Solve System of Equations using Matrix Inverses

Math Problem Statement

Write the given system of equations as a matrix equation and solve using inverses. Complete parts a through c. x 1 plus 2 x 2 equals k 1 x 2 plus x 3 equals k 2 5 x 1 minus x 2 minus 12 x 3 equals k 3 Question content area bottom Part 1 a. What are x 1, x 2, and x 3 when k 1equals3, k 2equals0, and k 3equals15?

Solution

To solve the given system of equations using matrix methods, we need to follow these steps:

Step 1: Write the System of Equations in Matrix Form

The given system of equations is:

x_1 + 2x_2 & = k_1 \quad \text{(Equation 1)} \\ x_2 + x_3 & = k_2 \quad \text{(Equation 2)} \\ 5x_1 - x_2 - 12x_3 & = k_3 \quad \text{(Equation 3)} \end{aligned}$$ This system can be written in matrix form as: $$\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 5 & -1 & -12 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} k_1 \\ k_2 \\ k_3 \end{bmatrix}$$ ### Step 2: Find the Inverse of the Coefficient Matrix The coefficient matrix $$A$$ is: $$A = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 5 & -1 & -12 \end{bmatrix}$$ We need to find the inverse of $$A$$, denoted as $$A^{-1}$$. ### Step 3: Multiply the Inverse by the Constant Matrix The constant matrix $$\mathbf{k}$$ is: $$\mathbf{k} = \begin{bmatrix} k_1 \\ k_2 \\ k_3 \end{bmatrix}$$ The solution to the system is given by: $$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = A^{-1} \begin{bmatrix} k_1 \\ k_2 \\ k_3 \end{bmatrix}$$ ### Step 4: Solve for $$x_1$$, $$x_2$$, and $$x_3$$ When $$k_1 = 3$$, $$k_2 = 0$$, and $$k_3 = 15$$ Substitute the values $$k_1 = 3$$, $$k_2 = 0$$, and $$k_3 = 15$$ into the constant matrix: $$\mathbf{k} = \begin{bmatrix} 3 \\ 0 \\ 15 \end{bmatrix}$$ Now, I'll calculate the inverse of the matrix $$A$$, then multiply it by the constant matrix $$\mathbf{k}$$ to find $$x_1$$, $$x_2$$, and $$x_3$$. Let's calculate this now.The solution to the system of equations when $$k_1 = 3$$, $$k_2 = 0$$, and $$k_3 = 15$$ is: $$\begin{aligned} x_1 & = 3 \\ x_2 & \approx 0 \quad \text{(close to zero due to numerical approximation)} \\ x_3 & \approx 0 \quad \text{(close to zero due to numerical approximation)} \end{aligned}$$ This means that $$x_1 = 3$$, $$x_2 = 0$$, and $$x_3 = 0$$ is the solution for the given values of $$k_1$$, $$k_2$$, and $$k_3$$. Would you like more details or have any questions? Here are some related questions you might consider: 1. How do you determine the inverse of a matrix? 2. What is the significance of the determinant in finding the inverse? 3. How would the solution change if $$k_1$$, $$k_2$$, and $$k_3$$ were different? 4. What if the determinant of the coefficient matrix $$A$$ was zero? 5. How can you verify the solution obtained from the matrix equation? **Tip:** When solving a system of equations using matrix inverses, always check the determinant first. If it’s zero, the matrix is non-invertible, meaning the system might have no solutions or infinitely many solutions.

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Math Problem Analysis

Mathematical Concepts

Matrix Algebra
System of Equations
Matrix Inverse

Formulas

Matrix equation solving using inverses

Theorems

Suitable Grade Level

Grades 11-12

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