To determine the value of [tex]\( x \)[/tex] that makes the matrix [tex]\( X \)[/tex] singular, we need to find the value of [tex]\( x \)[/tex] for which the determinant of the matrix [tex]\( X \)[/tex] equals zero. The matrix in question is:
[tex]\[
X = \begin{pmatrix}
2 & 0 & 1 \\
x & 2 & 4 \\
1 & 2 & x
\end{pmatrix}
\][/tex]
Let's compute the determinant of [tex]\( X \)[/tex]:
[tex]\[
\text{det}(X) = \begin{vmatrix}
2 & 0 & 1 \\
x & 2 & 4 \\
1 & 2 & x
\end{vmatrix}
\][/tex]
By expanding along the first row:
[tex]\[
\text{det}(X) = 2 \cdot \begin{vmatrix}
2 & 4 \\
2 & x
\end{vmatrix} - 0 \cdot \begin{vmatrix}
x & 4 \\
1 & x
\end{vmatrix} + 1 \cdot \begin{vmatrix}
x & 2 \\
1 & 2
\end{vmatrix}
\][/tex]
We know that if the matrix is singular, its determinant is zero. Let's expand and simplify these 2x2 determinants:
1. First determinant:
[tex]\[
\begin{vmatrix}
2 & 4 \\
2 & x
\end{vmatrix} = 2 \cdot x - 4 \cdot 2 = 2x - 8
\][/tex]
2. Second determinant (which includes the coeffiecient 0 and will disappear):
[tex]\[
\text{Not needed as coefficient is 0}
\][/tex]
3. Third determinant:
[tex]\[
\begin{vmatrix}
x & 2 \\
1 & 2
\end{vmatrix} = x \cdot 2 - 2 \cdot 1 = 2x - 2
\][/tex]
So, putting this all together:
[tex]\[
\text{det}(X) = 2 \cdot (2x - 8) + 1 \cdot (2x - 2)
\][/tex]
Simplify this expression to:
[tex]\[
\text{det}(X) = 4x - 16 + 2x - 2 = 6x - 18
\][/tex]
For the matrix to be singular, we set the determinant equal to zero:
[tex]\[
6x - 18 = 0
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
6x - 18 = 0 \implies 6x = 18 \implies x = 3
\][/tex]
Therefore, the value of [tex]\( x \)[/tex] that makes the matrix [tex]\( X \)[/tex] singular is:
[tex]\( \boxed{3} \)[/tex]
