# A matrix of a linear system of equations is Given

$$\left[ \begin{array}{ccc|c} 1 & 1 & 1&4\\ 0 & 0 & 1&2 \\ 0 & 0&a-4&a-2\\ \end{array} \right]$$

(a) Find all values for which the system is Consistent (b) find all values for which the system is inconsistent (c) find all values for which the system has infinite many solutions.

I tried to row reduce it and tried to compute the determinant and didn't help. I tried just to just write the values for which I believed to be correct and It was wrong. Would like help on how to go about this question. THank you.

– Raxy
Commented Oct 16, 2017 at 10:23
• What’s the urgency? Are you in the middle of an exam?
– amd
Commented Oct 16, 2017 at 19:05

$\det\left| \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & a-4 \\ \end{array} \right|=0$

for any $a$, thus the rank of the matrix is $2$

To be compatible also the rank of the completed matrix must be $2$

$\text{rank }\left( \begin{array}{ccc|c} 1 & 1 & 1 & 4 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & a-4 & a-2 \\ \end{array} \right)$

as we have

$\det \left| \begin{array}{ccc} 1 & 1 & 4 \\ 0 & 1 & 2 \\ 0 & a-4 & a-2 \\ \end{array} \right|=6-a$

(a)

the system is consistent for $a=6$

(b)

it's inconsistent for $a\ne 6$

(c)

if $a=6$ this determinant is $0$ and the rank of the completed matrix is $2$ so the system is compatible and has infinite solutions

You can answer all of those questions by inspecting the matrix as it is, but let’s go ahead and complete the row-reduction: $$\left[\begin{array}{ccc|c} 1&1&0&2 \\ 0&0&1&2 \\ 0&0&0&6-a\end{array}\right].$$ If you didn’t end up with this, you made a mistake in your calculations. It should be obvious at this point that the system is consistent iff $a=6$ and that it has an infinite number of solutions in that case since the last row will consist entirely of zeros.

Going back to the original augmented matrix, for the system to be consistent, the last row must be a scalar multiple of the second (do you see why?). This means that we must have $a-4=2(a-2)$, which again leads to $a=6$. Setting $a$ to $6$ makes the rows of the matrix linearly dependent, so it doesn’t have full rank and the system has an infinite number of solutions.