Problem 52 - 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: Determine the Determinant Expression

To solve a matrix determinant set to zero, we first write the expression for the determinant of the given matrix.\[\text{det}(A) = \begin{vmatrix} x & 1 & 1 \ 1 & 1 & x \ x & 1 & x \end{vmatrix}\]

Step 2: Apply the Cofactor Expansion

We apply cofactor expansion along the first row to find the determinant.\[= x \cdot \begin{vmatrix} 1 & x \ 1 & x \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & x \ x & x \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 1 \ x & 1 \end{vmatrix}\]

Step 3: Calculate Each 2x2 Determinant

Calculate each 2x2 determinant:1. \(\begin{vmatrix} 1 & x \ 1 & x \end{vmatrix} = 1 \cdot x - 1 \cdot x = 0\)2. \(\begin{vmatrix} 1 & x \ x & x \end{vmatrix} = 1 \cdot x - x \cdot x = x - x^2\)3. \(\begin{vmatrix} 1 & 1 \ x & 1 \end{vmatrix} = 1 \cdot 1 - 1 \cdot x = 1 - x\)

Step 4: Substitute into Cofactor Expansion

Substitute the results from the minor determinants back into the cofactor expansion expression:\[x \cdot 0 - 1 \cdot (x - x^2) + 1 \cdot (1 - x) = 0\]

Step 5: Simplify the Expression

Simplify the cofactors expression:\[0 - (x - x^2) + (1 - x) = 0\]This simplifies to:\[-x + x^2 + 1 - x = 0\]Combining like terms gives:\[x^2 - 2x + 1 = 0\]

Step 6: Solve the Quadratic Equation

Recognize the expression as a perfect square trinomial:\[(x - 1)^2 = 0\]Taking the square root of both sides, we find:\[x - 1 = 0\]Thus, \(x = 1\).

Key Concepts

Solving Quadratic Equations

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents an unknown variable. Solving these equations means finding the value(s) of \(x\) that make the equation true. There are several methods to solve quadratic equations, including:

• Factoring: This method involves expressing the quadratic in the form \((px + q)(rx + s) = 0\) and finding the values of \(x\) for which each factor equals zero.
• Using the Quadratic Formula: The formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) can solve any quadratic equation.
• Completing the Square: This involves rewriting the equation so that one side becomes a perfect square trinomial.

In this exercise, the quadratic \(x^2 - 2x + 1 = 0\) was solved using the fact that it is a perfect square, simplifying the process and yielding \(x = 1\). Understanding these methods and recognizing special forms, like a perfect square trinomial, can enhance problem-solving skills.

Cofactor Expansion

Cofactor expansion is a technique used to calculate the determinant of a matrix, especially useful for matrices larger than 2x2. The determinant of a matrix can be expanded by minors and cofactors along any row or column, but strategically choosing the row or column with the most zeros can make calculations easier.For a matrix \(A\), the determinant is given by selecting a row or column, then:

• Taking each element: Multiply each element by the determinant of its minor (the determinant of the matrix formed without the element's row and column).
• Considering sign: Each minor determinant is multiplied by \((-1)^{i+j}\), where \(i\) and \(j\) represent the element's row and column indices.
• Summing the products: Add or subtract these values as per their signs to find the total determinant.

In this exercise, cofactor expansion was used on the first row because it efficiently helped solve for \(x\) by creating simpler sub-determinants, ultimately simplifying the matrix determinant equation into a manageable quadratic form.

Perfect Square Trinomial

A perfect square trinomial is a quadratic expression that can be expressed as the square of a binomial. It takes the form \( (ax + b)^2 \), which expands to \( a^2x^2 + 2abx + b^2 \). Identifying perfect square trinomials is useful as they allow quick factoring and simplification of quadratic equations.In this exercise, the expression \(x^2 - 2x + 1 = 0\) was identified as a perfect square trinomial. This recognition is based on the pattern:

• The coefficient of \(x^2\) is a perfect square, \(1\), which is \((1)^2\).
• The constant term, \(1\), is also a perfect square \((1)^2\).
• The middle term, \(-2x\), equals \(2 \times 1 \times x\), verifying the standard form.

Recognizing these patterns allows us to rewrite the equation as \((x - 1)^2 = 0\). Consequently, solving it becomes straightforward: \(x - 1 = 0\), leading to \(x = 1\). Identifying perfect square trinomials is an efficient way to handle certain quadratic equations.