QuestionJuly 14, 2025

Use the inverse of the coefficient matrix to solve the following system of equations. 2x-y-2z=10 -2y-3z=6 3x-y-2z=14 x=square ,y= square , and z= square (Type integers or simplified fractions.)

Use the inverse of the coefficient matrix to solve the following system of equations. 2x-y-2z=10 -2y-3z=6 3x-y-2z=14 x=square ,y= square , and z= square (Type integers or simplified fractions.)
Use the inverse of the coefficient matrix to solve the following system of equations. 2x-y-2z=10 -2y-3z=6 3x-y-2z=14 x=square ,y= square , and z= square (Type integers or simplified fractions.)

Solution
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Answer

x = 4, y = -20, z = -10 Explanation 1. Write the system in matrix form The system can be written as AX = B, where A = \begin{bmatrix} 2 & -1 & -2 \\ 0 & -2 & -3 \\ 3 & -1 & -2 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, and B = \begin{bmatrix} 10 \\ 6 \\ 14 \end{bmatrix}. 2. Find the inverse of matrix A Calculate the inverse A^{-1} using the formula for a 3 \times 3 matrix. After computation, A^{-1} = \begin{bmatrix} 1 & -1 & 0 \\ -3 & -4 & -2 \\ -1 & -1 & -1 \end{bmatrix}. 3. Multiply the inverse by matrix B Use X = A^{-1}B. Compute X = \begin{bmatrix} 1 & -1 & 0 \\ -3 & -4 & -2 \\ -1 & -1 & -1 \end{bmatrix} \begin{bmatrix} 10 \\ 6 \\ 14 \end{bmatrix} = \begin{bmatrix} 4 \\ -20 \\ -10 \end{bmatrix}.

Explanation

1. Write the system in matrix form<br /> The system can be written as $AX = B$, where $A = \begin{bmatrix} 2 & -1 & -2 \\ 0 & -2 & -3 \\ 3 & -1 & -2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} 10 \\ 6 \\ 14 \end{bmatrix}$.<br /><br />2. Find the inverse of matrix A<br /> Calculate the inverse $A^{-1}$ using the formula for a $3 \times 3$ matrix. After computation, $A^{-1} = \begin{bmatrix} 1 & -1 & 0 \\ -3 & -4 & -2 \\ -1 & -1 & -1 \end{bmatrix}$.<br /><br />3. Multiply the inverse by matrix B<br /> Use $X = A^{-1}B$. Compute $X = \begin{bmatrix} 1 & -1 & 0 \\ -3 & -4 & -2 \\ -1 & -1 & -1 \end{bmatrix} \begin{bmatrix} 10 \\ 6 \\ 14 \end{bmatrix} = \begin{bmatrix} 4 \\ -20 \\ -10 \end{bmatrix}$.
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