Different values of $a$ in a linear system.

Question

For which real value(s) of $a$ does the system $$\left[\begin{array}{ccc|c}1&1&1&1\\ 1&2&a&2\\ 2&a&4&0\end{array}\right]$$ have (i) a unique solution, (ii) no solution, (iii) infinitely many solutions? Where there are infinitely many solutions, write the solutions in parametric form.

I am a little confused on how to approach this problem. I have tried to do the row reduction but I get stuck because I am not sure how to eliminate the $a$. Therefore, I am a little unsure on how I can determine for what numbers what solutions are present.

Any help would be highly appreciated!

Answer 1

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

Perform $R_2-R_1$ and $R_3-2R_1$

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

Now perform $R_3+(2-a)R_2,$

You matrix should then be in a row echelon form with a quadratic term as one of the entry. Consider cases, what if the quadratic term is equal to zero and what if it is not.

Answer 2

$\mathbf{Hint:}$ I think you can, as stated in the comments and by the other answer, first triangularize/row reduce the given augmented matrix.

For simplicity sake, suppose $[A \ | \ B]= \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 1 & 2 & a & 2 \\ 2 & a & 4 & 0 \\ \end{array} \right] $

That is, $A\in M_3(\mathbb{R})$ and $B\in \mathbb{R}^3$.

Let $[A' \ | \ B']$ be the triangularized/row reduced matrix obtained from the previous answer for $[A \ | \ B]$, and consider the following:

$\mathbf{1.}$ If all the row vectors of $A$ are linearly independent, then we can row reduce $A$ to the $3\times 3$ identity matrix $I_3$. Then, simply equate the bottom right entry of $A'$ to that of the bottom entry of $B'$. In fact, a full reduction of the augmented matrix $[A \ | \ B]$ is not required, and really, the form given by @Siong Thye Goh is good enough.

$\mathbf{2.}$ If two or more row vectors of $A$ are linearly dependent, then row reducing $A$ to its row reduced echelon form yields one or more $1\times 3$ zero rows. If this is the case, then there are two possibilities:

$\quad$ -If the bottom entry of $B'$ is zero, then there are infinitely many solutions. Why? Well we have $\quad$less pivot variables than columns.

$\quad$ -If the bottom entry of $B'$ is non-zero, then clearly we arrive at a contradiction, because the $\quad$equation given by the last row in the augmented matrix implies $0=c$, where $c\neq0$. Clearly this $\quad$implies there are no solutions in $\mathbb{R}$ (for this subcase).

Hopefully this intuition helps.