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A=$\begin{pmatrix} 1 & 3 & 1\\ 0 & 1 & 1\\ 2 & -2 & -6 \end{pmatrix}$

B=$\begin{pmatrix} 8\\ k\\ 8 \end{pmatrix}$

Discuss the solutions of the system S: Ax=B.

I have used the Rouche Capelli theorem wich implicate that S have solutions only if:

$rkA\leq$rk$A'$

where A'= A|B.

Well, $rkA$=2 still. But $rkA'=2$ only if $\begin{vmatrix} 1 & 3 & 8\\ 0 & 1 & k\\ 2 & -2 & 8 \end{vmatrix}$=$0$

This happens for $k=1$ wich is ok with my textbook but... ...if $k=1$ I can also calculate the other possible determinant:

$\begin{vmatrix} 1 & 1 & 8\\ 0 & 1 & k\\ 2 & -6 & 8 \end{vmatrix}$ wich for $k=1$ is not equal to $0$ $\Rightarrow $ for $k=1$, $rkA'$=$0$

My textbook still saying that for $k=1$, S have one solution. Why i'm wrong?

My solution will be to create a system made of the possible determinant equalized to 0 (in this case 2 deterimnants).

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    You may want to double-check Rouché-Capelli theorem: $\operatorname{rk}A\le\operatorname{rk}A'$ is always true. – 2017-01-12
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    @OsvaldoPaniccia: If you do RREF, you end up with a row that is $0, 0, 0, 8(k-1)$. For there to be a solution, $k$ must equal one, otherwise, there is no solution. Please note that the matrix is singular. – 2017-01-12
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    @G.Sassatelli Absolutely. – 2017-01-12
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    @Moo It's true. But why my method is wrong? – 2017-01-12
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    @OsvaldoPaniccia your method is fine, your execution is lacking. det $\begin{vmatrix} 1 & 1 & 8\\ 0 & 1 & 1\\ 2 & -6 & 8 \end{vmatrix} = 0$ – 2017-01-12

1 Answers 1

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Use row reduction for the augmented matrix: \begin{align} \begin{bmatrix}1&3&1&8\\0&1&1&k\\2&-2&-6&8\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&3&1&8\\0&1&1&k\\0&-8&-8&-8\end{bmatrix}\rightsquigarrow \begin{bmatrix}1&3&1&8\\0&1&1&k\\0&0&0&(k-1)8\end{bmatrix}\end{align} Thus $\operatorname{rank}A'=2$ if and only if $k=1$.

Moreover, the solutions are an affine subspace of dimension $1$: if we continue row reduction to obtain the reduced row echelon form, dropping the last (zero) row,we have $$\begin{bmatrix}1&3&1&8\\0&1&1&1\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&0&-2&5\\0&1&1&1\end{bmatrix}$$ whence the solutions: \begin{align} \begin{cases}x&=2z-5,\\ y&=-z-1,\end{cases}\quad\text{or, in vector form:}\quad \begin{bmatrix}x\\y\\z\end{bmatrix}=z\begin{bmatrix}2\\-1\\1\end{bmatrix}-\begin{bmatrix}5\\1\\0\end{bmatrix}. \end{align}