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Does the following system have exactly three solutions? $$\left\{ \begin{array}{l} 2x - y +3z = 1 \\ x + 4y - 2z = -7 \\ 3x + y -z = 4 \\ \end{array} \right.$$

I marked this answer as True. I proceeded to row reduce it and obtained the resultant matrix as -

$$\left[ \begin{array}{ccc|c} 1 & -1/2 & 3/2 & 4 \\ 0 & 1 & 9 & 13/9 \\ 0 & 0 & 1 & 10/28 \\ \end{array} \right]$$

I'm pretty sure I made some mistake while row-reducing it. I answered this question in an exam setting. Then I gave the explanation as follows -

Form row reduction, we know that the system is consistent and the rank of the matrix is 3 which is equal to the number of variables. So, the system of equations has 3 distinct solutions.

Can someone point out where I went wrong?

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    $\begingroup$ Perhaps, the wording of this question is confusing you. A single solution is a value for x, y, and z that satisfies your system of equations.Three solutions would mean you have 3 sets of values for x, y, and z that satisfy your system of equations. $\endgroup$ – Ralff Oct 19 '17 at 2:23
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    $\begingroup$ You have 3 planes... at how many points can all of them intersect at the same time? Surely not 3. $\endgroup$ – Mehrdad Oct 19 '17 at 5:51
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    $\begingroup$ @Mehrdad: 3 planes can all intersect at 3 points. They cannot all intersect at exactly 3 points, though. $\endgroup$ – Eric Duminil Oct 19 '17 at 12:31
  • $\begingroup$ Graphing this system indicates that each pair of planes intersect on a single line, but there is no line on which all three planes intersect. $\endgroup$ – DonielF Oct 19 '17 at 15:42
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HINT: The system of linear equations can have only 0, 1 or infinitely many solutions. Hence no calculations are needed.

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  • $\begingroup$ Oh right! Of course!! And the answer would be false. THANK YOU! $\endgroup$ – ChocolateAndMath Oct 19 '17 at 2:39
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    $\begingroup$ This is hardly a hint. If it can only be zero, one, infinity - it obviously can't be three and the answer is immediate. $\endgroup$ – Nij Oct 19 '17 at 4:14
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    $\begingroup$ @Nij I, personally, immediately assumed irony with the "HINT" there. The problem is so simple that the Przemysław's phrasing works perfectly IMO. $\endgroup$ – vaxquis Oct 19 '17 at 14:51
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Start with the augmented matrix:

\begin{eqnarray} \left(\begin{array}{cccc} 2&-1&3 &: 1\\ 1&4&2&:-7\\ 3&1&-1&:4\end{array}\right) \end{eqnarray}

First: Interchange rows 1 and 2 to get:

$$\left(\begin{array}{cccc} 1&4&-2&:-7\\ 2&-1&3 &: 1\\ 3&1&-1&:4\end{array}\right)$$

Second: Use the first element of the first row as a pivot to eliminate the 2 and 3 of the first column to get:

$$ \left(\begin{array}{cccc} 1&4&-2&:-7\\ 0&-9&7 &: 15\\ 0&-11&5&:25\end{array}\right) $$

Then you can replace Row2 by Row2-Row3 to get:

$$\left(\begin{array}{cccc} 1&4&-2&:-7\\ 0&2&2&: -10\\ 0&-11&5&:25\end{array}\right)$$

then you can divide by 2 the Row2 to get:

$$\left(\begin{array}{cccc} 1&4&-2&:-7\\ 0&1&1&: -5\\ 0&-11&5&:25\end{array}\right)$$

Then you can eliminate the -11 in the third row:

$$\left(\begin{array}{cccc} 1&4&-2&:-7\\ 0&1&1&: -5\\ 0&0&16&:-30\end{array}\right)$$

So $z=\frac{-30}{16}=\frac{-15}{8}$. Because you have the equivalent augmentedmatrix

$$\left(\begin{array}{cccc} 1&4&-2&:-7\\ 0&1&1&: -5\\ 0&0&1&:-\frac{15}{8}\end{array}\right)$$

To get $y$ you simply replace Row2 with Row2-Row3 and you will get:

$$\left(\begin{array}{cccc} 1&4&-2&:-7\\ 0&1&0&: -5+\frac{15}{8}\\ 0&0&1&:-\frac{15}{8}\end{array}\right)$$

and $y=-5+\frac{15}{8}=\frac{-40+15}{8}=\frac{-25}{8}$. I leave $x$ for you to compute.

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Using row reduction you can show that $x=7/4$, $y=-25/8$ and $z=-15/8$. You can use even Cramer rule to compute the unique solution using that the matrix $A$ given by $$A=\left(\begin{array}{ccc} 2&-1&3\\ 1&4&-2\\ 3&1&-1\end{array}\right)$$ and $\det(A)=-32\neq 0$ and this implies that the system has unique solution for any vector of independent terms.

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  • $\begingroup$ So, should I have done rref instead? $\endgroup$ – ChocolateAndMath Oct 19 '17 at 2:23
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    $\begingroup$ @ChocolateAndMath: You can use Gauss-Jordan method or Cramer formula because $det(A)=-32\neq 0$ where $A$ is the coefficients matrix showed above. $\endgroup$ – Hector Blandin Oct 19 '17 at 2:28
  • $\begingroup$ We hadn't touched determinants back then :( $\endgroup$ – ChocolateAndMath Oct 19 '17 at 2:38
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    $\begingroup$ @ChocolateAndMath: Ok, in that case you can reduce using Gauss method. $\endgroup$ – Hector Blandin Oct 19 '17 at 2:38
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    $\begingroup$ @ChocolateAndMath: Check my new answer below to see the idea of Gauss method $\endgroup$ – Hector Blandin Oct 19 '17 at 2:58

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