How can I determine what value of $h$ would make my system consistent. The augmented matrix of my system is:
$\begin{bmatrix} 1&4&-2\\ 3&h&6 \end{bmatrix}$
My first step is to add $-3R1+R2$ to get
$\begin{bmatrix} 1&4&-2\\ 0&-12+h&12 \end{bmatrix}$
The second step is $\frac{1}{h-12}$ R2 to get:
$\begin{bmatrix} 1&4&-2\\ 0&1&\frac{12}{h-12} \end{bmatrix}$
Then the next step is $-4R2+R1$
After much simplifying of the various fractions I get
$\begin{bmatrix} 1&0&\frac{(-2)(h-12)}{h-12}\\ 0&1&\frac{12}{h-12} \end{bmatrix}$
But I cannot seem to get the value of h.
$\endgroup$1 Answer
$\begingroup$Assuming that the matrices represent an augmented coefficient matrix, then the system will be consistent for any $h \neq 12$. That's the value at which there is no solution to the first nor to the second variables because the denominator would equal zero at $h = 12$ (e.g. at $h = 12$, $\dfrac{12}{h-12} $ is undefined.) E.g., the second variable, say, $y$ needs to be defined for a solution to the system to exist, since reading off the bottom row in the reduced matrix, we have that $y = \frac{12}{h-12}$.
Note that on the last row operation/reduction, $-4R_2 + R_1$ the entry in the first row, last column needs to be $$\frac{-4(12)}{h-12} + -2 = \frac{-48}{h-12} + \frac{-2(h-12)}{h-12}$$ $$ = \frac{-48 -2h+ 24}{h-12} = \frac{-2(h+12)}{h-12}$$ Correcting for sign issue in the numerator doesn't change the the answer in any way.
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