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
Best Answer
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.