Problem 39 - Precalculus with Limits

Question

Perform the sequence of row operations on the matrix. What did the operations accomplish? $$\left[\begin{array}{rrr}{1} & {2} & {3} \\ {2} & {-1} & {-4} \\ {3} & {1} & {-1}\end{array}\right]$$ (a) Add \(-2\) times \(R_{1}\) to \(R_{2}.\) (b) Add \(-3\) times \(R_{1}\) to \(R_{3}.\) (c) Add \(-1\) times \(R_{2}\) to \(R_{3}.\) (d) Multiply \(R_{2}\) by \(-\frac{1}{5.}\) (e) Add \(-2\) times \(R_{2}\) to \(R_{1}.\)

Step-by-Step Solution

Answer

The sequence of operations transforms the original matrix into upper triangular form, with leading entries of 1 in the first two rows. This is a structured form that is easier to work with when solving systems of linear equations. However, note that the last row suggests that the system may be inconsistent, which might require further investigation.

Step 1: Apply operation (a)

Add \(-2\) times the first row \(R_1\) to the second row \(R_2\). Which gives us a new matrix: \[ \left[\begin{array}{rrrr} 1 & 2 & 3\ 0 & -5 & -10 \ 3 & 1 & -1 \ \end{array}\right] \]

Step 2: Apply operation (b)

Add \(-3\) times the first row \(R_1\) to the third row \(R_3\). The matrix becomes: \[ \left[\begin{array}{rrrr} 1 & 2 & 3\ 0 & -5 & -10 \ 0 & -5 & -10 \ \end{array}\right] \]

Step 3: Apply operation (c)

Add \(-1\) times the second row \(R_2\) to the third row \(R_3\). This does not change the matrix, as these two rows are identical.

Step 4: Apply operation (d)

Multiply the second row \(R_2\) by \(-\frac{1}{5}\). Getting: \[ \left[\begin{array}{rrrr} 1 & 2 & 3\ 0 & 1 & 2 \ 0 & -5 & -10 \ \end{array}\right] \]

Step 5: Apply operation (e)

Add \(-2\) times the second row \(R_2\) to the first row \(R_1\). The final matrix is: \[ \left[\begin{array}{rrrr} 1 & 0 & -1\ 0 & 1 & 2 \ 0 & -5 & -10 \ \end{array}\right] \]