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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.

Thanks

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    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 = 8.2012-06-28

1 Answers 1

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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.

  • 0
    I am happy that my question has been answered. Thanks.2012-06-28