Say for what values of $a \in \mathbb {R} $ this matrix system has solutions

Question

Let $a \in \mathbb{R}$ and $$ A_a = \begin{pmatrix} 1 & a & 1 \\ a & 2 & 3 \\ 2 & 3 & 4 \end{pmatrix}. $$

1. Say for each values of $a \in \mathbb{R}$ the system: $$ A_a \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} $$ accepts one solution only and for what values have no solutions.
2. Resolve the system for $a = 1$.

I got this problem from this image.

I think I know how to resolve system of equations using inverse matrices $3 \times 3$, $2 \times 2$ and more by finding the determinant. Then, I can reflect the matrix about the diagonal and multiply it with the inverse to find the solution. However, here, I get a little lost, because it seems like I should work backwards. What should I do?

Answer 1

The determinant of $A_a$ is $\det(A_a)=-4a^2+9a-5$ so: $$ \det(A_a)=0 \iff a=1 \;\lor \; a=\frac{5}{4} $$ and, since the matrix is invertible only if its determinant is not null, for $a\ne 1$ and $a \ne \frac{5}{4}$ the system have only one solution. Now test these values, for which the determinant is null and the matrix is not invertible, for $A_a(x,y,z)^T=(1,0,1)^T$ and see when the system is impossible or have infinitely many solutions.

Answer 2

Writing $A_a$ short as $A$, and the inhomogenous system $A x = b$ as augmented matrix $[A|b]$ we get these transformations: $$ \left[ \begin{array}{rrr|r} 1 & a & 1 & 1 \\ a & 2 & 3 & 0 \\ 2 & 3 & 4 & 1 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & a & 1 & 1 \\ a-3 & 2-3a & 0 & -3 \\ -2 & 3-4a & 0 & -3 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & a & 1 & 1 \\ a-1 & a-1 & 0 & 0 \\ -2 & 3-4a & 0 & -3 \end{array} \right] $$ The second row would benefit from division by $a-1$, so it is important if $a-1$ is zero or not.

Case 1:

If $a = 1$ this augmented matrix turns into $$ \left[ \begin{array}{rrr|r} 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ -2 & 3 & 0 & -3 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 1 & 1 & 1 \\ -2 & 3 & 0 & -3 \\ 0 & 0 & 0 & 0 \\ \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 1 & 1 & 1 \\ 0 & 5 & 2 & -1 \\ 0 & 0 & 0 & 0 \\ \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 0 & 3/5 & 6/5 \\ 0 & 1 & 2/5 & -1/5 \\ 0 & 0 & 0 & 0 \\ \end{array} \right] $$ Only two equations for three variables, one variable stays free and in this case we have infinite many solutions $$ (x, y, z) = (6/5-3/5 z, -1/5 - 2/5z, z) \quad (z \in \mathbb{R}) $$ where we used $z$ as free variable.

Other Cases:

If $a \ne 1$ we are led to $$ \left[ \begin{array}{rrr|r} 1 & a & 1 & 1 \\ 1 & 1 & 0 & 0 \\ -2 & 3-4a & 0 & -3 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1-a & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ -2+4a & 3 & 0 & -3 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1-a & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ -5+4a & 0 & 0 & -3 \end{array} \right] $$ Here it makes a difference if $-5 + 4a$ is zero or not.

Case 2:

If $a = 5/4$ we have $$ \left[ \begin{array}{rrr|r} -1/4 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & -3 \end{array} \right] $$ and the equation $0 x + 0 y + 0 z = -3$ corresponding to the last row can not be fulfilled, this means no solution in this case.

Case 3:

If $a \ne 1$ and $a \ne 5/4$ we have: $$ \left[ \begin{array}{rrr|r} 1-a & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & -3/(-5+4a) \end{array} \right] $$ and the unique solution $$ (x, y, z) = \left( -\frac{3}{-5+4a}, \frac{3}{-5+4a}, 1+\frac{3}{(1-a)(-5+4a)} \right) $$

Answer 3

You can do the exercise in one swoop by using Gaussian elimination: \begin{align} \left[\begin{array}{ccc|c} 1 & a & 1 & 1\\ a & 2 & 3 & 0\\ 2 & 3 & 4 & 1 \end{array}\right] &\to \left[\begin{array}{ccc|c} 1 & a & 1 & 1\\ 0 & 2-a^2 & 3-a & -a\\ 0 & 3-2a & 2 & -1 \end{array}\right] && \begin{matrix}R_2\gets R_2-aR_1\\R_3\gets R_3-2R_1\end{matrix} \\[6px]&\to \left[\begin{array}{ccc|c} 1 & a & 1 & 1 \\ 0 & 3-2a & 2 & -1 \\ 0 & 2-a^2 & 3-a & -a \end{array}\right] && R_2\leftrightarrow R_3 \\[6px]\color{red}{(3-2a\ne0)}\quad&\to \left[\begin{array}{ccc|c} 1 & a & 1 & 1\\ 0 & 1 & 2/(3-2a) & 1/(2a-3)\\ 0 & 2-a^2 & 3-a & -a \end{array}\right] && R_2\gets (3-2a)^{-1}R_2 \\[6px]&\to \left[\begin{array}{ccc|c} 1 & a & 1 & 1 \\ 0 & 1 & 2/(3-2a) & 1/(2a-3) \\ 0 & 0 & \frac{(4a-5)(a-1)}{3-2a} & \frac{(a-2)(a-1)}{3-2a} \end{array}\right] && R_3\gets R_3-(2-a^2)R_2 \end{align}

If $a=1$, the last row is zero, which means the system has infinitely many solutions; the matrix $A_a$ has rank $2$.

If $a=5/4$, the matrix $A_a$ has rank $2$, and the system has no solution.

If $a\notin\{1,5/4,3/2\}$, the matrix $A_a$ has rank $3$ and the system has a single solution.

It remains to see the case when $a=3/2$. \begin{align} \left[\begin{array}{ccc|c} 1 & a & 1 & 1\\ 0 & 2-a^2 & 3-a & -a\\ 0 & 3-2a & 2 & -1 \end{array}\right] &= \left[\begin{array}{ccc|c} 1 & 3/2 & 1 & 1\\ 0 & -1/4 & 3/2 & -3/2\\ 0 & 0 & 2 & -1 \end{array}\right] \end{align} It's clear that the matrix $A_a$ has rank $3$ and the system has a single solution.

The solution for $a=1$ is readily available, because the last matrix in the elimination process is $$ \left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & -1 \\ 0 & 0 & 0 & 0 \end{array}\right] $$

Do the transformation $R_1\gets R_1-R_2$ to get $$ \left[\begin{array}{ccc|c} 1 & 0 & -1 & 2 \\ 0 & 1 & 2 & -1 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ so the system can be written $$ \begin{cases} x=2+z\\[4px] y=-2-2z \end{cases} $$ so the solutions are of the form $$ \begin{bmatrix}2\\2\\0\end{bmatrix}+ t\begin{bmatrix}1\\-2\\1\end{bmatrix} $$