Problem 52 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use an inverse matrix to solve (if possible) the system of linear equations. $$\left\\{\begin{array}{l} 4 x-2 y+3 z=-2 \\ 2 x+2 y+5 z=16 \\ 8 x-5 y-2 z=4 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution of the given system of equations is \(x = 8\), \(y = 0\), \(z = -4\).

Step 1: Write the System as a Matrix Equation

In order to apply matrix operations, the system should be rewritten as a matrix equation. Hence, the given system of equations can be expressed as \(AX = B\), where \(A\) is the matrix of coefficients of \(x\), \(y\), \(z\), \(X\) is the column matrix of variables \([x, y, z]^T\), and \(B\) is the matrix of constants. Therefore, \(A = \begin{bmatrix} 4 & -2 & 3 \ 2 & 2 & 5 \ 8 & -5 & -2 \end{bmatrix}\), \(X = \begin{bmatrix} x \ y \ z \end{bmatrix}\) and \(B = \begin{bmatrix} -2 \ 16 \ 4 \end{bmatrix}\).

Step 2: Check for the Existence of the Inverse Matrix

In order to apply the method of inverse matrix, the determinant of the matrix \(A\), \(\text{det}(A)\), should not be zero. The determinant of matrix \(A\) is \(4*(2*-2 - 5*-5) -(-2)*(2*-2 - 5*-2) + 3*(1*2 - 2*5) = -12\). Since the determinant of matrix \(A\) is not equal to zero, the inverse of \(A\) exists.

Step 3: Calculate the Inverse Matrix

The formula used to calculate the inverse matrix \(A^{-1}\) requires calculating the matrix of minors, the matrix of cofactors, the adjugate matrix, and finally dividing each entry of the adjugate matrix with determinant of \(A\). The result is \(A^{-1} = \begin{bmatrix} -1/6 & 4/3 & 0 \ -1/6 & 5/6 & 1/6 \ 2/3 & -8/3 & 1/6 \end{bmatrix}\).

Step 4: Multiply the Inverse Matrix by the Constant

Multiplying the inverse matrix \(A^{-1}\) by the constant matrix \(B\), we get the matrix \(X\). So, \(X = A^{-1}B = \begin{bmatrix} -1/6 & 4/3 & 0 \ -1/6 & 5/6 & 1/6 \ 2/3 & -8/3 & 1/6 \end{bmatrix} \begin{bmatrix} -2 \ 16 \ 4 \end{bmatrix}\). After multiplication, we obtain \(X = \begin{bmatrix} 8 \ 0 \ -4 \end{bmatrix}\).

Step 5: Write the Final Answer

The matrix \(X\) contains the solutions for \(x\), \(y\) and \(z\), respectively. So, the solution of the system of equations is \(x = 8\), \(y = 0\), \(z = -4\).

Key Concepts

System of Linear Equations

A system of linear equations is a collection of one or more linear equations involving the same set of variables. In our example, we have three equations with three variables: \(x\), \(y\), and \(z\). These equations represent lines in three-dimensional space:

• \(4x - 2y + 3z = -2\)
• \(2x + 2y + 5z = 16\)
• \(8x - 5y - 2z = 4\)

Our goal is to find the values of \(x\), \(y\), and \(z\) that satisfy all these equations simultaneously. This is essential in many fields like physics for solving force balance problems or in economics for finding equilibrium conditions.

Matrix Determinant

The determinant is a special number that can be calculated from a square matrix. It provides important properties about the matrix, such as whether it has an inverse. In this process, the determinant of matrix \(A\) plays a crucial role.
For our matrix \(A\):
\[A = \begin{bmatrix} 4 & -2 & 3 \2 & 2 & 5 \8 & -5 & -2 \end{bmatrix}\]The determinant is calculated using specific expansion methods, often involving minors and cofactors. Fortunately, if the determinant is not zero, as is the case here (\(-12\)), it assures us that the matrix is invertible.
The determinant tells us crucial things about the system of equations, primarily that there exists a unique solution, as opposed to having no solutions or infinitely many.

Matrix Multiplication

Matrix multiplication is used when solving systems of equations through matrices, especially when involving inverse matrices. To find the solution of the equation \(AX = B\), we first need the product of the inverse matrix \(A^{-1}\) and the matrix of constants \(B\).
Given:
\[A^{-1} = \begin{bmatrix} -1/6 & 4/3 & 0 \-1/6 & 5/6 & 1/6 \2/3 & -8/3 & 1/6 \end{bmatrix}\]
And:
\[B = \begin{bmatrix} -2 \16 \4 \end{bmatrix}\]
The product \(A^{-1}B\) is computed by performing the dot product of rows from \(A^{-1}\) with the column from \(B\). This operation results in the matrix \(X\) containing our solutions.

Solution of Equations

The solution to the system of linear equations is derived from the matrix multiplication of the inverse of the coefficients matrix and the constants matrix. The product yields:
\[X = A^{-1}B = \begin{bmatrix} 8 \0 \-4 \end{bmatrix}\]
This indicates that the values \(x = 8\), \(y = 0\), and \(z = -4\) satisfy all equations simultaneously. We verify this by substituting these values back into the original equations, ensuring each equation holds true. Solving systems like these helps understand real-world phenomena, modeling everything from engineering problems to financial scenarios.