MHB Matrix solutions with variable coefficient

[TheFallen018]

Hey guys,

I've got this question, that I think I have figured out, but I'm not completely sure.

Basically, I've got the following matrix to put into RREF. Not really a problem, but I'm not sure if I'm handling the variable incorrectly.
$\begin{bmatrix}
1 &-1 &|1 \\
3 &a &|3
\end{bmatrix}$

So, I subtracted 3*row1 from row2, which resulted in this
$\begin{bmatrix}
1 &-1 &|1 \\
0 &a+3 &|0
\end{bmatrix}$

The next thing I did is where I'm not sure about. I divided row2 by (a+3), and then added the resulting row into row 1 to get the matrix in RREF.
$\begin{bmatrix}
1 &0 &|1 \\
0 &1 &|0
\end{bmatrix}$

The remaining parts of the question ask me to find values of a where there is a unique solution, no solutions, and infinitely many solutions.

My thoughts were that there's no value a can take that will make the bottom row inconsistent, so there's no value where there's no solution. There can be infinitely many solutions if we go back to the part where we still have (a+3) in the second row. If we make a equal to -3, then $x_{2}$ becomes a free variable and gives us infinitely many solutions. If a is any other value, then there is a unique solution.

Does that seem right? Thanks

 

[Opalg]

Hey guys,

I've got this question, that I think I have figured out, but I'm not completely sure.

Basically, I've got the following matrix to put into RREF. Not really a problem, but I'm not sure if I'm handling the variable incorrectly.
$\begin{bmatrix}
1 &-1 &|1 \\
3 &a &|3
\end{bmatrix}$

So, I subtracted 3*row1 from row2, which resulted in this
$\begin{bmatrix}
1 &-1 &|1 \\
0 &a+3 &|0
\end{bmatrix}$

The next thing I did is where I'm not sure about. I divided row2 by (a+3), and then added the resulting row into row 1 to get the matrix in RREF.
$\begin{bmatrix}
1 &0 &|1 \\
0 &1 &|0
\end{bmatrix}$

The remaining parts of the question ask me to find values of a where there is a unique solution, no solutions, and infinitely many solutions.

My thoughts were that there's no value a can take that will make the bottom row inconsistent, so there's no value where there's no solution. There can be infinitely many solutions if we go back to the part where we still have (a+3) in the second row. If we make a equal to -3, then $x_{2}$ becomes a free variable and gives us infinitely many solutions. If a is any other value, then there is a unique solution.

Does that seem right? Thanks

Your conclusions are correct (solution is unique except when $a=-3$, when there are infinitely many solutions). But the RREF is only equal to $\begin{bmatrix}1 &0 &|1 \\ 0 &1 &|0 \end{bmatrix}$ when $a\ne-3$. In fact, if $a=-3$ then you cannot divide by $a+3$, so the final step of your calculation of the RREF is invalid in that case. If $a=-3$ then the previous step in the calculation becomes $\begin{bmatrix}1 &-1 &|1 \\ 0 &0 &|0 \end{bmatrix}$, and that is the RREF when $a=-3$.