How do I row-reduce this matrix?

Question

I've been trying for a while, but can't get the row reduced form. this is the matrix.

$$\begin{bmatrix} 4 & 8 & 1 & 1 & 6\\ 3 & 6 & 1 & 2 & 5\\ 2 & 4 & 1 & 9 & 10\\ 1 & 2 & 3 & 2 & 0\end{bmatrix}$$

steps I tried:

I subtracted the second row from the first. After doing that I can't get anything after the first row to be 0. I don't know if I should switch the rows or something...

Answer 1

The row reduction algorithm is as follows:

1. Swap rows if first row first column entry is zero.
2. Divide everything in the first row by the first row first column entry.
3. Zero out each other first column entry by subtracting an appropriate multiple of the first row from each row.
4. Repeat this process for each additional pivot point (skipping a column and maintaining a row if necessary)

Using the notation $\frac{1}{4}R_1\rightarrow R_1$ to denote multiplying the first row by $\frac{1}{4}$ and storing that as the new first row, $R_2\leftrightarrow R_1$ to denote swapping the first and second rows, and $R_2-3R_1\rightarrow R_2$ to denote subtracting 3 times the first row from the second row to store as the new second row, the first several steps in the row reduction process is as follows:

$$\frac{1}{4}R_1\rightarrow R_1\\ R_2-3R_1\rightarrow R_2\\ R_3-2R_1\rightarrow R_3\\ R_4-R_1\rightarrow R_4$$

At this point the matrix should look like this (assuming I haven't made any mental arithmetic mistakes):

$$\begin{bmatrix} 1 & 2 & \frac{1}{4} & \frac{1}{4} & \frac{3}{2}\\ 0 & 0 & \frac{1}{4} & \frac{5}{4} & \frac{7}{4}\\ 0 & 0 & \frac{1}{2} & \frac{17}{2} & 7\\ 0 & 0 & \frac{11}{4} & \frac{7}{4} & -\frac{3}{2}\end{bmatrix}$$

Notice that there is no way to swap rows with any to get the next diagonal entry to be nonzero. That is okay though! According to our algorithm, if faced with this situation, we may move to the next available pivot point, which will be the second row, third column entry.

Continue the process then with: $4R_2\rightarrow R_2$ and zero out the rest of the third column entries, etcetera.

Answer 2

$$ \begin{pmatrix} 4 & 8 & 1 & 1 & 6 \\ 3 & 6 & 1 & 2 & 5 \\ 2 & 4 & 1 & 9 & 10 \\ 1 & 2 & 3 & 2 & 0 \end{pmatrix} $$ Take matrix to echelon form:

• multiply row 1 by $1/4$
• add -3 times row 1 to row 2
• add -2 times row 1 to row 3
• add -1 times row 1 to row 4
• multiply row 2 by 4
• add $-1/2$ times row 2 to row 3
• add $-11/4$ times row 2 to row 4
• multiply row 3 by $1/6$
• add 12 times row 3 to row 4
• multiply row 4 by $1/5$

$$ \begin{pmatrix} 1 & 2 & 0.25 & 0.25 & 1.5 \\ 0 & 0 & 1 & 5 & 2 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} $$

Take matrix to reduced echelon form:

• add $-1/4$ times row 2 to row 1
• add 1 times row 3 to row 1
• add -5 times row 3 to row 2
• add -2 times row 4 to row 1
• add 3 times row 4 to row 2
• add -1 times row 4 to row 3

Reduced echelon form:

$$ \begin{pmatrix} 1 & 2 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} $$