Problem 40 - Precalculus with Limits

Question

Perform the sequence of row operations on the matrix. What did the operations accomplish? $$\left[\begin{array}{rr}{7} & {1} \\ {0} & {2} \\ {-3} & {4} \\ {4} & {1}\end{array}\right]$$ (a) Add \(R_{3}\) to \(R_{4}.\) (b) Interchange \(R_{1}\) and \(R_{4}\) . (c) Add 3 times \(R_{1}\) to \(R_{3}.\) (d) Add \(-7\) times \(R_{1}\) to \(R_{4}.\) (e) Multiply \(R_{2}\) by \(\frac{1}{2}.\) (f) Add the appropriate multiples of \(R_{2}\) to \(R_{1}, R_{3},\) and \(R_{4}\) .

Step-by-Step Solution

Answer

The row operations transformed the original matrix into the 2x2 identity matrix with two additional rows of zeros.

Step 1: Add \(R_3\) to \(R_4\)

By adding the third row to the fourth row, the fourth row will become : [1, 5]. The matrix becomes: \[ \left[\begin{array}{rr} 7 & 1 \ 0 & 2 \ -3 & 4 \ 1 & 5\ \end{array}\right] \]

Step 2: Interchange \(R_1\) to \(R_4\)

Interchanging the first row and the fourth row will give: \[ \left[\begin{array}{rr} 1 & 5 \ 0 & 2 \ -3 & 4 \ 7 & 1\ \end{array}\right] \]

Step 3: Add \(3\) times \(R_1\) to \(R_3\)

Adding 3 times the first row to the third row will give: \[ \left[\begin{array}{rr} 1 & 5 \ 0 & 2 \ 0 & 19 \ 7 & 1\ \end{array}\right] \]

Step 4: Add \(-7\) times \(R_1\) to \(R_4\)

Adding negative 7 times the first row to the fourth row will give: \[ \left[\begin{array}{rr} 1 & 5 \ 0 & 2 \ 0 & 19 \ 0 & -34\ \end{array}\right] \]

Step 5: Multiply \(R_2\) by \(\frac{1}{2}\)

Multiplying the second row by 1/2 will give: \[ \left[\begin{array}{rr} 1 & 5 \ 0 & 1 \ 0 & 19 \ 0 & -34\ \end{array}\right] \]

Step 6: Add the appropriate multiples of \(R_2\) to \(R_1\), \(R_3\) and \(R_4\)

Adding -5 times the second row to the first row, -19 times to the third row, and 34 times to the fourth row will yield the final matrix: \[ \left[\begin{array}{rr} 1 & 0 \ 0 & 1 \ 0 & 0 \ 0 & 0\ \end{array}\right] \]