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Solve the system of equations using the inverse of the coefficient matrix of the equivalent matrix equation. $$\begin{aligned}x &+z=1 \\\2 x+y &=3 \\\x-y+z &=4\end{aligned}$$

Short Answer

Expert verified
The solution is \( x = -1, y = -8, z = -6 \).

Step by step solution

01

Write the system of equations in matrix form

The system of equations can be written in the form \textbf{AX} = \textbf{B}, where \textbf{A} is the matrix of coefficients, \textbf{X} is the column matrix of variables, and \textbf{B} is the column matrix of constants. Identify these matrices: \[ \begin{bmatrix} 1 & 0 & 1 \ 2 & 1 & 0 \ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 1 \ 3 \ 4 \end{bmatrix} \]
02

Calculate the inverse of the coefficient matrix

To solve for \textbf{X}, find the inverse of matrix \textbf{A} (denoted as \textbf{A}^{-1}). The inverse is calculated using standard methods for finding the inverse of a matrix. Calculate the determinant of \textbf{A} and use the adjugate method to find \textbf{A}^{-1}. First, determinant of \textbf{A} is \[ \text{det}(A) = 1(1(1) - (-1)(0)) - 0(2(1) - 0(1)) + 1(2(-1) - 1(1)) = 1 + 0 - 3 = -2 \]
03

Find the adjugate of matrix A

The adjugate of \textbf{A} is the transpose of the cofactor matrix. Compute each cofactor and then the transpose: \[ \text{cofactor matrix of A} = \begin{bmatrix} 1-0 & -(0-2) & 0-(-2) \ -(1-0) & 1-0 & 0-(-2) \ 1-0 & 1-(-2) & 1-2 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 2 \ -1 & 1 & 2 \ 1 & 3 & -1 \end{bmatrix} \] \(\text{Adjugate of A} = \begin{bmatrix} 1 & -1 & 1 \ 2 & 1 & 3 \ 2 & 2 & -1 \end{bmatrix}\)
04

Compute the inverse of A

Use the formula \( A^{-1} = \frac{1}{\det(A)} \text{Adj}(A) \) to find the inverse of A. \[ A^{-1} = \frac{1}{-2} \begin{bmatrix} 1 & -1 & 1 \ 2 & 1 & 3 \ 2 & 2 & -1 \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \ -1 & -\frac{1}{2} & -\frac{3}{2} \ -1 & -1 & \frac{1}{2} \end{bmatrix} \]
05

Multiply the inverse of A by matrix B

To find the solution matrix \textbf{X}, multiply \textbf{A}^{-1} by \textbf{B}: \[ X = A^{-1}B = \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \ -1 & -\frac{1}{2} & -\frac{3}{2} \ -1 & -1 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} 1 \ 3 \ 4 \end{bmatrix} \]
06

Compute the resulting matrix

Multiply the matrices to obtain \[ X = \begin{bmatrix} -\frac{1}{2}(1) + \frac{1}{2}(3) - \frac{1}{2}(4) \ -1(1) - \frac{1}{2}(3) - \frac{3}{2}(4) \ -1(1) - 1(3) + \frac{1}{2}(4) \end{bmatrix} = \begin{bmatrix} -1 \ -8 \ -6 \end{bmatrix} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

inverse matrix
An inverse matrix, denoted as \(\textbf{A}^{-1}\), is a matrix that, when multiplied by the original matrix \(\textbf{A}\), results in an identity matrix. The identity matrix \(\textbf{I}\) is a square matrix with ones on the diagonal and zeros elsewhere. Finding the inverse of a matrix is essential in solving systems of linear equations through matrix algebra. \
\
For a matrix to have an inverse, it must be square (same number of rows and columns) and its determinant should not be zero. The formula for finding the inverse of a 3x3 matrix is based on its determinant and adjugate matrix. Here, we calculated the inverse of the matrix \(\textbf{A}\) as: \(\textbf{A}^{-1} \). \
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To find \(\textbf{A}^{-1}\), we used the formula: \[A^{-1} = \frac{1}{\text{det}(\textbf{A})} \text{Adj}(\text{A})\]. Multiplying the inverse matrix by a constants matrix allows us to solve for the variables in a system of equations.
system of equations
A system of equations is a set of two or more equations with the same set of unknowns. Solving these equations means finding the values of the variables that satisfy all equations simultaneously. Matrix algebra provides a streamlined approach for solving these systems. \
\
In the given exercise, our system of linear equations was written in matrix form as \(\textbf{AX} = \textbf{B}\), where \(\textbf{A}\) is the coefficient matrix, \(\textbf{X}\) is the column matrix of variables, and \(\textbf{B}\) is the column matrix of constants. By transforming the system into matrix form, we could employ algebraic techniques to find the solution efficiently. \
\
The inverse matrix technique, specifically, allows us to isolate \(\textbf{X}\) by multiplying both sides of the equation by \(\textbf{A}^{-1}\), yielding \[X = A^{-1}B\]. This approach is highly methodical and works well for consistent systems.
determinant
The determinant of a matrix is a special number that is computed from its elements. It provides essential information about the matrix, such as whether it is invertible. If the determinant is zero, the matrix does not have an inverse. \
\
In our exercise, we calculated the determinant of matrix \(\textbf{A}\) as follows: \[ \text{det}(\textbf{A}) = 1(1(1) - (-1)(0)) - 0(2(1) - 0(1)) + 1(2(-1) - 1(1)) = 1 + 0 - 3 = -2 \]. This non-zero determinant (\text{-2}) indicated that \(\textbf{A}\) is invertible, allowing us to further calculate its inverse. \
\
Knowing how to compute the determinant is crucial for various applications in mathematics and helps ensure the matrix operations we perform are valid.
adjugate matrix
The adjugate matrix, often called the adjoint, is derived from the matrix of cofactors. It plays a vital role in finding the inverse of a matrix. To compute the adjugate, follow these steps: \
    \
  • Calculate the cofactor of each element in the matrix.
  • \
  • Form the cofactor matrix.
  • \
  • Take the transpose of the cofactor matrix.
  • \
\
\
In our example, after computing the cofactor matrix, the adjugate of matrix \(\textbf{A}\) was derived, as shown: \[ \text{Adj}(\textbf{A}) = \begin{bmatrix} 1 & -1 & 1 \ 2 & 1 & 3 \ 2 & 2 & -1 \end{bmatrix} \]. This step is essential in determining the inverse matrix, as seen in the formula \[ A^{-1} = \frac{1}{\text{det}(\textbf{A})} \text{Adj}(\textbf{A}) \]. \
\
The adjugate matrix method ensures systematic computation and validity of the matrix inversion.

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Most popular questions from this chapter

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