For which values of $a$ does this system of equations $\mathbf{{not}}$ have a unique solution?

Question

Here's my system of linear equations:

$\begin{cases} x + 2y + 2z = 1\\x + ay + 3z = 3\\x + 11y +az = 0\\ \end{cases}$

Thus I have the augmented matrix $\left[\begin{array}{ccc|c}1&2&2&1\\1&a&3&3\\1&11&a&0\end{array}\right]$

By row reduction, I obtain:

$\left[\begin{array} {ccc|c}1&2&2&1\\0&a-2&1&2\\0&9&a-2&-1\end{array}\right]$

Unfortunately, I am stuck at this stage. I have tried swapping rows around but I didn't have much luck.

Update: I have managed to solve this with the use of the determinant.

Matrix of minors: $\left[\begin{array} {ccc}a^2-33&a-3&11-a\\2a-22&a-2&9\\6-2a&1&a-2\end{array}\right]$

Matrix of cofactors: $\left[\begin{array} {ccc}a^2-33&3-a&11-a\\22-a&9&a-2\\6-2a&-1&a-2\end{array}\right]$

Adjugate matrix: $\left[\begin{array} {ccc}a^2-33&22-2a&6-2a\\3-a&a-2&-1\\11-a&-9&a-2\end{array}\right]$

Det(A) = $1(a^2 - 33) + 2(3 - a) + 2(11 - a) = a^2 - 4a - 5$

$(a - 5)(a + 1) = 0$

Thank you all for your help!

Answer 1

Let's rearrange the augmented matrix representing the system of linear equations, by swapping $R_2$ and $R_3$ (to avoid division by $a-2$ while row-reduction), to get

$\left[\begin{array}{ccc|c}1&2&2&1\\1&11&a&0\\1&a&3&3\end{array}\right]$ and a subsequent row-reduction will give you

$\left[\begin{array}{ccc|c}1&2&2&1\\0&9&a-2&-1\\0&0&1-\frac{(a-2)^2}{9}&2+\frac{a-2}{9}\end{array}\right]$

Unique solution: in order for the system to have a unique solution, the coefficient and the augmented matrix both must be of full rank i.e., both must have rank $n=3$. To make that happen, we must have $1\neq\frac{(a-2)^2}{9} \implies a \notin \{-1,5\}$.

No solution: for this to happen, we need to have the augment matrix rank higher than the coefficient matrix rank. This can happen if the coefficient matrix has rank < 3, e.g., if $1=\frac{(a-2)^2}{9} \implies a \in \{-1,5\}$ and augmented matrix has rank $n=3$ simultaneously, which means $2+\frac{a-2}{9} \neq 0 \implies a \neq -16$, i.e., for $a=-1,5$ the system will have no solution.

Infinitely many solutions: it can happen iff both the coefficient and the augmented matrix are rank-deficient, i.e. for both of them the rank is $<3$. For this the last row of the augmented matrix needs to be zero. Which implies both $1-\frac{(a-2)^2}{9}=0 \implies a \in \{-1,5\}$ and $2+\frac{a-2}{9}=0 \implies a=-16$ simultaneously. Which is clearly impossible.

Answer 2

Hint: Set the determinant equal to zero and solve for $a$.

Answer 3

The easiest thing is to use determinants: the system has a unique solution if and only if the matrix is invertible, i.e., if and only if the matrix has determinant not equal to $0$. So you can compute the determinant (either in your original or row-reduced matrix) in terms of $a$, set it equal to $0$, and solve for $a$. But if you do not know this yet, you can proceed as follows:

If $a=2$, then the system clearly has a unique solution, so you may assume $a\neq 2$. The reason to do this is so that we can continue the row reduction by multiplying by $a-2$. Multiply the second row by $9$ and the last row by $a-2$ to get: $$\left(\begin{array}{ccc|c} 1 & 2 & 2 & 1\\ 0 & 9(a-2) & 9 & 18\\ 0 & 9(a-2) & (a-2)^2 & 2-a \end{array}\right).$$ Subtracting the second row from the third row, we get $$\left(\begin{array}{ccc|c} 1 & 2 & 2 & 1\\ 0 & 9(a-2) & 9 & 18\\ 0 & 0 & (a-2)^2-9 & -16-a \end{array}\right).$$ So: you will have a unique solution if $(a-2)^2-9\neq 0$. If $(a-2)^2-9=0$, you either get infinitely many solutions (if $16+a=0$ as well), or no solutions (if $-16-a\neq 0$). So now we just need to figure out when that bottom right entry of the coefficient matrix is $0$.

Answer 4

Well if $a = 2$ you have $z=2; y=-\frac 19$ and $x = whatever$.

But if $a-2\ne 0$ keep at it.

Multiply row 2 by $\frac 9{a-2}$ and subtract getting rows.

$\begin{array}\\ 1&2&2&|\,\,1\\ 0&a-2&1&|\,\,2\\ 0&0&(a-2)-\frac 9{a-2}&|-\frac 9{a-2}\\ \end{array}$

Now if $(a-2)-\frac 9{a-2} = 0$ we have a case with not solutions.

That is if $(a-2)^2 = 9$ of $a = 5,-1$.

If we assume $a\ne 2,5,-1$ then we can reduce

$\begin{array}\\ 1&2&2&|\,\,1\\ 0&1&\frac 1{a-2}&|\,\,\frac 2{a-2}\\ 0&0&1&|-\frac 9{a-2}\cdot\frac 1{(a-2)-\frac 9{a-2}}=-\frac{9}{(a-2)^2-9}\\ \end{array}$

..... and that's in the whatchamacallit form "upper triangle"?) so we know this results in a unique solution.

So $a = 5$ or $1$ are the cases without solutions.

(because we'd get $((a-2)-\frac 9{a-2})z=-\frac 9{a-2}$ which if $a = 5$ is

$(3-3)z = -3$ or $0 = 3$ and if $a=-1$ we get $(-3 + 3)z = 3$ or $0 = 3$.