Math  /  Algebra

QuestionPerform the elementary row operation 4R1+R3R3-4 R_{1}+R_{3} \rightarrow R_{3} on the given matrix. Write numbers as integers or simplified fractions. [231623172301]\left[\begin{array}{lll:l} 2 & 3 & 1 & 6 \\ 2 & 3 & 1 & 7 \\ 2 & 3 & 0 & 1 \end{array}\right]
Resulting matrix: []\left[\begin{array}{cc:c} \square & \square & \square \\ \square & \square & \square \\ \square & \square & \square \\ \square & \square & \square \end{array}\right]

Studdy Solution

STEP 1

1. We are given a matrix and an elementary row operation to perform.
2. The operation involves modifying the third row by adding 4-4 times the first row to it.
3. The operation will only affect the third row, leaving the first and second rows unchanged.

STEP 2

1. Identify the elements of the first row R1 R_1 and the third row R3 R_3 .
2. Multiply each element of the first row by 4-4.
3. Add the resulting values to the corresponding elements of the third row.
4. Write the resulting matrix.

STEP 3

Identify the elements of the first row R1 R_1 and the third row R3 R_3 :
First row R1 R_1 : [2,3,1,6][2, 3, 1, 6]
Third row R3 R_3 : [2,3,0,1][2, 3, 0, 1]

STEP 4

Multiply each element of the first row by 4-4:
4×2=84×3=124×1=44×6=24-4 \times 2 = -8 \\ -4 \times 3 = -12 \\ -4 \times 1 = -4 \\ -4 \times 6 = -24

STEP 5

Add the results from Step 2 to the corresponding elements of the third row R3 R_3 :
2+(8)=63+(12)=90+(4)=41+(24)=232 + (-8) = -6 \\ 3 + (-12) = -9 \\ 0 + (-4) = -4 \\ 1 + (-24) = -23

STEP 6

Write the resulting matrix, keeping the first and second rows unchanged:
[2316231769423]\begin{bmatrix} 2 & 3 & 1 & 6 \\ 2 & 3 & 1 & 7 \\ -6 & -9 & -4 & -23 \end{bmatrix}
The resulting matrix is: [2316231769423]\begin{bmatrix} 2 & 3 & 1 & 6 \\ 2 & 3 & 1 & 7 \\ -6 & -9 & -4 & -23 \end{bmatrix}

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