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Use matrix multiplication to get an equation for \\(k\\) which is then to be solved.

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We have

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\\(\\begin{pmatrix} k & 1 &1\\end{pmatrix}\\var{A}\\begin{pmatrix}k\\\\1\\\\1\\end{pmatrix}= \\begin{pmatrix} k & 1 &1\\end{pmatrix}\\begin{pmatrix}k+1\\\\k+2\\\\-1\\end{pmatrix}\\)

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\\(=k(k+1)+k+2-1=k^2+2k+1=(k+1)^2\\).

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So this is zero when \\(k=-1\\).

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For which \\(k\\) is

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\\(\\begin{pmatrix} k & 1 &1\\end{pmatrix}\\var{A}\\begin{pmatrix}k\\\\1\\\\1\\end{pmatrix}=0\\)?

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\\(k= \\)[[0]]

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Calculate the matrix multiplication in two steps:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\\(\\var{A}\\begin{pmatrix}k\\\\1\\\\1\\end{pmatrix} = \\left(\\begin{matrix}\\phantom{.}\\\\\\phantom{.}\\\\\\phantom{.}\\\\\\phantom{.}\\end{matrix}\\right.\\)[[0]]\\(\\left.\\begin{matrix}\\phantom{.}\\\\\\phantom{.}\\\\\\phantom{.}\\\\\\phantom{.}\\end{matrix}\\right)\\)
[[1]]
[[2]]
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Then calculate \\(\\begin{pmatrix} k & 1 &1\\end{pmatrix} \\) times your answer, to get: [[3]]

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Now set this expression \\(=0\\) and resolve for \\(k\\). \\(k= \\)[[4]]

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