Invertible Matrix 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? 
 A: $$\begin{vmatrix}
1&0&1&2\\
2&3&\beta&4\\
4&0&-\beta&-8\\
\beta&0&\beta&\beta
\end{vmatrix}
=
\beta\begin{vmatrix}
1&0&1&2\\
2&3&\beta&4\\
4&0&-\beta&-8\\
1&0&1&1
\end{vmatrix}
=
\beta\begin{vmatrix}
0&0&0&1\\
2&3&\beta&4\\
4&0&-\beta&-8\\
1&0&1&1
\end{vmatrix}
=
-\beta\begin{vmatrix}
2&3&\beta\\
4&0&-\beta\\
1&0&1
\end{vmatrix}
=3\beta
\begin{vmatrix}
4&-\beta\\
1&1
\end{vmatrix}
=3\beta(4+\beta)$$
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:


*

*Effect of Elementary Row Operations on Determinant

*Laplace's Expansion Theorem

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.
A: 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.
A: 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 $0$. 
A: 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.
