Problem 102 - Algebra and Trigonometry

Question

Consider the system $$ \left\\{\begin{array}{r} {x-y+z=-3} \\ {-2 y+z=-6} \\ {-2 x-3 y=-10} \end{array}\right. $$ a. Write the system as a matrix equation in the form \(A X=B\) b. Solve the system using the fact that the inverse of $$ \left[\begin{array}{rrr} {1} & {-1} & {1} \\ {0} & {-2} & {1} \\ {-2} & {-3} & {0} \end{array}\right] \text { is }\left[\begin{array}{rrr} {3} & {-3} & {1} \\ {-2} & {2} & {-1} \\ {-4} & {5} & {-2} \end{array}\right] $$

Step-by-Step Solution

Answer

\(x = -7\), \(y = 4\), and \(z = -2\)

Step 1: Write the system as a matrix equation

In order to write the system as a matrix equation in the form \( A X=B \), identify A as the matrix of coefficients, X as the matrix of variables, and B as the matrix of results. From the given equations, we have\[A = \begin{bmatrix} 1 & -1 & 1 \ 0 & -2 & 1 \ -2 & -3 & 0 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \ z \end{bmatrix}, \quad \text{and} \quad B = \begin{bmatrix} -3 \ -6 \ -10 \end{bmatrix}\]So, in matrix form, \(AX = B\), becomes \[\left[ \begin{array}{ccc} 1 & -1 & 1 \ 0 & -2 & 1 \ -2 & -3 & 0 \end{array} \right] \left[ \begin{array}{c} x \ y \ z \end{array} \right] = \left[ \begin{array}{c} -3 \ -6 \ -10 \end{array} \right]\]

Step 2: Multiply both sides by the inverse of A

We are given the inverse matrix of A, which is \[A^{-1} = \begin{bmatrix} 3 & -3 & 1 \ -2 & 2 & -1 \ -4 & 5 & -2 \end{bmatrix}\]So we multiply both sides of the equation by \(A^{-1}\) to isolate X\[ A^{-1}AX = A^{-1}B \]

Step 3: Simplify and obtain the solution

\(\(A^{-1}AX\) simplifies to X as when a matrix is multiplied by its inverse, it gives the Identity matrix. Hence, the equation becomes\[X = A^{-1}B \]Now, substitute \(A^{-1}B\) with the respective matrix values.\[X = \begin{bmatrix} 3 & -3 & 1 \ -2 & 2 & -1 \ -4 & 5 & -2 \end{bmatrix} \begin{bmatrix} -3 \ -6 \ -10 \end{bmatrix}\]Solving this matrix multiplication gives the matrix X as the solution\[X = \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} -7 \ 4 \ -2 \end{bmatrix}\]Therefore, the solution to the system of equations is \(x = -7\), \(y = 4\), and \(z = -2\).

Key Concepts

Matrix Inverse

The concept of the matrix inverse is crucial in solving matrix equations, especially in linear algebra. An inverse of a matrix \( A \) is another matrix \( A^{-1} \) such that when the two are multiplied, they yield the identity matrix. For a matrix to have an inverse, it must be a square matrix (same number of rows and columns) and its determinant must not be zero.
If \( A \) is a matrix, and \( A^{-1} \) is its inverse, then:

• The operation satisfies: \( A \, A^{-1} = I \) and \( A^{-1} \, A = I \)
• \( I \) is the identity matrix, which looks like \( \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \) for a 3x3 matrix.
• Not every matrix has an inverse. A matrix without an inverse is called singular or non-invertible.

The inverse matrix is used in matrix equations to solve for variables, as shown in the solution process. Multiplying both sides of the matrix equation by the inverse of Matrix \( A \) helps isolate matrix \( X \), the matrix of variables.

Systems of Equations

Systems of equations consist of multiple equations that are solved together. This exercise involved a system of three equations with three variables. These systems can often be represented in matrix form, which makes them easier to solve using linear algebra techniques.

• A system of equations can be expressed as \( AX = B \), where \( A \) is the matrix of coefficients, \( X \) is the column matrix of variables, and \( B \) is the matrix of constants on the right side of the equations.
• The goal is to find the values of the variables that satisfy all equations at once.
• If possible, one effective technique to solve a system of equations is using the matrix inverse, turning \( AX = B \) into \( X = A^{-1}B \).

Understanding systems of equations is essential in various fields such as engineering, physics, computer science, and economics. Matrices provide a structured way to manage and solve these equations.

Matrix Multiplication

Matrix multiplication is key when working with matrix equations. To multiply two matrices together, the number of columns in the first matrix must equal the number of rows in the second matrix. Moreover, each element is calculated as the sum of the products of the elements in the respective rows and columns.

• For matrices \( A \) and \( B \), to multiply AB, sum the products of the elements of each row in \( A \) with the corresponding elements of each column in \( B \).
• The resulting matrix's size will be determined by the number of rows in the first matrix and the number of columns in the second matrix.
• An important property of matrix multiplication is that it is associative but not commutative, meaning \( AB eq BA \) in general.

In the step-by-step solution, matrix multiplication is used to find \( X = A^{-1}B \), by multiplying the inverse matrix with the constants matrix \( B \).

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, matrices, and linear transformations. It is used to solve linear equations and understand vector spaces. In mathematical applications such as computer algorithms, physics simulations, and data analysis, linear algebra is an integral part.

• In this exercise, linear algebra helps transform a complex system of equations into a solvable matrix equation.
• The concepts of matrix inverses and multiplication are fundamental components of linear algebra.
• Understanding the properties of matrices and how to manipulate them is crucial. It helps in the efficient solving of equations, modeling of systems, and analysis of multidimensional data.

Linear algebra provides the tools and techniques required to solve real-world problems efficiently by simplifying and structuring the problem into a mathematical format that can be analysed systematically.