Decide h so that the linear system has infinite solutions

[mpdancow]

Hi! I'm need some help with this question:

Decide $h$ so that the linear system $Ax=b$ has infinite solutions.

$$A=\pmatrix{
5 & 6 & 7 \cr
-7 & -4 & 1 \cr
-4 & 4 & 16 \cr}$$

$$b=\pmatrix{
6 \cr
30 \cr
h \cr}$$

I solved a similar question before but with A being a 2x2 matrix (and B a 2x1) and the equations multiples of each other, so it was easier. I don't even know how to start with this one. Any help I can get is appreciated!

 

[Guest2]

You've to put it in the augmented matrix form then row reduce it (not to echelon form necessarily).

$\begin{aligned} \begin{pmatrix}\begin{array}{rrr|r}
5&6&7&6 \\
-7 & -4 & 1 & 30 \\
-4 & 4 &16 &h
\end{array}\end{pmatrix} & \xrightarrow{R_1 \to R_1+R_2}\begin{pmatrix}\begin{array}{rrr|r}
-2&2&8&36 \\
-7 & -4 & 1 & 30 \\
-4 & 4 &16 &h
\end{array}\end{pmatrix} \xrightarrow{R_3 \to R_3-2R_1}\begin{pmatrix}\begin{array}{rrr|r}
-2&2&8&36 \\
-7 & -4 & 1 & 30 \\
0 & 0 &0 &h -72
\end{array}\end{pmatrix}\end{aligned}$

If $h = 72$ we have the last row of all zeroes, therefore the system has infinite number of solutions.

On the other hand for $h \ne 72$ we can divide the last row by $h-72$ to get $0=1$ (so no solutions).