Chapter 10: Problem 64
Solve for \(x\) $$\left|\begin{array}{lll} x & 1 & 1 \\ 1 & 1 & x \\ x & 1 & x \end{array}\right|=0$$
Short Answer
Expert verified
The solution is \( x = 1 \).
Step by step solution
01
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.
02
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:\[\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)\]
03
Substitute the matrix elements into the determinant formula
Substitute the values from the given matrix into the formula:\[\det(A) = x(1 \cdot x - x \cdot 1) - 1(1 \cdot x - x \cdot x) + 1(1 \cdot 1 - 1 \cdot x)\]
04
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:\[0 + x^2 - x + 1 - x = x^2 - 2x + 1\]
05
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 \(x\).
06
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 \( x \):\[x - 1 = 0 \x = 1\]
07
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
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: \(x=1\)