Determining solutions of a matrix equation with variables in matrix

Question

given the matrix equation $Ax=b$ (shown as $[A|b]$ )

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

I've used Gaussian elimination to find that this equals:

$$\left[\begin{array}{ccc|c} 1&1 &0 &1 \\ 0 &1 & \frac1k & \frac1k \\ 0&0 &1& \frac{k-1}{k^2-1}\end{array}\right]$$

I need to find what value/s of $k$ give a) infinite solutions, b) a unique solution and c) no solution.

I'm guessing that 'no solution' is given when $k=0$ (row $2$) or $k=1$ (row $3$) as you cant divide by $0$.

Is this correct? And please advise how I determine value/s of $k$ to give a unique or infinite solutions.

Thank you

Answer 1

Note that \begin{bmatrix} 2&2&0|2\\0& k&1|1\\1&2& k|2\end{bmatrix}

$R_1=R_1/2\implies $\begin{bmatrix} 1&1&0|1\\0& k&1|1\\1&2& k|2\end{bmatrix}

$R_3=R_3-R_1\implies $\begin{bmatrix} 1&1&0|1\\0& k&1|1\\0&1& k|1\end{bmatrix}

Now $\det A$

where $$A=$$ \begin{bmatrix} 1&1&0\\0& k&1\\1&2& k\end{bmatrix}

$\neq 0\iff k\neq \pm 1$

Hence we have unique solution for $k\neq \pm 1$

For $k=\pm 1$ we have infinitely many solutions since $\text{rank A}=2$

Answer 2

Guide:

You have been dividing numbers. Before we perform division, make sure that we are not dividing number by $0$.

Hence I will check substitute $k=0$, $k=1$, $k=-1$ into the original matrix and check what do I get.

Reduce the matrix to row echelon form for those values of $k$ as well as $k \notin \{1,0,-1\}$.

Suppose the row echelon form is of the form of $[R_1|r].$

If there is a zero row on the left, that is $R_1$ has a zero row, but the corresponding entry for $r$ is non-zero, then there is no solution.

Otherwise, if there is a non-pivot columns in $R_1$, then there are infinitely many solutions.

Otherwise, the solution is unique.