Problem 31 - Algebra and Trigonometry

Question

Use a calculator that can perform matrix operations to solve the system, as in Example 7. \(\left\\{\begin{aligned} x+y-2 z &=3 \\ 2 x+& 5 z=11 \\ 2 x+3 y &=12 \end{aligned}\right.\)

Step-by-Step Solution

Answer

The solution is \(x = 3\), \(y = 1\), \(z = 2\).

Step 1: Write the system in matrix form

Express the system of equations as a matrix equation of the form \(A\mathbf{x} = \mathbf{b}\). Here, \(\mathbf{x}\) is the column matrix of variables \([x, y, z]^T\), \(A\) is the coefficient matrix, and \(\mathbf{b}\) is the constant column matrix. For the given system:\[\begin{bmatrix} 1 & 1 & -2 \ 2 & 0 & 5 \ 2 & 3 & 0 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 3 \ 11 \ 12 \end{bmatrix}\]

Step 2: Find the inverse of the coefficient matrix

Calculate the inverse of matrix \(A\). This step assumes that the matrix is invertible. Use a calculator capable of matrix operations to compute the inverse:\[A^{-1} = \begin{bmatrix} 0 & \frac{3}{5} & -\frac{3}{5} \ 2 & -1 & 0 \ \frac{2}{5} & -\frac{1}{5} & \frac{1}{5} \end{bmatrix}\]

Step 3: Multiply the inverse by the constant matrix

Multiply \(A^{-1}\) by \(\mathbf{b}\) to solve for \(\mathbf{x}\) using the formula \(\mathbf{x} = A^{-1}\mathbf{b}\). Perform the matrix multiplication:\[\begin{bmatrix} 0 & \frac{3}{5} & -\frac{3}{5} \ 2 & -1 & 0 \ \frac{2}{5} & -\frac{1}{5} & \frac{1}{5} \end{bmatrix} \begin{bmatrix} 3 \ 11 \ 12 \end{bmatrix} = \begin{bmatrix} 3 \ 1 \ 2 \end{bmatrix}\]

Step 4: Interpret the result

The resulting column matrix \(\begin{bmatrix} 3 \ 1 \ 2 \end{bmatrix}\) represents the solution to the system of equations, which means \(x = 3\), \(y = 1\), and \(z = 2\).

Key Concepts

Understanding Systems of Equations

A system of equations is a collection of one or more equations involving the same set of variables. In the context of linear algebra, these equations are usually linear, meaning each term is either a constant or the product of a constant with a single variable. The goal is to find the values for the variables that satisfy all equations simultaneously. For example, in the given problem, we have a system with three equations and three variables:

• \(x + y - 2z = 3\)
• \(2x + 5z = 11\)
• \(2x + 3y = 12\)

These kinds of problems can be represented and solved using matrices, which simplifies the process, especially when dealing with more complex or larger systems.

Inverse Matrix in Solving Equations

The inverse of a matrix is akin to the reciprocal of a number. If we have a matrix \(A\), its inverse is denoted \(A^{-1}\). When multiplied, they yield the identity matrix, similar to how a number multiplied by its reciprocal equals one. However, not all matrices have an inverse. A matrix must be square (having the same number of rows and columns) and have a non-zero determinant to be invertible. In the context of solving systems of equations, finding the inverse is critical. By expressing the system in the form \(A\mathbf{x} = \mathbf{b}\), we can solve for \(\mathbf{x}\) by computing \(\mathbf{x} = A^{-1}\mathbf{b}\). This step requires that \(A\) has an inverse; otherwise, the system may not have a unique solution.

The Role of Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra. It involves combining rows from one matrix and columns from another to produce a new matrix. When solving a system of equations using matrices, matrix multiplication plays a major role in finding the solution.After obtaining the inverse matrix \(A^{-1}\), the next step is to multiply it by the constant matrix \(\mathbf{b}\). The formula \(\mathbf{x} = A^{-1}\mathbf{b}\) uses this principle. For our example,

• Matrix \(A^{-1}\):
• \(\begin{bmatrix} 0 & \frac{3}{5} & -\frac{3}{5} \ 2 & -1 & 0 \ \frac{2}{5} & -\frac{1}{5} & \frac{1}{5} \end{bmatrix}\)
• Constant matrix \(\mathbf{b}\):
• \(\begin{bmatrix} 3 \ 11 \ 12 \end{bmatrix}\)

Multiplying them provides the solution vector \(\mathbf{x}\), which contains the values of \(x\), \(y\), and \(z\) that satisfy all equations in the system.

The Coefficient Matrix and Its Importance

The coefficient matrix is derived from the system of equations. It is composed of the coefficients of the variables in the equations. This matrix is crucial because it encapsulates all the information about the linear relationships between the variables. For the given problem, the coefficient matrix \(A\) is:\[\begin{bmatrix}1 & 1 & -2 \2 & 0 & 5 \2 & 3 & 0\end{bmatrix}\]This matrix acts as the bridge between the variables and their solutions. Inverting the coefficient matrix \(A\) is the key step in solving the system using the formula \(\mathbf{x} = A^{-1}\mathbf{b}\). If the matrix cannot be inverted, it implies that the system is either dependent or inconsistent, meaning not all equations are linearly independent or there is no solution that satisfies all equations.