end{array}] \to [\begin{array}{rrr} 1 & \frac{1}{3} & 1 \\ 0 & ? & - 3 \end{array}]$

Select one:
a. $\frac{4}{3}$
b. 4
c. -4
d. $- \frac{4}{3}$
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Final Answer: The missing entry is $\frac{4}{3}$ .

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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 $\frac{4}{3}$ .

Question 2Not yet answeredMarked out of 1P Flag questionRecognize the elementary row operation used and determine the missing entry.[113122−1]→[11310?−3] Select one:a. 43b. 4c. -4d. −43Previous pageNext page

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