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Math  /  Algebra

QuestionPerform the elementary row operation 13R2→R2\frac{1}{3} R_{2} \rightarrow R_{2} on the given matrix. [132−693]\left[\begin{array}{cc:c} 1 & 3 & 2 \\ -6 & 9 & 3 \end{array}\right]
Resulting matrix: □\square □\square □\square □\square □\square □\square

Studdy Solution

STEP 1

1. We are given a matrix and need to perform a specific elementary row operation.
2. The operation 13R2→R2\frac{1}{3} R_{2} \rightarrow R_{2} means to multiply every element in the second row by 13\frac{1}{3}.
3. The matrix is a 2×32 \times 3 matrix, with the second row being [−6,9,3][-6, 9, 3].

STEP 2

1. Identify the elements in the second row.
2. Apply the row operation to each element in the second row.
3. Write the resulting matrix after the operation.

STEP 3

Identify the elements in the second row of the matrix:
The second row is [−6,9,3][-6, 9, 3].

STEP 4

Apply the row operation 13R2→R2\frac{1}{3} R_{2} \rightarrow R_{2} to each element in the second row:
- Multiply −6-6 by 13\frac{1}{3}: $ -6 \times \frac{1}{3} = -2 \]
- Multiply 99 by 13\frac{1}{3}: $ 9 \times \frac{1}{3} = 3 \]
- Multiply 33 by 13\frac{1}{3}: $ 3 \times \frac{1}{3} = 1 \]

STEP 5

Write the resulting matrix after performing the operation:
The first row remains unchanged: [1,3,2][1, 3, 2].
The new second row is: [−2,3,1][-2, 3, 1].
Thus, the resulting matrix is: [132−231]\left[\begin{array}{cc:c} 1 & 3 & 2 \\ -2 & 3 & 1 \end{array}\right]
The resulting matrix is: [132−231]\left[\begin{array}{cc:c} 1 & 3 & 2 \\ -2 & 3 & 1 \end{array}\right]

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