Problem 66 - Precalculus: Mathematics 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 \(0\) and \(a + b\).

Step 1: Recall the Formula for the Determinant of a 3x3 Matrix

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

Step 2: Identify Elements in the Given Matrix

For the given matrix:\[\begin{bmatrix}a & b & x-a \x & x+b & x \0 & 1 & 1 \end{bmatrix}\]Identify: \(a = a\), \(b = b\), \(c = x-a\), \(d = x\), \(e = x+b\), \(f = x\), \(g = 0\), \(h = 1\), and \(i = 1\).

Step 3: Plug Elements into the Determinant Formula

Apply the elements to the determinant formula:\[det(A) = a(x \cdot 1 - x \cdot 1) - b(x \cdot 1 - 0 \cdot x) + (x-a)(x \cdot 1 - 0 \cdot (x+b))\]This simplifies to:\[det(A) = 0 - b(x) + (x-a)(x)\]

Step 4: Simplify the Determinant Equation

Simplify further:\[0 - bx + x^2 - ax\]Which results in:\[x^2 - ax - bx = 0\]

Step 5: Factor the Quadratic Equation

Combine like terms:\[x^2 - (a+b)x = 0\]Factor out \(x\):\[x(x - (a+b)) = 0\]

Step 6: Solve for x

Set each factor equal to zero:1. \(x = 0\)2. \(x = a+b\)Thus, the possible solutions for \(x\) are \(x = 0\) or \(x = a+b\).

Key Concepts

Matrix Algebra

Matrix algebra is a cornerstone of linear algebra and involves operations with matrices, such as addition, multiplication, and finding determinants. This exercise involves finding the determinant of a 3x3 matrix. The determinant is a scalar value that can be calculated from a square matrix, providing important properties about the matrix, like whether it is invertible or not. To calculate the determinant of a 3x3 matrix, you use a formula involving the elements of the matrix: \[ det(A) = a(ei-fh) - b(di- fg) + c(dh- eg) \] In this exercise, using the given matrix:\ \[\begin{bmatrix}\ a & b & x-a \ x & x+b & x \ 0 & 1 & 1 \end{bmatrix}\] We identify the elements:\ \(a, b, c, d, e, f, g, h,\) and \(i\). We plug these into the formula, simplify the equation to zero (since the determinant equals zero), and solve. This foundational matrix algebra skill is essential for tasks like solving systems of linear equations or transformations.

Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\). They are called 'quadratic' due to the term \(x^2\). In the exercise given, after calculating the determinant of a matrix, we end up with the equation \(x^2 - (a+b)x = 0\), which is a quadratic equation.- **Structure:** A standard quadratic equation comes in the form of \(ax^2 + bx + c = 0\). In our case: - \(a = 1\) (coefficient of \(x^2\)) - \(b = -(a+b)\) (coefficient of \(x\)) - \(c = 0\) (constant term)- **Solving:** Quadratic equations can often be solved by factoring, using the quadratic formula, or graphically. Here, the equation is set to zero, which simplifies factoring.

Factoring Polynomials

Factoring polynomials is a method used to solve polynomial equations and involves expressing a polynomial as a product of its factors. In our problem, after simplifying the determinant, we had the quadratic equation \(x^2 - (a+b)x = 0\). Factoring simplifies the equation by identifying common terms.- **Steps in Factoring:** 1. Look for common factors: The term \(x\) is common, so factor it out: \[x(x - (a+b)) = 0\] 2. Once factored, set each factor equal to zero. Solutions are the values of \(x\) making each factor zero.- **Solving Factored Forms:** - If \(x(x - (a+b)) = 0\), then: - \(x = 0\) - \(x = a+b\)Factoring is a key skill in algebra and can greatly simplify solving equations, especially when dealing with quadratic polynomials.