Certainly! Let's solve the problem step-by-step.
We are given the inequality involving the determinant of a 3x3 matrix and must solve it for [tex]\( x \)[/tex]:
[tex]\[
\left|\begin{array}{ccc}
3 & 2 & 1 \\
x & 2 & 4 \\
x-1 & 4 & 7
\end{array}\right|
< \frac{x-7}{3}
\][/tex]
### Step 1: Calculate the Determinant of the Matrix
To find the determinant of the given matrix, let's use the standard method of calculating determinants for 3x3 matrices. The matrix we have is:
[tex]\[
\begin{vmatrix}
3 & 2 & 1 \\
x & 2 & 4 \\
x-1 & 4 & 7
\end{vmatrix}
\][/tex]
The formula for the determinant is:
[tex]\[
\text{det}(A) = a(ei − fh) − b(di − fg) + c(dh − eg)
\][/tex]
Here:
- [tex]\(a = 3\)[/tex]
- [tex]\(b = 2\)[/tex]
- [tex]\(c = 1\)[/tex]
- [tex]\(d = x\)[/tex]
- [tex]\(e = 2\)[/tex]
- [tex]\(f = 4\)[/tex]
- [tex]\(g = x-1\)[/tex]
- [tex]\(h = 4\)[/tex]
- [tex]\(i = 7\)[/tex]
Let's compute each term:
- [tex]\(ei - fh = 2 \cdot 7 - 4 \cdot 4 = 14 - 16 = -2\)[/tex]
- [tex]\(di - fg = x \cdot 7 - 4 \cdot (x - 1) = 7x - 4x + 4 = 3x + 4\)[/tex]
- [tex]\(dh - eg = x \cdot 4 - 2 \cdot (x - 1) = 4x - 2x + 2 = 2x + 2\)[/tex]
Substitute these back into the determinant formula:
[tex]\[
\text{det}(A) = 3(-2) - 2(3x + 4) + 1(2x + 2)
\][/tex]
Simplifying this:
[tex]\[
\text{det}(A) = -6 - 6x - 8 + 2x + 2 = -6x - 12
\][/tex]
Thus:
[tex]\[
\text{det}(A) = -4x - 12
\][/tex]
### Step 2: Setting Up the Inequality
Now we substitute the determinant value into the inequality:
[tex]\[
-4x - 12 < \frac{x-7}{3}
\][/tex]
### Step 3: Solving the Inequality
First, we clear the fraction by multiplying both sides by 3:
[tex]\[
3(-4x - 12) < x - 7
\][/tex]
Simplify:
[tex]\[
-12x - 36 < x - 7
\][/tex]
Now, move all terms involving [tex]\( x \)[/tex] to one side and constants to the other:
[tex]\[
-12x - x < -7 + 36
\][/tex]
This gives:
[tex]\[
-13x < 29
\][/tex]
Divide both sides by [tex]\(-13\)[/tex] (remember to reverse the inequality sign when dividing by a negative number):
[tex]\[
x > \frac{29}{13}
\][/tex]
### Step 4: Expressing the Solution Set
Finally, the solution in interval notation is:
[tex]\[
x > \frac{29}{13}
\][/tex]
Therefore, the solution set is:
[tex]\[
\left(\frac{29}{13}, \infty\right)
\][/tex]
So, the inequality [tex]\(\left|\begin{array}{ccc}3 & 2 & 1 \\ x & 2 & 4 \\ x-1 & 4 & 7\end{array}\right|<\frac{x-7}{3}\)[/tex] holds true for:
[tex]\[
x > \frac{29}{13}
\][/tex]
