Which one of the following matrices has an inverse?

The correct option is:
$\left[\begin{array}{lll}1& 4& 8\\ 0& 4& 2\\ 1& 2& 4\end{array}\right]$

Underlying Concept:
A square matrix has an inverse (is invertible or non-singular) if and only if its determinant is non-zero. If the determinant of a matrix is equal to zero, the matrix is singular and does not have an inverse.

Let us evaluate the determinant of the correct option to verify that it is non-zero. Let the matrix be defined as:
$A=\left[\begin{array}{ccc}1& 4& 8\\ 0& 4& 2\\ 1& 2& 4\end{array}\right]$

We calculate the determinant of matrix $A$, denoted as $\det \left(A\right)$ or $\left|A\right|$, by expanding along the first column (which contains a convenient zero):
$\left|A\right|=1\cdot \left|\begin{array}{cc}4& 2\\ 2& 4\end{array}\right|-0\cdot \left|\begin{array}{cc}4& 8\\ 2& 4\end{array}\right|+1\cdot \left|\begin{array}{cc}4& 8\\ 4& 2\end{array}\right|$

Now, we calculate the determinants of the $2\times 2$ submatrices:
1. For the first term:
$\left|\begin{array}{cc}4& 2\\ 2& 4\end{array}\right|=(4\cdot 4)-(2\cdot 2)=16-4=12$

2. For the third term:
$\left|\begin{array}{cc}4& 8\\ 4& 2\end{array}\right|=(4\cdot 2)-(8\cdot 4)=8-32=-24$

Substituting these values back into the determinant expression:
$\left|A\right|=1\cdot \left(12\right)-0+1\cdot (-24)=12-24=-12$

Since $\left|A\right|=-12\ne 0$, this matrix has a non-zero determinant, which guarantees that it is invertible (has an inverse).

Why the other options do not have an inverse:
Let's analyze why the other options have a determinant of zero:
1. First Option:
$\left[\begin{array}{ccc}1& 4& 8\\ 0& 4& 2\\ 0.5& 2& 4\end{array}\right]$
Notice that Row 3 $[0.5,2,4]$ is exactly half of Row 1 $[1,4,8]$. Since the rows are linearly dependent, the determinant is $0$.

2. Second Option:
$\left[\begin{array}{ccc}1& 2& 3\\ 2& 4& 6\\ 3& 2& 9\end{array}\right]$
Here, Row 2 $[2,4,6]$ is exactly twice Row 1 $[1,2,3]$. Thus, the rows are linearly dependent, and the determinant is $0$.

3. Fourth Option:
$\left[\begin{array}{ccc}1& 4& 8\\ 0& 4& 2\\ 3& 12& 24\end{array}\right]$
In this matrix, Row 3 $[3,12,24]$ is exactly three times Row 1 $[1,4,8]$. Hence, the rows are linearly dependent, and the determinant is $0$.

Therefore, the only matrix that does not have linearly dependent rows, and consequently has a non-zero determinant and a well-defined inverse, is the third option.