Answer :
Sure! Let's solve the problem step-by-step using the given matrix:
Original Matrix:
[tex]\[ \left[\begin{array}{ccc|c} 4 & -1 & -4 & -7 \\ 0 & -40 & 0 & 40 \\ 0 & 0 & 1 & 3 \end{array}\right] \][/tex]
We need to multiply the second row by the reciprocal of -40. The reciprocal of -40 is [tex]\(-\frac{1}{40}\)[/tex].
By performing this operation, we will update each element in the second row as follows:
1. Element in column 1: [tex]\(0 \times -\frac{1}{40} = 0\)[/tex]
2. Element in column 2: [tex]\(-40 \times -\frac{1}{40} = 1\)[/tex]
3. Element in column 3: [tex]\(0 \times -\frac{1}{40} = 0\)[/tex]
4. Element in column 4: [tex]\(40 \times -\frac{1}{40} = -1\)[/tex]
Updating the second row with these results, the matrix becomes:
[tex]\[ \left[\begin{array}{ccc|c} 4 & -1 & -4 & -7 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{array}\right] \][/tex]
So, the final matrix after multiplying the second row by the reciprocal of -40 is:
[tex]\[ \left[\begin{array}{ccc|c} 4 & -1 & -4 & -7 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{array}\right] \][/tex]
Original Matrix:
[tex]\[ \left[\begin{array}{ccc|c} 4 & -1 & -4 & -7 \\ 0 & -40 & 0 & 40 \\ 0 & 0 & 1 & 3 \end{array}\right] \][/tex]
We need to multiply the second row by the reciprocal of -40. The reciprocal of -40 is [tex]\(-\frac{1}{40}\)[/tex].
By performing this operation, we will update each element in the second row as follows:
1. Element in column 1: [tex]\(0 \times -\frac{1}{40} = 0\)[/tex]
2. Element in column 2: [tex]\(-40 \times -\frac{1}{40} = 1\)[/tex]
3. Element in column 3: [tex]\(0 \times -\frac{1}{40} = 0\)[/tex]
4. Element in column 4: [tex]\(40 \times -\frac{1}{40} = -1\)[/tex]
Updating the second row with these results, the matrix becomes:
[tex]\[ \left[\begin{array}{ccc|c} 4 & -1 & -4 & -7 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{array}\right] \][/tex]
So, the final matrix after multiplying the second row by the reciprocal of -40 is:
[tex]\[ \left[\begin{array}{ccc|c} 4 & -1 & -4 & -7 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{array}\right] \][/tex]
