In Algebra we've been given this matrix:

$P=\begin{pmatrix} 3 & 2 & 6 & 10 \\ 8 & 4 & 9 & 5 \\ 7 & 3 & 12 & 4 \end{pmatrix}$

I'm asked to row-reduce it with Gauss-Jordan, pivotting on $p_{3,2}$, $p_{2,4}$ and $p_{1,3}$. Now, I understand how to row-reduce it without being required to pivot certain matrix elements. But this beats me.

I tried this approach:

$R_3 = \frac13R_3$

$\to R_2 = -2R_1 + R_2$

$\to R_1 = -2R_3 + R_1$

$\to R_2 = -\frac{1}{15}R_2$

$\to R_1 = -\frac{11}{2}R_3$

$\to R_3 = -\frac43R_2$

$\to R_1 = -\frac{1}{24}R_1$

$\to R_2 = -\frac15R_1 + R_2$

Ultimately, I get this matrix:

$P=\begin{pmatrix} \frac{29}{48} & \frac{11}{48} & 1 & 0 \\ \frac{-2}{15} & 0 & \frac15 & 1 \\ \frac{113}{15} & 1 & \frac{56}{15} & 0 \end{pmatrix}$

How should I proceed? Is there an algorithm to solve these type of exercises where the pivots you must work with aren't the usual ones?

  • 1
    $\begingroup$ It isn't yet row reduced at the three prearranged pivot spots. The rest of the elements in those three columns should be zero. That can be done with more row operations. $\endgroup$ – coffeemath May 18 '14 at 22:17

Technically pivoting on the three positions $(3,2),(2,4),(1,3)$ doesn't give a row reduced matrix because the $1$'s don't wind up going down the diagonal, which should happen in a case like this where the original matrix is $3 \times 4$ and has rank $3$.

However the procedure which seems to be suggested by the assignment is to pivot at the given three spots, in that order, and then stop. Each pivot should consist of two steps: Multiply the pivot row by the reciprocal of the pivot element, to get a $1$ at that position, then use row operations of the type (constant)*(pivot row) + (other row) --> (other row) so as to obtain zeroes in the rest of the column containing the given pivot position.

It is important to complete each pivot operation, getting the $1$ and also the $0$'s in the rest of that column, before going on to the next pivot position. Otherwise further pivot operations will interfere with the pivot column already finished. So in your case you begin with the pivot at $(3,2)$ by multiplying that row by $(1/3)$ which you did above. But then you didn't follow up with the next two steps to "zero out" the 4 and 2 lying in the pivot column of the $(3,2)$ position. Once that is done, one turns to the next pre-given pivot position at $(2,4)$, multiplies that row by the reciprocal of whatever happens to be there now, etc.

When I did this I got the final result to have first row $[21/156,0,1,0]$ and two other rows (each with only one $1$ and a fraction as first entry).

  • $\begingroup$ Thanks for your help, the problem was I didn't "zero out" the column before moving on to the next pivot. $\endgroup$ – Joel Alejandro May 18 '14 at 23:12
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    $\begingroup$ @JoelAlejandro Yes, that causes problems, and is one reason usual procedure is to do that "zeroing out", so one doesn't double back and have to do things twice. Once a column has only the 1 and others are 0's it cannot be affected by other pivots, which naturally will not be in the same row as the pivot already done, and also they won't change the zeroes already there. $\endgroup$ – coffeemath May 18 '14 at 23:28

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