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If there are an infinite number of solutions to the system

$$\left|\begin{array}{cc|c} -5 & 6 & h\\ -8 & k & -7\end{array}\right|$$

then what do $h$ and $k$ equal?

I know that for a system to have infinity many solutions then both of the rows must be equal.

George V. Williams
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ground.clouds1
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  • I don't understand what you're asking. I thought I wrote my system in the question. – ground.clouds1 Jan 19 '13 at 18:37
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    @ground.clouds1 The way you have written down is unclear. If you meant $$\begin{bmatrix} -5 & 6\ -8 & k\end{bmatrix} \begin{bmatrix}x_1\ x_2 \end{bmatrix} = \begin{bmatrix} h \ -7\end{bmatrix}$$ then it is preferable to separate the second column and third column with a vertical straight line. Currently, it looks like just a $2 \times 3$ matrix. –  Jan 19 '13 at 18:40

4 Answers4

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I assume the system you have is $$\begin{bmatrix} -5 & 6\\ -8 & k\end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix} = \begin{bmatrix} h \\ -7\end{bmatrix}$$ For this to have infinite solutions, the two rows must be linearly dependent i.e. the second row must be a scalar multiple of the first row.

Note that $- 8 = - 5 \times \dfrac85$. Hence, $k = 6 \times \dfrac85 = \dfrac{48}5$ and $-7 = \dfrac85 \times h \implies h = - \dfrac{35}8$.

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the rows of matrix are proportional $$(-5):(-8)=6:k=h:(-7)$$ $$-5k=-48,h=35/-8$$ $$k=\frac{48}{5},h=-\frac{35}{8}$$

Adi Dani
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If you have the system of two lines: $$\begin{cases} ax+by=c \\ a'x+b'y=c' \end{cases}$$ then the system is infinite solutions iff we have: $$\frac{a}{a'}=\frac{b}{b'}=\frac{c}{c'}$$ Of course we have some restrictions on the coefficients.

Mikasa
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As you say, the rows must be equal. They should give you the same information, which occurs only if one is a scalar multiple of the other. Therefore

$\frac{-5}{-8}=\frac{6}{k}=\frac{h}{-7}$. This gives you $k=\frac{48}{5}$ and $h=\frac{-35}{8}$.

mathemagician
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