Question

URGENT 100 POINTS HELP

Wilma performs the elementary row operation represented by 2R1−R3 on matrix A.

Answer 1

[ -8 1]

[ 0 4 ]

[6 3 ]

[-3 -4 ]

Task:

• To find the resultant of the elementary row operation represented by 2R1 - R3 on Matrix A.

Solution :

Given Matrix A =

[ - 1 2 ]

[ 0 4 ]

[6 3 ]

[-3 -4 ]

In the Matrix given,

• R1 = (-1 ,2)
• R3 = (6,3)

Performing the operation,

• = 2(-1,2) - (6,3)
• = (-2,4) - (6,3)
• = (-8,1)

Therefore,the new matrix would be :

[ -8 1]

[ 0 4 ]

[6 3 ]

[-3 -4 ]

`\:`

Answer 2

`\left[\begin{array}{rr}-8&1\\0&4\\6&3\\-3&-4\end{array}\right]`

Explanation:

Given matrix:

`A=\left[\begin{array}{rr}-1&2\\0&4\\6&3\\-3&-4\end{array}\right]`

The operation 2R₁ - R₃ means that we multiply row 1 (R₁) by 2 and then subtract row 3 (R₃) from the result. After performing this operation, the original row 1 is replaced with the new result, which is the outcome of this operation. This operation modifies row 1 and leaves the other rows unchanged.

`\left[\begin{array}{rr}-1&2\\0&4\\6&3\\-3&-4\end{array}\right]2R_1-R_3 \rightarrow R_1\left[\begin{array}{rr}(2(-1)-6)&(2(2)-3)\\0&4\\6&3\\-3&-4\end{array}\right]`

`\left[\begin{array}{rr}-1&2\\0&4\\6&3\\-3&-4\end{array}\right]2R_1-R_3 \rightarrow R_1\left[\begin{array}{rr}-8&1\\0&4\\6&3\\-3&-4\end{array}\right]`

Therefore, the matrix that results from the transformation is:

`\left[\begin{array}{rr}-8&1\\0&4\\6&3\\-3&-4\end{array}\right]`