Problem 48 - Precalculus: Mathematical 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: Recognize the Problem

We need to solve the determinant of a 3x3 matrix set equal to zero. The determinant is given by:\[\left|\begin{array}{ccc} x & 1 & 1 \ 1 & 1 & x \ x & 1 & x \end{array}\right| = 0.\] Our goal is to find the values of \(x\) that satisfy this equation.

Step 2: Calculate the Determinant

The determinant of a 3x3 matrix can be calculated using the rule of Sarrus or cofactor expansion. We begin by expanding along the first row:\[\det = x \cdot \left|\begin{array}{cc} 1 & x \ 1 & x \end{array}\right| - 1 \cdot \left|\begin{array}{cc} 1 & x \ x & x \end{array}\right| + 1 \cdot \left|\begin{array}{cc} 1 & 1 \ x & 1 \end{array}\right|.\]

Step 3: Calculate Cofactor Determinants

Now, calculate each of the 2x2 determinants:\[\left|\begin{array}{cc} 1 & x \ 1 & x \end{array}\right| = 1 \cdot x - x \cdot 1 = 0, \\left|\begin{array}{cc} 1 & x \ x & x \end{array}\right| = 1 \cdot x - x \cdot x = x-x^2 = x(1-x), \\left|\begin{array}{cc} 1 & 1 \ x & 1 \end{array}\right| = 1 \cdot 1 - 1 \cdot x = 1-x.\]

Step 4: Substitute and Simplify

Substitute the cofactor values back into the main determinant equation:\[\det = x \cdot 0 - 1 \cdot (x(1-x)) + 1 \cdot (1-x).\]Simplify:\[\det = 0 - x + x^2 + 1 - x = x^2 - 2x + 1.\]

Step 5: Solve the Quadratic Equation

The determinant simplifies to the quadratic equation:\[x^2 - 2x + 1 = 0.\] Factor the quadratic:\[(x - 1)^2 = 0.\]So, the solution is: \[x = 1.\] The only root, repeated twice, is \(x = 1\).

Key Concepts

3x3 matrix

Understanding a 3x3 matrix is crucial for solving problems involving matrices, such as finding determinants. A 3x3 matrix is simply a grid consisting of three rows and three columns of numbers or variables, written as:\[\left(\begin{array}{ccc}a & b & c \d & e & f \g & h & i \\end{array}\right)\]In our given exercise, the 3x3 matrix is: \[\begin{array}{ccc}x & 1 & 1 \1 & 1 & x \x & 1 & x \\end{array}\]Each element within this matrix can interact to form different calculations, such as determinants. When solving these, you might use specific methods like Sarrus’s rule or cofactor expansion to simplify calculations. Matrices like this are typically used in linear algebra to represent systems of equations or transformations in space. To enhance understanding, it's useful to construct and deconstruct simpler matrices and calculate their determinants using these techniques.

quadratic equation

A quadratic equation is a polynomial equation of degree two. It takes the general form:\[ ax^2 + bx + c = 0 \]Quadratic equations are significant because they frequently appear in different areas of algebra and applied mathematics. In our solution, the quadratic equation derived from the determinant of a 3x3 matrix was:\[ x^2 - 2x + 1 = 0 \]To solve this, we often use methods like factoring, completing the square, or the quadratic formula. In this specific problem, the quadratic was easy to factor:\[ (x - 1)^2 = 0 \]This tells us that the solution, or root, of the equation is \( x = 1 \). Understanding how to work with quadratic equations is a foundational skill in algebra. They can always be solved, because even when factoring is difficult, other methods like the quadratic formula (\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)) can provide the solution.

cofactor expansion

Cofactor expansion is a vital method used to calculate the determinant of a matrix, especially when dealing with a 3x3 matrix or larger. It involves breaking down a larger matrix into smaller parts.For a 3x3 matrix, as seen in this problem:\[det = x \cdot \left|\begin{array}{cc} 1 & x \1 & x \end{array}\right| - 1 \cdot \left|\begin{array}{cc} 1 & x \x & x \end{array}\right| + 1 \cdot \left|\begin{array}{cc} 1 & 1 \x & 1 \end{array}\right|\]This process expands the original matrix into several, simpler 2x2 matrices (called minors). Subsequently, the determinant of each minor is computed. These determinants are then multiplied by their respective cofactors and summed.The cofactor is found by assigning a sign (- or +) depending on the position of the element in its row and column, often decided by an alternating pattern starting with + for the top-left. In our example, the expansions involved: - A minor: \( \left|\begin{array}{cc} 1 & x \1 & x \end{array}\right| \)- Another minor: \( \left|\begin{array}{cc} 1 & x \x & x \end{array}\right| \)- A final minor: \( \left|\begin{array}{cc} 1 & 1 \x & 1 \end{array}\right| \)Mastering cofactor expansion enables one to find determinants of complex matrices efficiently, which is fundamental in solving systems of linear equations, inverting matrices, and determining matrix properties such as rank.