To solve the given determinant equation for [tex]\( x \)[/tex]:
[tex]\[
\left|\begin{array}{ccc}
7 & 2 & -1 \\
3x & 2 & 2 \\
-3 & 1 & -7
\end{array}\right| = 65
\][/tex]
we need to follow these steps:
1. Calculate the Determinant of the Matrix:
We start by calculating the determinant of the [tex]\( 3 \times 3 \)[/tex] matrix:
[tex]\[
A = \begin{vmatrix}
7 & 2 & -1 \\
3x & 2 & 2 \\
-3 & 1 & -7
\end{vmatrix}
\][/tex]
Using the rule for finding the determinant of a [tex]\( 3 \times 3 \)[/tex] matrix, we have:
[tex]\[
\text{det}(A) = 7 \begin{vmatrix} 2 & 2 \\ 1 & -7 \end{vmatrix} - 2 \begin{vmatrix} 3x & 2 \\ -3 & -7 \end{vmatrix} - 1 \begin{vmatrix} 3x & 2 \\ -3 & 1 \end{vmatrix}
\][/tex]
Now, calculate the [tex]\( 2 \times 2 \)[/tex] determinants:
[tex]\[
\begin{vmatrix} 2 & 2 \\ 1 & -7 \end{vmatrix} = (2)(-7) - (2)(1) = -14 - 2 = -16
\][/tex]
[tex]\[
\begin{vmatrix} 3x & 2 \\ -3 & -7 \end{vmatrix} = (3x)(-7) - (2)(-3) = -21x + 6
\][/tex]
[tex]\[
\begin{vmatrix} 3x & 2 \\ -3 & 1 \end{vmatrix} = (3x)(1) - (2)(-3) = 3x + 6
\][/tex]
Substitute these back into the determinant formula:
[tex]\[
\text{det}(A) = 7(-16) - 2(-21x + 6) - 1(3x + 6)
\][/tex]
Simplify:
[tex]\[
\text{det}(A) = -112 + 42x - 12 - 3x - 6
\][/tex]
Combine like terms:
[tex]\[
\text{det}(A) = 39x - 130
\][/tex]
2. Set the Determinant Equal to 65:
We are given that the determinant is equal to 65. So, we set up the equation:
[tex]\[
39x - 130 = 65
\][/tex]
3. Solve for [tex]\( x \)[/tex]:
To isolate [tex]\( x \)[/tex], add 130 to both sides of the equation:
[tex]\[
39x = 195
\][/tex]
Now, divide by 39:
[tex]\[
x = \frac{195}{39} = 5
\][/tex]
Thus, the solution to the equation is:
[tex]\[
x = 5
\][/tex]
So, we have calculated that the determinant of the matrix equals [tex]\( 39x - 130 \)[/tex], and solving for [tex]\( x \)[/tex] when this determinant is set to 65, we find:
[tex]\[
x = 5
\][/tex]
