Questions: A system of equations is given, together with the inverse of the coefficient matrix. Use the inverse of the coefficient matrix to solve the system of equations. -x-y = -1 3x+y+2z = -1 2x-y+z = 2 A^(-1) = 1/4 [ -3 -1 2 -1 1 -2 5 3 -2 ] The solution to the system is (Type an ordered triple. Type integers or fractions.)

A system of equations is given, together with the inverse of the coefficient matrix. Use the inverse of the coefficient matrix to solve the system of equations.

-x-y = -1
3x+y+2z = -1
2x-y+z = 2

A^(-1) = 1/4
[ -3 -1 2
-1 1 -2
5 3 -2 ]

The solution to the system is 
(Type an ordered triple. Type integers or fractions.)
Transcript text: Points: 0 of 1 A system of equations is given, together with the inverse of the coefficient matrix. Use the inverse of the coefficient matrix to solve the system of equations. \[ \begin{array}{ll} -x-y & =-1 \\ 3 x+y+2 z & =-1 \\ 2 x-y+z & =2 \end{array} \] \[ A^{-1}=\frac{1}{4}\left[\begin{array}{rrr} -3 & -1 & 2 \\ -1 & 1 & -2 \\ 5 & 3 & -2 \end{array}\right] \] The solution to the system is $\square$ $\square$ $\square$ (Type an ordered triple. Type integers or fractions.)
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Solution

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Solution Steps

Step 1: Represent the System of Equations

The given system of equations can be represented in matrix form as follows: \[ A = \begin{bmatrix} -1 & -1 & 0 \\ 3 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} -1 \\ -1 \\ 2 \end{bmatrix} \] This leads to the equation \( AX = B \).

Step 2: Use the Inverse Matrix

The inverse of the coefficient matrix \( A \) is given as: \[ A^{-1} = \frac{1}{4} \begin{bmatrix} -3 & -1 & 2 \\ -1 & 1 & -2 \\ 5 & 3 & -2 \end{bmatrix} = \begin{bmatrix} -0.75 & -0.25 & 0.5 \\ -0.25 & 0.25 & -0.5 \\ 1.25 & 0.75 & -0.5 \end{bmatrix} \] To find the solution \( X \), we calculate \( X = A^{-1}B \).

Step 3: Calculate the Solution

Performing the matrix multiplication yields: \[ X = A^{-1}B = \begin{bmatrix} 2 \\ -1 \\ -3 \end{bmatrix} \] Thus, the values of the variables are \( x = 2 \), \( y = -1 \), and \( z = -3 \).

Final Answer

The solution to the system of equations is \[ \boxed{(2, -1, -3)} \]

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