if$\begin{cases}2x-y=1\\x-2y=a\\x-ay=122\end{cases}$ has solution(s), find $a$

Question

My attempt:

it can be written as $$\left[ \begin{array}{ccc} 2&-1\\ 1&-2\\ 1&-a\\ \end{array}\right]\left[ \begin{array}{ccc} x\\ y\\ \end{array}\right]=\left[ \begin{array}{ccc} 1\\ a\\ 122\\ \end{array}\right]$$

I'm trying to solve it by using the idea of vector transformation, but the matrices are nonsquare so I'm stuck

Answer 1

$\left(\begin {array}{rr|r}2&-1&1\\1&-2&a\\1&-a&122\end {array}\right)\to\left(\begin {array}{rr|r}1&-2&a\\0&-a+2&122-a\\0&3&1-2a\end{array}\right)\to\left (\begin{array}{rr|r}1&-2&a\\0&1&\dfrac{122-a}{2-a}\\0&0&\dfrac{-366+3a}{2-a}+1-2a\end{array}\right) $.

Thus we have $-366+3a=(1-2a)(a-2)$.

So, $366-3a=2a^2-5a+2$. Thus $a^2-a-182=0$. So $a=-13,14$.

Answer 2

For the system to be consistent, the augmented matrix $$\left[\begin{array}{cc|c}2&-1&1\\1&-2&a\\1&-a&122\end{array}\right]$$ must be rank-deficient. Its determinant is equal to $2a^2-2a-364$. Set this equal to zero and solve the resulting quadratic equation for $a$. This isn’t a sufficient condition, so you’ll still need to check that the two roots do lead to consistent systems, but that’s fairly easy to do.

You could instead row-reduce this matrix directly to obtain $$\left[\begin{array}{cc|c}1&-\frac12&\frac12\\0&1&{1-2a\over3}\\0&0&-{2a^2-2a-364\over3}\end{array}\right]$$ and extract an equivalent equation from the last row, but I find the first method a bit less error-prone.