Let $$A=\left(\begin{smallmatrix}2&-1&0&1\\0&0&a&3\\0&0&0&b\end{smallmatrix}\right)$$
For what choices (if any) of real numbers a and b are the rows of A linearly dependent? Justify your response.
I figured out that the free variables in this matrix are x2 and x4. However, I got the matrixes [-1/2 , -1 , -3/a , 1] and [-2, 1 , 0 , 0 ]. I'm not sure what my next step would be.
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$\begingroup$To make them linearly dependent, you must find nontrivial $\lambda_1, \lambda_2, \lambda_3$ such that $$\lambda_1 \pmatrix{2\\-1\\0\\1} + \lambda_2 \pmatrix{0\\0\\a\\3} + \lambda_3 \pmatrix{0\\0\\0\\b} = \pmatrix{0\\0\\0\\0}$$ The first two equations enforce $\lambda_1 = 0$ and the third equation enforces $\lambda_2 a = 0$, the fourth $3\lambda_2 + b\lambda_3 = 0$. To get a solution with $\lambda_2 \ne 0$ or $\lambda_3 \ne 0$ there are two possibilities
- $b\ne 0$ then $\lambda_2 = -\lambda_3$ but this implies for nonzero $\lambda_3$. that $a = 0$:
$a=0, b\ne 0$ - $b=0$ then we can chose $\lambda_2 = 0$ and $\lambda_3$ as well as $a$ arbitrary
Summing up: $a$ is arbitrary. If $a\ne0$ we must chose $b=0$, else $b$ is arbitrary as well.
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