Answer :

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]