vault backup: 2026-10-05 08:52:57
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- There can also be non-trivial solutions
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- There can also be non-trivial solutions
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To find when the system $$\begin{gathered} 3x_{1} + 6x_{2} = 2 \\ hx_{1} + 12x_{2} = k \end{gathered}$$ has no solutions
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$$\begin{bmatrix}
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$$\begin{bmatrix}
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3 & 6 & 2 \\ h & 12 & k
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3 & 6 & 2 \\ h & 12 & k
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\end{bmatrix} \sim \begin{bmatrix}
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\end{bmatrix} \sim \begin{bmatrix}
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h & 2h & \frac{2h}{3} \\ h & 12 & k
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\end{bmatrix} \sim \begin{bmatrix}
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h & 2h & \frac{2h}{3} \\ 0 & 12 - 2h & k - \frac{2h}{3}
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\end{bmatrix} $$
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\end{bmatrix} $$
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There are no solutions the system of equations when there is no pivot in a row and the value on the right is not equal to zero (if the value on the right is also equal to zero than that equation is essentially $0=0$ which is always true).
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This means that there are no solutions when $$12-2h = 0 \to h=6$$ and $$k-\frac{2h}{3} \ne 0$$
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