Let's tackle question 26:
Given matrix [tex]\( A \)[/tex] is:
[tex]\[
A = \begin{pmatrix} 3 & x-1 \\ 1 & x \end{pmatrix}
\][/tex]
We need to find the values of [tex]\( x \)[/tex] such that [tex]\(\operatorname{det}(A) = 0\)[/tex].
Step-by-Step Solution:
1. Write the Determinant Formula:
The determinant of a [tex]\( 2 \times 2 \)[/tex] matrix
[tex]\[
\begin{pmatrix} a & b \\ c & d \end{pmatrix}
\][/tex]
is calculated using the formula:
[tex]\[
\text{det} = ad - bc
\][/tex]
2. Apply the Formula to Our Matrix:
Given our matrix [tex]\( A = \begin{pmatrix} 3 & x-1 \\ 1 & x \end{pmatrix} \)[/tex], we identify:
[tex]\[
a = 3, \quad b = x-1, \quad c = 1, \quad d = x
\][/tex]
The determinant of [tex]\( A \)[/tex] is therefore:
[tex]\[
\text{det}(A) = (3)(x) - (1)(x-1)
\][/tex]
3. Simplify the Determinant Expression:
[tex]\[
\text{det}(A) = 3x - (x - 1) = 3x - x + 1 = 2x + 1
\][/tex]
4. Set the Determinant to Zero:
[tex]\[
2x + 1 = 0
\][/tex]
5. Solve for [tex]\( x \)[/tex]:
[tex]\[
2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}
\][/tex]
Thus, the value of [tex]\( x \)[/tex] that makes the determinant of matrix [tex]\( A \)[/tex] equal to zero is:
[tex]\[
x = -\frac{1}{2}
\][/tex]
