To solve the determinant of the given 3x3 matrix
[tex]\[
\left|\begin{array}{ccc}
3 & -2 & 1 \\
3 & 1 & 5 \\
3 & 4 & 5
\end{array}\right|,
\][/tex]
we use the method of cofactor expansion along the first row. The determinant of a 3x3 matrix
[tex]\[
\left|\begin{array}{ccc}
a & b & c \\
d & e & f \\
g & h & i
\end{array}\right|
\][/tex]
is calculated as follows:
[tex]\[
a \left|\begin{array}{cc}
e & f \\
h & i
\end{array}\right|
- b \left|\begin{array}{cc}
d & f \\
g & i
\end{array}\right|
+ c \left|\begin{array}{cc}
d & e \\
g & h
\end{array}\right|.
\][/tex]
In our matrix:
[tex]\[
\left|\begin{array}{ccc}
3 & -2 & 1 \\
3 & 1 & 5 \\
3 & 4 & 5
\end{array}\right|,
\][/tex]
we identify [tex]\( a = 3 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = 1 \)[/tex], and the corresponding 2x2 submatrices:
1. For [tex]\( a = 3 \)[/tex]:
[tex]\[
\left|\begin{array}{cc}
1 & 5 \\
4 & 5
\end{array}\right|.
\][/tex]
The determinant of this 2x2 matrix is calculated as
[tex]\[
\left( (1 \cdot 5) - (5 \cdot 4) \right) = 5 - 20 = -15.
\][/tex]
So, the term is
[tex]\[
3 \cdot (-15) = -45.
\][/tex]
2. For [tex]\( b = -2 \)[/tex]:
[tex]\[
\left|\begin{array}{cc}
3 & 5 \\
3 & 5
\end{array}\right|.
\][/tex]
The determinant of this 2x2 matrix is calculated as
[tex]\[
\left( (3 \cdot 5) - (5 \cdot 3) \right) = 15 - 15 = 0.
\][/tex]
So, the term is
[tex]\[
-2 \cdot 0 = 0.
\][/tex]
3. For [tex]\( c = 1 \)[/tex]:
[tex]\[
\left|\begin{array}{cc}
3 & 1 \\
3 & 4
\end{array}\right|.
\][/tex]
The determinant of this 2x2 matrix is calculated as
[tex]\[
\left( (3 \cdot 4) - (1 \cdot 3) \right) = 12 - 3 = 9.
\][/tex]
So, the term is
[tex]\[
1 \cdot 9 = 9.
\][/tex]
Adding up these contributions, we have:
[tex]\[
-45 + 0 + 9 = -36.
\][/tex]
Thus, the determinant of the matrix is:
[tex]\[
\left|\begin{array}{ccc}
3 & -2 & 1 \\
3 & 1 & 5 \\
3 & 4 & 5
\end{array}\right| = -36.
\][/tex]
