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I've been given the following matrix $X$: $$X= \begin{bmatrix} 1 & 4 & 8 & 1 \\ 0 & 30 &1 & 0 \\ 0 &2& 0& 0 \\ 1 &2 & 9 & x \\ \end{bmatrix} $$

and I have to determine for which value of $x$ the matrix is invertible.
Any hints/suggestions would be greatly appreciated.

So far I have substituted different values for $x$, but there seems to be an inverse for all values of $x$ above $1$.

HK Lee
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Kevin
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    Hint: a matrix is invertible whenever its determinant is non-zero. – TZakrevskiy Oct 06 '14 at 12:36
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    If you have studied linear algebra, another hint: $X$ is invertible iff its columns are linearly independent. – Nighty Oct 06 '14 at 12:40
  • A third hint: If you use row operations to reduce the matrix to an upper or lower triangular matrix, then the determinant will just be the product of the main diagonal, then you will get invertible iff that product is nonzero – Alan Oct 06 '14 at 12:42

2 Answers2

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$$X= \begin{pmatrix} 1 & 4 & 8 & 1 \\ 0 & 30 &1 & 0 \\ 0 &2& 0& 0 \\ 1 &2 & 9 & x \\ \end{pmatrix}\stackrel{ R_4-R_1}\longrightarrow\;\; \begin{pmatrix} 1 & 4 & 8 & 1 \\ 0 & 30 &1 & 0 \\ 0 &2& 0& 0 \\ 0 &\!\!-2 & 1 & x-1 \\ \end{pmatrix}\stackrel{R_3-\frac1{15}R_2}{\stackrel{R_4+\frac1{15}R_2\;,\;R_2}\longrightarrow} $$

$$\begin{pmatrix} 1 & 4 & 8 & 1 \\ 0 & 30 &1 & 0 \\ 0 &0& \!\! -\frac1{15}& 0 \\ 0 &0 & \frac{16}{15} & x-1 \\ \end{pmatrix}\stackrel{R_4+16R_3}\longrightarrow\begin{pmatrix} 1 & 4 & 8 & 1 \\ 0 & 30 &1 & 0 \\ 0 &0& \!\! -\frac1{15}& 0 \\ 0 &0 & 0 & x-1 \\ \end{pmatrix}$$

and the above matrix (and thus the original one as well) is singular iff the last row is all zeros, so...

Timbuc
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Hint: You may use the well-known relation: $$ \operatorname{tr}(X) = \log\Big(\det\big(\exp(X)\big)\Big) $$

nullgeppetto
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