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.)
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.)
Solution
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
\[
\boxed{(2, -1, -3)}
\]