Consider the following system and find the values of b for which the system has a solution

Question

So I have this system:

$$\left\{\begin{array}{c} x_1 &− x_2 &+ 2x_3 &= 2 \\ x_1 &+ 2x_2 &− x_3 &= 2 \\ x_1 &+ x_2 & &= 2 \\ x_1 & & +x_3 &= α \end{array}\right.$$

And we are asked to find the values of $\alpha$ for which the system has a solution. When I do the coefficient matrix and a Gauss-Jordan elimination i get two entirely zero rows with one of the augmented matrix solutions being $\alpha-2$ so am I correct in saying that the system will have an infinite number of solutions if $\alpha=2$?

The thing that concerns me is that we are then asked to find the value of $\alpha$ if the degree of freedom is $2$?

Any help would be greatly appreciated!

PS. Sorry for any formatting errors!

Answer 1

Yes you are correct that if $\alpha=2$, then there would be two zero rows. Therefore, (in this 4x4 augmented matrix case) one of the variables, say $x_3$ can be chosen as a "free" variable.

This can be seen by letting $t:=x_3$. Now you can see that $x_2$ and $x_1$ are fully defined by the equations of your row-reduced matrix and $t$. Thus, you only have one free variable, and your degrees of freedom is 1.

Alternatively, you can reason that because you have 3 variables, and the rank of your matrix is 2 (a 4x3 matrix with two non-zero rows), then you have one free variable or a degree of freedom of 1.

Answer 2

I'd argue like the following.
From the first two rows it follows, that whatever value $x_1$ has, $x_2$ must equal $x_3$.
If we insert this knowledge into the last two rows, (replacing $x_3$ by $x_2$) then we see, that $\alpha$ must equal $2$. And if we insert also that, $$\left\{\begin{array}{c} x_1 &− x_2 &+ 2x_3 &= 2 \\ x_1 &+ 2x_2 &− x_3 &= 2 \\ x_1 &+ x_2 & &= 2 \\ x_1 & +x_2 &&= 2&=α \end{array}\right.$$

then it follows $x_1 = 2 - x_2$ and (unfortunately) nothing more - and thus we have an infinitude of solutions.

Answer 3

HINT: use Rouché–Capelli theorem

A system of linear equations with $\,n\,$ variables has a solution if and only if the rank of its coefficient matrix $\,A\,$ is equal to the rank of its augmented matrix $\,\left[\,A \mid b\,\right].\,$

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Let us rewrite your system in matrix form $\,\mathbf A\,\vec{\boldsymbol{x}} = \vec{\boldsymbol{b}}$:

\begin{align} \left\lbrace \begin{array}{ccccccc} x_1 & - & x_2 & +& 2\, x_3 & = & 2 \\ x_1 & + & 2\,x_2 & -& x_3 & = & 2 \\ x_1 & + & 2\,x_2 & & & = & 2 \\ x_1 & & &+& x_3 & = & \alpha \\ \end{array} \right. \iff %\underbrace{ \begin{pmatrix} 1 & -1 & 2 \\ 1 & 2 & -1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \\ \end{pmatrix} %}_{\;\ \mathbf A} %\overbrace{ \begin{matrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}\\ \phantom{x_4}\end{matrix} %}^{\vec{\boldsymbol{b}}} = %\underbrace{ \begin{pmatrix} 2 \\ 2 \\ 2 \\ \alpha \end{pmatrix} %}_{\vec{\boldsymbol{b}}} \end{align}

$$ \mathbf{A} = \begin{pmatrix} 1 & -1 & 2 \\ 1 & 2 & -1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \\ \end{pmatrix} , \quad b =\begin{pmatrix} 2 \\ 2 \\ 2 \\ \alpha \end{pmatrix} \implies \mathbf{B} := \left[ \, \mathbf A \,\left\lvert \;\, \vec{\boldsymbol{b}} \right.\, \right] = \left[ \begin{array}{ccc|c} 1 & -1 & 2 & 2 \\ 1 & 2 & -1 & 2 \\ 1 & 1 & 0 & 2 \\ 1 & 0 & 1 & \alpha \\ \end{array} \right] $$

It is easy to see that rank $\,\mathbf A = 2\,$ because its first row is linear combination of the second and the third rows: $\,\mathbf A\left[1\right] = 3\mathbf A\left[2\right] - 2\mathbf A\left[2\right]\,$ $$ \mathbf A\left[1\right] = 3\mathbf A\left[2\right] - 2\mathbf A\left[2\right] \quad\iff\qquad \begin{array}{ccc} & \begin{pmatrix} 1 & \phantom{-}1 & \phantom{-}0 \end{pmatrix} & \times \,3 \\ - & \begin{pmatrix} 1 & \phantom{-}2 & -1 \end{pmatrix} & \times \,2 \\ \hline & \begin{pmatrix} 1 & -1 & \phantom{-}2 \end{pmatrix} & \\ \end{array} $$

Therefore we need to find values of $\,\alpha\,$ for which rank $\,\mathbf B = 2$. I hope you can pick it from here.

Answer 4

Gaussian elimination.

Downwards: $$\tag 1 \begin{array} {l} x_1 & -x_2 & + 2 x_3 & 0 & = 2 \\ x_1 & +2x_2 & - x_3 & 0 & = 2 \\ x_1 & +1x_2 & 0 & 0 & = 2 \\ x_1 & 0 &+1 x_3 & -\alpha & = 0 \\ \end{array} $$ Subtract row 1 from rows 2 to 4: $$\tag 2 \begin{array} {l} x_1 & -x_2 & + 2 x_3 & 0 & = 2 \\ \hline 0 & 3x_2 & - 3 x_3 & 0 & = 0 \\ 0 & 2x_2 & -2x_3 & 0 & = 0 \\ 0 & 1x_2 & -1x_3 & -1\alpha & = -2 \\ \end{array} $$ Cancel the 3 and the 2 because on the rhs are zeros in row 2 and 3: $$\tag 3 \begin{array} {l} x_1 & -x_2 & + 2 x_3 & 0 & = 2 \\ \hline 0 & x_2 & - x_3 & 0 & = 0 \\ 0 & x_2 & -x_3 & 0 & = 0 \\ 0 & x_2 & -x_3 & -1\alpha & = -2 \\ \end{array} $$ Subtract row 2 from rows 3 and 4: $$ \begin{array} {l} x_1 & -x_2 & + 2 x_3 & 0 & = 2 \\ 0 & x_2 & - x_3 & 0 & = 0 \\ \hline 0 & 0 & 0 & 0 & = 0 \\ 0 & 0 & 0 & -1\alpha & = -2 \\ \end{array} $$ It follows $\alpha = 2$. Also, from row 2 that $x_2=x_3$
From the remaining gaussian-eiminate upwards: add row 2 to row 1 $$ \begin{array} {l} x_1 & 0 & + x_3 & 0 & = 2 \\ \hline 0 & x_2 & - x_3 & 0 & = 0 \\ \hline 0 & 0 & 0 & 0 & = 0 \\ 0 & 0 & 0 & -1\alpha & = -2 \\ \end{array} $$ We get $x_1 =2-x_3$ with arbitrary many solutions.