Problem 56 - Algebra 2

Question

Solve each matrix equation or system of equations by using inverse matrices. $$ \left[\begin{array}{rr}{3} & {6} \\ {2} & {-1}\end{array}\right] \cdot\left[\begin{array}{l}{a} \\\ {b}\end{array}\right]=\left[\begin{array}{c}{-3} \\ {18}\end{array}\right] $$

Step-by-Step Solution

Answer

The solution is \( a = 2.2 \) and \( b = -4.2 \).

Step 1: Identify the Matrices

Recognize the given matrices in the equation. The matrix on the left is \( A = \begin{bmatrix} 3 & 6 \ 2 & -1 \end{bmatrix} \). The unknown matrix is \( X = \begin{bmatrix} a \ b \end{bmatrix} \), and the result matrix is \( B = \begin{bmatrix} -3 \ 18 \end{bmatrix} \). The equation is \( A \cdot X = B \).

Step 2: Find the Inverse of Matrix A

To solve for \( X \), calculate the inverse of matrix \( A \), denoted \( A^{-1} \). For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is \( \dfrac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \) if \( ad - bc eq 0 \). For matrix \( A \), \( ad - bc = 3(-1) - 2(6) = -3 - 12 = -15 \). Thus, the inverse is \( \dfrac{1}{-15} \begin{bmatrix} -1 & -6 \ -2 & 3 \end{bmatrix} = \begin{bmatrix} \frac{1}{15} & \frac{2}{5} \ \frac{1}{5} & -\frac{1}{5} \end{bmatrix} \).

Step 3: Multiply by the Inverse

Multiply both sides of the equation \( A \cdot X = B \) by \( A^{-1} \) to isolate \( X \). The equation becomes \( A^{-1} \cdot A \cdot X = A^{-1} \cdot B \). Since \( A^{-1} \cdot A = I \) (the identity matrix), \( X = A^{-1} \cdot B \).

Step 4: Calculate the Product

Compute \( X = A^{-1} \cdot B \) using the inverse matrix:\[ X = \begin{bmatrix} \frac{1}{15} & \frac{2}{5} \ \frac{1}{5} & -\frac{1}{5} \end{bmatrix} \cdot \begin{bmatrix} -3 \ 18 \end{bmatrix} \ = \begin{bmatrix} \frac{1}{15}(-3) + \frac{2}{5}(18) \ \frac{1}{5}(-3) + -\frac{1}{5}(18) \end{bmatrix} = \begin{bmatrix} \frac{-3}{15} + \frac{36}{15} \ \frac{-3}{5} + \frac{-18}{5} \end{bmatrix} = \begin{bmatrix} \frac{33}{15} \ \frac{-21}{5} \end{bmatrix} = \begin{bmatrix} \frac{11}{5} \ -\frac{21}{5} \end{bmatrix} \].

Step 5: Interpret the Solution

From the calculations we find that \( a = \frac{11}{5} \) and \( b = -\frac{21}{5} \). Thus, the solution to the system of equations is \( a = 2.2 \) and \( b = -4.2 \).

Key Concepts

matrix equations

Matrix equations provide a great way to solve systems that involve multiple variables. By using matrices, we can represent these complex systems with a simple equation, usually written in the form \( A \cdot X = B \). Here, \( A \) is a known matrix that contains coefficients, \( X \) is an unknown matrix typically containing the variables like \( a \) and \( b \), and \( B \) is a known result matrix that contains the outcomes.

• This format allows us to handle multiple linear equations as a single mathematical entity.
• It simplifies the process of solving for unknowns by compressing detailed arithmetic into high-level manipulations.

To solve such equations, the primary goal is to isolate \( X \). Once \( X \) is isolated, it becomes straightforward to identify the values of unknowns through calculations, precisely as demonstrated in the exercise above. However, isolating \( X \) involves leveraging matrix inverses, which is a cornerstone concept in solving matrix equations. Understanding these basics builds a robust foundation for tackling more intricate problems involving matrices.

matrix inversion

Matrix inversion is a crucial tool for solving matrix equations. The inverse of a matrix \( A \) is denoted as \( A^{-1} \), and it has the special property such that \( A \cdot A^{-1} = I \), where \( I \) is the identity matrix. The identity matrix plays a role similar to the number 1 in regular arithmetic, meaning any matrix multiplied by the identity matrix remains unchanged.

• Not all matrices have inverses; a matrix must be square (having the same number of rows and columns) and its determinant must be non-zero.
• For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is \( \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \) provided \( ad-bc eq 0 \).

In the given exercise, finding the inverse of matrix \( A \) allows us to multiply both sides of the equation \( A \cdot X = B \) by \( A^{-1} \), effectively isolating \( X \), and thus solving for the variables \( a \) and \( b \). This powerful technique often simplifies solving large systems of equations, albeit requiring careful calculation to ensure accuracy in matrix inversion.

systems of equations

Systems of equations are sets of multiple equations that share the same set of unknowns. Solving these systems is about finding the values of variables that satisfy all equations involved. Using matrices provides a convenient way to handle systems of linear equations by transforming them into a matrix equation.

• This transformation is beneficial for clarity and computational efficiency.
• By applying matrix inversion, we simplify the process of solving systems with multiple variables.

In the exercise, the system of equations can be represented through matrix calculation leading to the solution \( a = 2.2 \) and \( b = -4.2 \). Each entry in the solutions directly corresponds to satisfying the original equations. This shows the effectiveness of using matrices for solving complex algebraic equations. Understanding how to manipulate these matrix-based equations opens pathways to tackle diverse problem domains, from engineering to computer science.