Problem 30 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

(a) write each system of equations as a matrix equation and (b) solve the system of equations by using the inverse of the coefficient matrix. \(\begin{aligned} 2 x_{1}+x_{2}+x_{3}=b_{1} \\ x_{1}-3 x_{2}+4 x_{3}=b_{2} \\\\-x_{1}+x_{3}=b_{3} \\ \text { where } & \text { (i) } b_{1}=1, b_{2}=4, b_{3}=-3 \\ \text { and } & \text { (ii) } b_{1}=2, b_{2}=-5, b_{3}=0 \end{aligned}\)

Step-by-Step Solution

Answer

For the given system of equations, the matrix equation AX = B is: \(AX = \begin{pmatrix} 2 & 1 & 1 \\ 1 & -3 & 4 \\ -1 & 0 & 1 \end{pmatrix}\begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{pmatrix}=\begin{pmatrix} b_{1} \\ b_{2} \\ b_{3} \\ \end{pmatrix}\) The inverse of matrix A is: \(A^{-1}=\begin{pmatrix} -\frac{3}{14}& -\frac{3}{7} & -\frac{3}{14} \\ \frac{1}{7} & \frac{3}{14}& \frac{1}{14}\\ \frac{3}{14} & \frac{1}{7} & \frac{1}{14} \end{pmatrix}\) (i) For \(b_1=1, b_2=4, b_3=-3\), The solution is: \(X=\begin{pmatrix} -1 \\ 2 \\ -1 \end{pmatrix}\), so \(x_1 = -1, x_2 = 2, x_3 = -1\). (ii) For \(b_1=2, b_2=-5, b_3=0\), The solution is: \(X=\begin{pmatrix} -3 \\ 4 \\ -2 \end{pmatrix}\), so \(x_1 = -3, x_2 = 4, x_3 = -2\).

Step 1: Write the matrix equation

We are given the system of equations: \(2x_1 + x_2 + x_3 = b_1\) \(x_1 - 3x_2 + 4x_3 = b_2\) \(-x_1 + x_3 = b_3\) We can write this system of equations as the matrix equation AX = B, where: A = \(\begin{pmatrix} 2 & 1 & 1 \\ 1 & -3 & 4 \\ -1 & 0 & 1 \end{pmatrix}\), X = \(\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}\), B = \(\begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}\) So, AX = B is: \(\begin{pmatrix} 2 & 1 & 1 \\ 1 & -3 & 4 \\ -1 & 0 & 1 \end{pmatrix}\begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{pmatrix}=\begin{pmatrix} b_{1} \\ b_{2} \\ b_{3} \\ \end{pmatrix}\) Now, we need to find the inverse of matrix A and use it to solve the system of equations.

Step 2: Find the inverse of the coefficient matrix (A)

To find the inverse of matrix A, we need to find the determinant of A, and then find the adjugate of A, finally, we divide each element of adjugate of A by the determinant of A. Matrix A: \(\begin{pmatrix} 2 & 1 & 1 \\ 1 & -3 & 4 \\ -1 & 0 & 1 \end{pmatrix}\) Determinant of A: |A| = \(2(-3(1) - 4(0)) - 1(1(1) - 4(-1)) + 1((-1)(0) + 1(1)) = 14\) We know |A| is non-zero, so the matrix A is invertible. Now find the adjugate of A: adj(A) = \(\begin{pmatrix} -3 & -6 & -3 \\ 2 & 3 & 1 \\ 3 & 2 & 1\end{pmatrix}\) Finally, find the inverse of A: A⁻¹ = (1/14) * adj(A) = \(\begin{pmatrix} -\frac{3}{14}& -\frac{3}{7} & -\frac{3}{14} \\ \frac{1}{7} & \frac{3}{14}& \frac{1}{14}\\ \frac{3}{14} & \frac{1}{7} & \frac{1}{14} \end{pmatrix}\)

Step 3: Solve the system of equations using the inverse of A

Now that we have the inverse of the coefficient matrix A⁻¹, we can use it to solve the system of equations for (i) and (ii). (i) For \(b_1=1, b_2=4, b_3=-3\): X = A⁻¹B = \(\begin{pmatrix} -\frac{3}{14} & -\frac{3}{7} & -\frac{3}{14} \\ \frac{1}{7} & \frac{3}{14} & \frac{1}{14} \\ \frac{3}{14} & \frac{1}{7} & \frac{1}{14} \end{pmatrix}\) \(\begin{pmatrix} 1 \\ 4 \\ -3 \\ \end{pmatrix}\) = \(\begin{pmatrix} -1 \\ 2 \\ -1 \end{pmatrix}\) So, \(x_1 = -1, x_2 = 2, x_3 = -1\) (ii) For \(b_1=2, b_2=-5, b_3=0\): X = A⁻¹B = \(\begin{pmatrix} -\frac{3}{14} & -\frac{3}{7} & -\frac{3}{14} \\ \frac{1}{7} & \frac{3}{14} & \frac{1}{14} \\ \frac{3}{14} & \frac{1}{7} & \frac{1}{14} \end{pmatrix}\) \(\begin{pmatrix} 2 \\ -5 \\ 0 \\ \end{pmatrix}\) = \(\begin{pmatrix} -3 \\ 4 \\ -2 \end{pmatrix}\) So, \(x_1 = -3, x_2 = 4, x_3 = -2\)

Key Concepts

Matrix Equation

A matrix equation is a mathematical expression where variables and constants are arranged in a rectangular grid format. When we work with systems of linear equations, we can rewrite them as matrix equations to leverage matrix algebra for finding solutions.

Consider the system of linear equations from our exercise. By expressing the system as a matrix equation AX = B, we neatly package the coefficients into matrix A, the unknowns into vector X, and the constant terms into vector B. This compact form is not just visually tidy but also ready for powerful matrix operations, which can provide the solution to the system if matrix A is invertible.

Inverse of a Matrix

The inverse of a matrix plays a critical role in solving matrix equations. Just as dividing both sides of an algebraic equation by the same non-zero number can help isolate the variable, multiplying both sides of a matrix equation by the inverse of the coefficient matrix can help us solve for the unknown vector X.

The inverse of a matrix A, denoted as A^-1, has the property that A × A^-1 = I, where I is the identity matrix. However, not all matrices are invertible. A matrix has an inverse if and only if its determinant is not zero. Calculating the inverse involves finding the determinant and then the adjugate matrix, which is then scaled by the reciprocal of the determinant.

Determinant of a Matrix

The determinant of a matrix is a special scalar value that provides important information about the matrix. It can tell us whether a matrix is invertible, and it also gives a sense of the matrix's 'scaling factor' for areas or volumes.

In our example, the determinant of matrix A is calculated using a method (such as expansion by minors) that culminates in a single number, 14 in this case. Since the determinant is non-zero, it confirms that matrix A is indeed invertible, which means we have a green light to proceed with finding its inverse to solve the equation.

Matrix Algebra

The field of matrix algebra is an extension of traditional algebra into multiple dimensions. Instead of dealing with single numbers, we are manipulating arrays of numbers, which requires new rules and operations.

Matrix addition, subtraction, and multiplication are some of the basics, but when it comes to solving systems of equations, matrix multiplication is particularly important. We use it to multiply both sides of the equation by the inverse matrix (if it exists), which isolates the variable vector X. This elegant method of solving systems is powerful and often more efficient than other techniques, such as substitution or elimination, especially for larger systems.