Problem 50 - Precalculus: Mathematical for Calculus

Question

Solve for \(x\) $$\left|\begin{array}{ccc} a & b & x-a \\ x & x+b & x \\ 0 & 1 & 1 \end{array}\right|=0$$

Step-by-Step Solution

Answer

The solutions for \(x\) are \(x = 0\) or \(x = a+b\).

Step 1: Identify the Matrix

The given matrix is a 3x3 matrix with rows: \([a, b, x-a]\), \([x, x+b, x]\), and \([0, 1, 1]\). We need to find its determinant.

Step 2: Calculate the Determinant

The determinant of a 3x3 matrix \( \begin{pmatrix} a & b & c \ d & e & f \ g & h & i \end{pmatrix} \) is calculated as \( a(ei - fh) - b(di - fg) + c(dh - eg) \). Apply this to the given matrix: \[|A| = a((x)(1) - (x)(1)) - b((x)(1) - (0)(x)) + (x-a)((x)(1) - (x+b)(0))\]This simplifies to:\[|A| = a(0) - b(x) + (x-a)x\]Thus, \[|A| = -bx + x^2 - ax\]This simplifies to \[|A| = x^2 - (a+b)x\].

Step 3: Set Determinant to Zero

To find the value of \(x\), solve \( x^2 - (a+b)x = 0 \) for \(x\).

Step 4: Solve the Quadratic Equation

The equation \( x^2 - (a+b)x = 0 \) can be factored as \( x(x - (a+b)) = 0 \). Therefore, the solutions for \(x\) are \(x = 0\) or \(x = a+b\).

Key Concepts

Determinant Calculation

Calculating the determinant of a matrix, especially a 3x3 matrix, is a foundational skill in matrix algebra. The determinant provides vital information about the matrix's properties, such as whether it's invertible. For a 3x3 matrix \[ A = \begin{pmatrix} a & b & c \ d & e & f \ g & h & i \end{pmatrix} \],the determinant \(|A|\) is calculated using the formula:\[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \].Breaking this down:

• Products: Calculate the products within each group: \(ei, fh, di, fg, dh,\) and \(eg\).
• Apply the sign rules: Each product group is subject to specific signs (positive or negative) based on its original position in the matrix.
• Sum and subtract: Finally, you sum up with subtraction to get the determinant.

In practice, knowing how to expand along rows or columns efficiently, especially by picking ones with zeros, can simplify the calculation.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation in a single variable, commonly written as \( ax^2 + bx + c = 0 \). Solving these equations is crucial in various branches of mathematics and engineering. Here are the methods you can use:

• Factoring: One common method is to factor the quadratic into two binomials, which are then set to zero to find the values of \(x\).
• Quadratic Formula: When the equation is tough to factor, use the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where the discriminant \(b^2 - 4ac\) tells you the nature of the roots.
• Completing the Square: You can also transform the equation into a perfect square trinomial, making it straightforward to solve for \(x\).

In the context of our problem, solving \(x^2 - (a+b)x = 0\) by factoring involves setting the equation as \(x(x - (a+b)) = 0\), giving solutions \(x = 0\) or \(x = a+b\). This shows the power of recognizing simple patterns.

Matrix Algebra

Matrix algebra is the study of matrices and the various operations that can be performed on them. It is a key tool for solving systems of linear equations and performing complex calculations in different fields. Some of the fundamental operations in matrix algebra include:

• Addition and Subtraction: Combine matrices of the same dimensions by adding or subtracting corresponding elements.
• Multiplication: Multiply matrices using dot products of rows and columns, essential for transformations and composition of linear functions.
• Row Reduction: Simplify matrices to row-echelon form to solve systems of equations using techniques like Gaussian elimination.

Matrix algebra also extends to concepts like eigenvalues and eigenvectors, useful for more advanced mathematical modeling. Understanding how matrices represent systems and transformations is crucial for applying these concepts to real-world problems.