Problem 64 - Precalculus: Mathematics for Calculus

Question

Solve for \(x\) $$\left|\begin{array}{lll} x & 1 & 1 \\ 1 & 1 & x \\ x & 1 & x \end{array}\right|=0$$

Step-by-Step Solution

Answer

The solution is \( x = 1 \).

Step 1: Identify the determinant of the matrix

Given the matrix as: \[\begin{bmatrix} x & 1 & 1 \1 & 1 & x \x & 1 & x \end{bmatrix}\]we must first compute its determinant.

Step 2: Apply the formula for the determinant of a 3x3 matrix

The determinant of a 3x3 matrix \( A = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} \) is calculated using the formula:\[\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)\]

Step 3: Substitute the matrix elements into the determinant formula

Substitute the values from the given matrix into the formula:\[\det(A) = x(1 \cdot x - x \cdot 1) - 1(1 \cdot x - x \cdot x) + 1(1 \cdot 1 - 1 \cdot x)\]

Step 4: Simplify the determinant expression

Simplifying step-by-step, the expression becomes:- First term: \(x(x - x) = x \cdot 0 = 0\)- Second term: \(-1(x - x^2) = -1(x-x^2) = x^2 - x \)- Third term: \(+1(1 - x) = 1 - x\)Combine these:\[0 + x^2 - x + 1 - x = x^2 - 2x + 1\]

Step 5: Set the determinant to zero and solve

The determinant is given as zero in the problem statement:\[x^2 - 2x + 1 = 0\]This is a quadratic equation that can be solved for \(x\).

Step 6: Solve the quadratic equation

The equation \(x^2 - 2x + 1 = 0\) can be recognized as a perfect square:\[(x-1)^2 = 0\]Thus, solve for \( x \):\[x - 1 = 0 \x = 1\]

Step 7: Confirm the solution

Check whether substituting \( x = 1 \) results in the determinant being zero. When \( x = 1 \), each term of the determinant calculation becomes zero, confirming the correctness of the solution.

Key Concepts

3x3 matrix

A 3x3 matrix is a square matrix that consists of three rows and three columns. Each element in this matrix is typically represented by a variable or a constant arranged in a structured format. In this exercise, the matrix is:\[\begin{bmatrix}x & 1 & 1 \1 & 1 & x \x & 1 & x\end{bmatrix}\]To find solutions linked to 3x3 matrices, like their determinants, we use specific mathematical formulas. Determinants are scalars that provide useful properties of matrices. Calculating the determinant of a 3x3 matrix involves using a specific formula involving the elements of the matrix. This formula helps us explore the matrix's properties.Key aspects when dealing with a 3x3 matrix include:

• Identifying each element in the matrix.
• Applying matrix operations like addition, subtraction, and multiplication.
• Calculating the determinant using the standard formula.

quadratic equations

Quadratic equations appear frequently in mathematics and are characterized by their general form: \ax^2 + bx + c = 0. These are second-degree equations and can have two solutions, one solution, or no real solutions depending on their discriminants.In our exercise, we encountered a quadratic equation formed from the determinant of the 3x3 matrix:\[x^2 - 2x + 1 = 0\]Solving quadratic equations generally involves different methods such as:

• Factoring: Useful when the trinomial can be easily decomposed.
• Completing the square: Converting the equation into a perfect square trinomial.
• Using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Each method provides a clear path to finding the values of \(x\) that satisfy the equation. Recognizing these equations helps solve diverse mathematical challenges involving areas, trajectories, and shapes.

perfect square trinomial

A perfect square trinomial is a quadratic expression that can be expressed as \\((ax + b)^2\). This means the quadratic is derived from the square of a binomial, simplifying the process of solving the equation since it can be directly equated to zero.In the determinant step of this exercise, we simplified the quadratic equation to:\[x^2 - 2x + 1 = 0\]This expression perfectly fits the pattern \\((x-1)^2\), making it a perfect square trinomial. To solve a perfect square trinomial, you simply take the square root of both sides, equate it to zero, and solve for \(x\):

• Recognize the perfect square form: \((x-1)^2 = 0\)
• Set the base of the square to zero: \(x-1=0\)
• Find the solution: \(x=1\)

Understanding perfect square trinomials allows for quick and effective solutions to certain types of quadratic equations, reducing the complexity of algebraic calculations.