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I have the matrix

$$\ \left( \begin{array}{ccc} 2 & -6d & 3 \\ 3d & -1 & 1.5 \\\end{array} \right) $$

for which the reduction (before back substitution) is equal to

$$\ \left( \begin{array}{ccc} 3d & -1 & 1.5 \\ 0 & -9d^2 + 1 & 9d - 1.5 \\\end{array} \right) $$

For a matrix with infinite solutions I know that the bottom row must equal zero, so I have to find $d$ such that I have a row of zeros at the bottom. But I don't think there is a solution. Is that true?

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  • $\begingroup$ you have a typo in the second row. it should read $0, \ \ 1 - 9d^2, \ \ 4.5d - 1.5$ $\endgroup$ – abel Jan 11 '15 at 1:58
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reducing the matrix gives $\begin{pmatrix} 1 &-3d & \frac{3}{2}\\0 & 9d^2-1 & \frac{3}{2}(1-3d)\end{pmatrix}$, so there will be infinitely many solutions for

the associated linear system when $9d^2-1=0 \text{ and }1-3d=0$, so when $d=\frac{1}{3}$.

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If $d=1/3$, then the first row equals two times the second row(in the unreduced matrix), so you can easily create a row of zeroes and there are infinitely many solutions.

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  • $\begingroup$ How $2\cdot 0$ equals $3d = 3\cdot \frac{1}{3} = 1$ ? $\endgroup$ – Integral Jan 11 '15 at 1:12
  • $\begingroup$ I guess you mean the first column equal $-1$ times the second column. Therefore the columns are linearly dependent. $\endgroup$ – Integral Jan 11 '15 at 1:16

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