Question 2
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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. $-\frac{4}{3}$
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Answer
Final Answer: The missing entry is \(\boxed{\frac{4}{3}}\).
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 \(\boxed{\frac{4}{3}}\).
