Determine the amount of solutions for every real number "a" to the following system of linear equations.

Question

I'm studying linear algebra and I have received the following problem to solve:

"Determine the amount of solutions for every real number "a" to the following system of linear equations."

$ (4-a)x_1+2x_2-x_3=1 \\ -2x_1+(1-a)x_2+2x_3=2\\ -x_1+2x_2+(4-a)x_3=1 $

I have a hard time interpreting the problem I think. According to my literature, for a system of linear equations, there will be either

• One unique solution,
• An infinite amount of solutions or
• No solution

for every real number "a".

How do you check for each real number "a" which one of these three cases that is true, without actually testing for each real number?

Any input is appreciated. Thanks

Answer 1

The determinant of the coefficient matrix in terms of $a$ factors nicely as $-(a-1)(a-3)(a-5)$, so for $a\ne1,3,5$ the system has exactly one solution.

For $a=1$ and $a=3$ the rank of the coefficient matrix drops to $2$ but the augmented matrix (with the right-hand sides added) still has rank $3$, so there are no solutions in these cases. For $a=5$ we get $2$ and $2$ respectively, so there are infinitely many solutions.

Answer 2

It is more convenient to use matrix notation to express this system.

$$\begin{pmatrix}4 -a & 2 & -1 &|&1\\ -2 & 1-a & 2 &|& 2 \\ -1 & 2 & 4 -a &|& 1 \end{pmatrix}$$

Applying elementary row operations, we can make $R_1 = R_1 - R_3$, $R_3 = R_3 - R1$ which results in

$$\begin{pmatrix} 5 -a & 0 & -5+a &|&0\\ -2 & 1-a & 2 &|& 2 \\ -5+a & 0 & 5 -a &|& 0 \end{pmatrix}$$

We can divide $R_1, R_3$ both by $5-a$ (assuming $a \neq 5$) to obtain

$$\begin{pmatrix} 1 & 0 & -1 &|&0\\ -2 & 1-a & 2 &|& 2 \\ -1 & 0 & 1 &|& 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & -1 &|&0\\ -2 & 1-a & 2 &|& 2 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & -1 &|&0\\ 0 & 1-a & 0 &|& 0 \end{pmatrix}$$

The system to which we have arrived has solutions for any $a$. Namely, it will be solved by any coefficients $(x_1, x_2, x_3)$ such that $x_1 = x_3, x_2=0$.

These already tells you the number of solutions the system has for $a$ any real number except $5$. Can you see what happens if $a = 5$?