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Let $A$ be the matrix

$$ A=\left(\begin{array}{cccc}1&0&1&2\\2&3&\beta&4\\4&0&-\beta&-8\\ \beta&0&\beta&\beta \end{array}\right). $$

For what values of $\beta$ is the matrix invertible?

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When is a matrix invertible ? What did you try ? –  Belgi Oct 23 '12 at 15:19
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$$\begin{vmatrix} 1&0&1&2\\ 2&3&\beta&4\\ 4&0&-\beta&-8\\ \beta&0&\beta&\beta \end{vmatrix} = \beta\begin{vmatrix} 1&0&1&2\\ 2&3&\beta&4\\ 4&0&-\beta&-8\\ 1&0&1&1 \end{vmatrix} = \beta\begin{vmatrix} 0&0&0&1\\ 2&3&\beta&4\\ 4&0&-\beta&-8\\ 1&0&1&1 \end{vmatrix} = -\beta\begin{vmatrix} 2&3&\beta\\ 4&0&-\beta\\ 1&0&1 \end{vmatrix} =3\beta \begin{vmatrix} 4&-\beta\\ 1&1 \end{vmatrix} =3\beta(4+\beta)$$

So the determinant is zero only for $\beta=0$ and $\beta=-4$. (i.e.: these are the only cases when the matrix is singular.)

We have used the following:


If - for some reason - you want to avoid determinants, you can simply do the elementary row operations. (You mentioned in a comment that you haven't learned about determinants yet. It would have been better to mention this in your post.) $$\begin{pmatrix} 1&0&1&2\\ 2&3&\beta&4\\ 4&0&-\beta&-8\\ \beta&0&\beta&\beta \end{pmatrix} \overset{(1)}\sim \begin{pmatrix} 1&0&1&2\\ 2&3&\beta&4\\ 4&0&-\beta&-8\\ 1&0&1&1 \end{pmatrix} \sim \begin{pmatrix} 0&0&0&1\\ 2&3&\beta&4\\ 4&0&-\beta&-8\\ 1&0&1&1 \end{pmatrix} \sim \begin{pmatrix} 1&0&1&1\\ 2&3&\beta&4\\ 4&0&-\beta&-8\\ 0&0&0&1 \end{pmatrix} \sim \begin{pmatrix} 1&0&1&1\\ 0&3&\beta-2&2\\ 0&0&-\beta-4&-12\\ 0&0&0&1 \end{pmatrix} \sim \begin{pmatrix} 1&0&1&1\\ 0&3&\beta-2&0\\ 0&0&-\beta-4&0\\ 0&0&0&1 \end{pmatrix} \overset{(2)}\sim \begin{pmatrix} 1&0&1&0\\ 0&3&\beta-2&0\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix} \sim \begin{pmatrix} 1&0&0&0\\ 0&3&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix} \sim \begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix} $$ You see that the matrix is invertible. But we have to be careful about steps $(1)$ and $(2)$. In $(1)$ we have divided by $\beta$ and in $(2)$ we have divided by $-\beta-4$. So these steps are valid only for $\beta\ne 0,4$. If you try elimination for $\beta=0$ and $\beta=4$, you will find out that for these values the matrix is singular.

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Thankyou for the explanation. Until now, the only method that I've learned was the one that you used. I'm sorry for the fact I didn't write in my post that I dindn't learn detemrinants. –  user44133 Oct 27 '12 at 20:39
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Recall that $A$ is invertible iff $\rm{det}(A)\neq 0$. Can you calculate $\rm{det}(A)$ as a function of $\beta$? If you can, solve $\rm{det}(A)(\beta)=0$ and get a solution set. Then the values of $\beta$ for which $A$ is invertible are precisely those in the complement of the solution set.

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You know that a square matrix $A$ is invertible only when its determinant is non-zero, so this question is just solving for those $\beta$ such that $\det A(\beta) \neq 0$. Finding the determinant of a $4 \times 4$ matrix is a little tedious, but not difficult; you can use LaPlace Expansion twice until you're evaluating the determinants of $2 \times 2$ sub-matrices.

Spoiler: I found that $\det(A(\beta))= 3 \beta^2 + 12 \beta$, so then the roots of this parabola are the only places that make the determinant zero, i.e., the matrix is invertible if $\beta \neq -4$ or $0$.

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Compute the determinat of $A$ with laplace expansion expanding at second column it will ease the computations because there are there many zeros

$|A|=(-1)^{2+2}3 \left|\begin{array}{ccc}1&1&2\\4&-\beta&-8\\ \beta&\beta&\beta \end{array}\right|$

For $\beta=0 \Rightarrow |A|=0$ Now for $\beta \not =0$ $$|A| \not= 0 \Leftrightarrow \left|\begin{array}{ccc}1&1&2\\4&-\beta&-8\\ 1&1&1 \end{array}\right| \not =0$$

$$\left|\begin{array}{ccc}1&1&2\\4&-\beta&-8\\ 1&1&1 \end{array}\right|=\left|\begin{array}{ccc}1&1&2\\4&-\beta&-8\\ 0&0&-1 \end{array}\right|=\left|\begin{array}{ccc}1&1&2\\0&-\beta-4&-16\\ 0&0&-1 \end{array}\right|=\beta +4$$ Therefore for $\beta \not = 4,0$ $A$ is invertible.

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That's minor, but you should have got $2(\beta-4)$ –  Dennis Gulko Oct 23 '12 at 16:00
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