I am currently completing a first year Linear Algebra course via correspondence. I am busy with the chapter on Gaussian Elimination. My textbook had the following exercise:

For which values of a will the following system have no solutions? Exactly one solution? Infinitely many solutions?

$x + 2y - 3z = 4$

$3x - y + 5z = 2$

$4x + y + (a^2 - 14)z = (a + 2)$

I then create this augmented matrix:

$\begin{bmatrix}1 & 2 & -3 & 4 \\3 & -1 & 5 & 2\\4 & 1 & (a^2 - 14) & (a + 2)\end{bmatrix}$

And perform the following row operations:

$-3R_1 + R_2 \rightarrow R_2$

$-1/7R_2 \rightarrow R_2$

$-4R_1 + R_3 \rightarrow R_3$

$7R_2 + R_3 \rightarrow R_3$

to arrive at this matrix:

$\begin{bmatrix}1 & 2 & -3 & 4 \\0 & 1 & -2 & 10/7\\0 & 0 & (a^2 - 16) & (a - 4)\end{bmatrix}$

My understanding is that this is Gaussian elimination, and to do Gauss-Jordan elimination I need to do a backward phase to introduce zeros above all leading ones. However the solution in my textbook says:

The Gauss-Jordan process will reduce this system to the equations

$x + 2y – 3z = 4$

$y – 2z = 10/7$

$(a^2 – 16)z = a – 4$

I am not sure if this is a printing error or if I am misunderstanding something. The solution appears to be derived from the matrix I arrived at (i.e. not using Gauss-Jordan elimination).

Any help would be much appreciated.


  • $\begingroup$ Yes. This is Gaussian Elimination. Gauss Jordan would require a reduced row echelon form (Identity in case of full rank) rather than an upper triangular. Whatever it be, the questions you need to answer about the number of solutions can be answered from both eliminations. $\endgroup$ – Inquest Jun 27 '12 at 15:25
  • $\begingroup$ You're right. The reduced row echelon form via Gauss-Jordan will look something like this: $$ \left[ \begin {array}{cccc} 1&0&0&1/7\,{\frac {8\,a+25}{a+4}} \\ 0&1&0&2/7\,{\frac {5\,a+27}{a+4}} \\ 0&0&1& \left( a+4 \right) ^{-1}\end {array} \right] $$ It seems that Maple symbolic simplifier assumed $a\neq -4$ for some reason... $\endgroup$ – user2468 Jun 27 '12 at 15:26
  • $\begingroup$ Thanks J.D. I know part of the solution to this problem was that the equations have no solution (inconsistent) for a = -4 since the last equation would be 0 = 8. $\endgroup$ – King_Nothing Jun 28 '12 at 7:41

You last matrix is the result of the gauss-Jordan algorithm.

It contain all coeffcient from the linear equationsystem Ax=b.

Since it is not clear if the entry in last row is zero you cannot divide it to get a leading 1.

But you are right. Normally there is a leading 1 in every row.

  • $\begingroup$ I am happy that my question has been answered. Thanks. $\endgroup$ – King_Nothing Jun 28 '12 at 7:42

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