Problem 54 - Algebra and Trigonometry

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

\(x = a + \sqrt{a^2 - ab}\) or \(x = a - \sqrt{a^2 - ab}\).

Step 1: Identify the determinant structure

The problem involves a 3x3 determinant. We need to solve for \( x \) where this determinant equals zero. The matrix is:\[\begin{bmatrix}a & b & x-a \x & x+b & x \0 & 1 & 1\end{bmatrix}\]

Step 2: Apply cofactor expansion along the third row

We choose the third row for expansion because it contains a zero element, making the calculation simpler:The formula for the determinant using the third row is:\[0 \cdot \text{det}(M_{11}) - 1 \cdot \text{det}(M_{12}) + 1 \cdot \text{det}(M_{13}) = 0\]We only need to calculate the determinants of the 2x2 submatrices corresponding to the second and third elements of the third row.

Step 3: Calculate 2x2 determinants

The necessary cofactors are:1. \( \text{det}(M_{12}) = \begin{vmatrix}a & x-a \x & x\end{vmatrix} = a(x) - (x-a)x = ax - x^2 + ax \)2. \( \text{det}(M_{13}) = \begin{vmatrix}a & b \x & x+b\end{vmatrix} = a(x+b) - b(x) = ax + ab - bx \)These simplify to:1. \( ext{det}(M_{12}) = 2ax - x^2 \)2. \( ext{det}(M_{13}) = ab \) because \( ax - bx \) cancels out to zero.

Step 4: Solve the determinant equation

Substitute the determinants back into the equation from Step 2:\(-1(2ax - x^2) + 1(ab) = 0\)Simplify to get:\[x^2 - 2ax + ab = 0\]This is a quadratic equation in standard form \(x^2 + bx + c = 0\).

Step 5: Solve the quadratic equation

Use the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For the equation \(x^2 - 2ax + ab = 0\), the coefficients are \(a = 1\), \(b = -2a\), \(c = ab\).Plug into the quadratic formula:\[x = \frac{-(-2a) \pm \sqrt{(-2a)^2 - 4 \cdot 1 \cdot ab}}{2 \cdot 1}\]\[x = \frac{2a \pm \sqrt{4a^2 - 4ab}}{2}\]\[x = \frac{2a \pm 2\sqrt{a^2 - ab}}{2}\]\[x = a \pm \sqrt{a^2 - ab}\]

Step 6: Final expression for x

Thus, the solutions for \(x\) are:\[x = a + \sqrt{a^2 - ab} \quad \text{and} \quad x = a - \sqrt{a^2 - ab}\]

Key Concepts

Understanding Quadratic Equations

Quadratic equations are polynomials of degree two. They take on the general form: \( ax^2 + bx + c = 0 \). In this equation, \( a \), \( b \), and \( c \) are constants where \( a eq 0 \). A quadratic equation can have two solutions, which are known as the roots of the equation.
To solve a quadratic, we often use the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula helps in finding the two potential solutions (roots) derived from the discriminant, \( b^2 - 4ac \).
Depending on the value of the discriminant:

• If the discriminant is positive, there are two distinct real solutions.
• If it is zero, both solutions are the same, indicating one real double root.
• If negative, the solutions involve complex numbers.

In our exercise, the quadratic equation derived was \( x^2 - 2ax + ab = 0 \). Using the quadratic formula, the solutions \( x = a + \sqrt{a^2 - ab} \) and \( x = a - \sqrt{a^2 - ab} \) were found. This highlights the essence of quadratic equations in finding unknown values.

Applying Matrix Operations

Matrix operations involve several key processes including addition, subtraction, scalar multiplication, and finding determinants. A matrix determinant is a special number that can be calculated from its elements. Determinants help us understand properties like invertibility and solution characteristics of linear equations. For a 3x3 matrix, the determinant is calculated using cofactor expansion, which involves the minors and cofactors of its elements.
In our given exercise, we dealt with a 3x3 matrix:

• \[\begin{bmatrix} a & b & x-a \ x & x+b & x \ 0 & 1 & 1 \end{bmatrix}\]

To solve for \( x \) when the determinant equals zero, matrix operations were used. This involved expanding across the row with a zero for efficient calculation. Evaluating matrix determinants helps solve for unknown variables in a system, such as \( x \) in our problem.

Cofactor Expansion in Determinants

Cofactor expansion, also known as Laplace expansion, is a method used to calculate the determinant of a matrix. This technique involves breaking down a larger matrix into smaller components for easier calculation.
Specifically, you choose a row or column, multiply each element by its cofactor, and sum the results. The cofactor is the determinant of a smaller matrix formed by removing the row and column of a given element, often alternating signs.
In our exercise, we performed cofactor expansion on the third row of our 3x3 matrix, which included a zero:

• \( 0 \cdot \text{det}(M_{11}) - 1 \cdot \text{det}(M_{12}) + 1 \cdot \text{det}(M_{13}) = 0 \)

The use of a row with zero simplifies the process, reducing computations. The final 2x2 matrices' determinants were used to derive the solutions. Understanding cofactor expansion is essential for handling complex matrices and solving systems that involve determinants.