Problem 15 - Elementary and Intermediate Algebra

Question

Solve using matrices. $$ \begin{aligned} a-2 b-3 c &=3 \\ 2 a-b-2 c &=4 \\ 4 a+5 b+6 c &=4 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \( a = -1 \), \( b = 1 \), and \( c = 1 \).

Step 1: Write the system of equations in matrix form

Express the system \(\begin{aligned} a-2b-3c &= 3 \ 2a-b-2c &= 4 \ 4a+5b+6c &= 4 \end{aligned}\) as a matrix equation \(\textbf{Ax} = \textbf{B}\).

Step 2: Form the coefficient matrix \( \textbf{A} \)

The coefficient matrix is created from the coefficients of \(a\), \(b\), and \(c\): \[ \textbf{A} = \begin{pmatrix} 1 & -2 & -3 \ 2 & -1 & -2 \ 4 & 5 & 6 \end{pmatrix} \]

Step 3: Form the variable matrix \(\textbf{x}\)

The variable matrix consists of the variables \(a\), \(b\), and \(c\):\[ \textbf{x} = \begin{pmatrix} a \ b \ c \end{pmatrix} \]

Step 4: Form the constant matrix \(\textbf{B}\)

The constant matrix consists of the constants from the right-hand side of each equation: \[ \textbf{B} = \begin{pmatrix} 3 \ 4 \ 4 \end{pmatrix} \]

Step 5: Solve for \(\textbf{x}\) using the inverse of \(\textbf{A}\)

Compute the inverse of \(\textbf{A}\) if it exists, then solve \( \textbf{x} = \textbf{A}^{-1} \textbf{B} \).

Step 6: Calculate the inverse of \(\textbf{A}\)

Find the inverse of \(\textbf{A}\) such that \(\textbf{A}^{-1} \textbf{A} = \textbf{I}\). The inverse of \( \textbf{A} \) is:\[ \textbf{A}^{-1} = \begin{pmatrix} -0.05 & -0.15 & 0.2 \ 0.1 & -0.1 & 0 \ -0.1 & 0.2 & 0 \end{pmatrix} \]

Step 7: Compute \( \textbf{x} = \textbf{A}^{-1} \textbf{B} \)

Multiply \( \textbf{A}^{-1} \) by \( \textbf{B} \) to solve for \( \textbf{x} \):\[ \textbf{x} = \begin{pmatrix} -0.05 & -0.15 & 0.2 \ 0.1 & -0.1 & 0 \ -0.1 & 0.2 & 0 \end{pmatrix} \begin{pmatrix} 3 \ 4 \ 4 \end{pmatrix} = \begin{pmatrix} -1 \ 1 \ 1 \end{pmatrix} \]

Step 8: Interpret the solution

The values for \(a\), \(b\), and \(c\) are the components of the resulting matrix \( \textbf{x} \). Therefore, the solution to the system is \( a = -1 \), \( b = 1 \), and \( c = 1 \).

Key Concepts

Coefficient Matrix

Understanding the coefficient matrix is vital when solving systems of equations using matrices. The coefficient matrix, denoted as \(\textbf{A}\), is created by extracting the coefficients of each variable from the system of equations. For instance, given the system: \(\begin{aligned} a-2b-3c &= 3 \ 2a-b-2c &= 4 \ 4a+5b+6c &= 4 \end{aligned}\), the coefficient matrix will be: \(\textbf{A} = \begin{pmatrix} 1 & -2 & -3 \ 2 & -1 & -2 \ 4 & 5 & 6 \end{pmatrix}\). This matrix represents the linear relationships of the variables in the given equations.

Variable Matrix

The variable matrix, denoted as \(\textbf{x}\), consists of the variables that we need to solve. In our example, these variables are a, b, and c. The variable matrix is represented vertically to align with the coefficients in the coefficient matrix. It's expressed as: \(\textbf{x} = \begin{pmatrix} a \ b \ c \end{pmatrix}\). This format is crucial because it allows us to use matrix multiplication effectively when working through the problem. It aligns each variable with its corresponding coefficient in the equations.

Constant Matrix

Next, we need to construct the constant matrix, denoted as \(\textbf{B}\). This matrix consists of the constants from the right side of the equations. For our given system, these constants are 3, 4, and 4. So, the constant matrix is written as: \(\textbf{B} = \begin{pmatrix} 3 \ 4 \ 4 \end{pmatrix}\). This matrix is critical because it represents the values we are aiming to balance or equalize using our coefficients and variables.

Inverse Matrix

The inverse matrix is a powerful tool necessary for solving systems of equations through matrix methods. To solve for the variable matrix \(\textbf{x}\), we need to find the inverse of the coefficient matrix \(\textbf{A}^{-1}\). For a matrix \(\textbf{A}\), the inverse is a matrix such that when \(\textbf{A}\) is multiplied by its inverse \(\textbf{A}^{-1}\), it results in the identity matrix \(\textbf{I}\). In our example, \(\textbf{A}^{-1} = \begin{pmatrix} -0.05 & -0.15 & 0.2 \ 0.1 & -0.1 & 0 \ -0.1 & 0.2 & 0 \end{pmatrix}\). With \(\textbf{A}^{-1}\), we solve for the variables by multiplying \(\textbf{A}^{-1}\) with the constant matrix \(\textbf{B}\).

Matrix Multiplication

Lastly, matrix multiplication is used to find the solution to the system of equations. We compute \(\textbf{x} = \textbf{A}^{-1} \textbf{B}\). This process involves multiplying each row of the inverse matrix \(\textbf{A}^{-1}\) with the corresponding element of \(\textbf{B}\), then summing the results. In our problem, this step looks like: \(\textbf{x} = \begin{pmatrix} -0.05 & -0.15 & 0.2 \ 0.1 & -0.1 & 0 \ -0.1 & 0.2 & 0 \end{pmatrix} \begin{pmatrix} 3 \ 4 \ 4 \end{pmatrix} = \begin{pmatrix} -1 \ 1 \ 1 \end{pmatrix}\). By performing this multiplication, we can determine the values for a, b, and c.