Solve the system by inverting the coefficient matrix and using the following theorem: If $$ A $$ is an invertible $$ n imes n $$ matrix, then for each $$ n imes 1 $$ matrix $$ \mathbf{b} $$, the system of equations $$ A \mathbf{x}=\mathbf{b} $$ has exactly one solution, namely, $$ \mathbf{x}=A^{-1} \mathbf{b} $$. $$ x_{1}+x_{2}=5 $$ $$ 2 x_{1}+3 x_{2}=7 $$ $$ x_{1}=\mathbf{i} $$ $$ x_{2}=\mathbf{i} $$

Question

Solve the system by inverting the coefficient matrix and using the following theorem: If $$ A $$ is an invertible $$ n 	imes n $$ matrix, then for each $$ n 	imes 1 $$ matrix $$ \mathbf{b} $$, the system of equations $$ A \mathbf{x}=\mathbf{b} $$ has exactly one solution, namely, $$ \mathbf{x}=A^{-1} \mathbf{b} $$. $$ x_{1}+x_{2}=5 $$ $$ 2 x_{1}+3 x_{2}=7 $$ $$ x_{1}=\mathbf{i} $$ $$ x_{2}=\mathbf{i} $$

UpStudy AI · Accepted Answer

To solve the given system of linear equations using matrix inversion, we'll follow these steps:

$$
\begin{cases}
x_{1} + x_{2} = 5 \
2x_{1} + 3x_{2} = 7
\end{cases}
$$

### Step 1: Represent the System in Matrix Form

First, express the system $$ A\mathbf{x} = \mathbf{b} $$ where:

$$
A = \begin{bmatrix} 
1 & 1 \ 
2 & 3 
\end{bmatrix}, \quad 
\mathbf{x} = \begin{bmatrix} 
x_{1} \ 
x_{2} 
\end{bmatrix}, \quad 
\mathbf{b} = \begin{bmatrix} 
5 \ 
7 
\end{bmatrix}
$$

### Step 2: Calculate the Inverse of Matrix $$ A $$

To find $$ A^{-1} $$, first calculate the determinant of $$ A $$:

$$
\det(A) = (1)(3) - (1)(2) = 3 - 2 = 1
$$

Since the determinant is non-zero, $$ A $$ is invertible. The inverse of a $$ 2 \times 2 $$ matrix is given by:

$$
A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} 
d & -b \ 
-c & a 
\end{bmatrix}
$$

Applying this to our matrix $$ A $$:

$$
A^{-1} = \frac{1}{1} \begin{bmatrix} 
3 & -1 \ 
-2 & 1 
\end{bmatrix} = \begin{bmatrix} 
3 & -1 \ 
-2 & 1 
\end{bmatrix}
$$

### Step 3: Multiply $$ A^{-1} $$ by $$ \mathbf{b} $$ to Find $$ \mathbf{x} $$

$$
\mathbf{x} = A^{-1}\mathbf{b} = \begin{bmatrix} 
3 & -1 \ 
-2 & 1 
\end{bmatrix} \begin{bmatrix} 
5 \ 
7 
\end{bmatrix}
$$

Perform the matrix multiplication:

$$
x_{1} = (3)(5) + (-1)(7) = 15 - 7 = 8
$$
$$
x_{2} = (-2)(5) + (1)(7) = -10 + 7 = -3
$$

### Step 4: Verify the Solution

Plugging $$ x_{1} = 8 $$ and $$ x_{2} = -3 $$ back into the original equations:

1. $$ 8 + (-3) = 5 $$ ✔️
2. $$ 2(8) + 3(-3) = 16 - 9 = 7 $$ ✔️

### Final Solution

$$
\boxed{ \begin{cases}
x_{1} = 8 \
x_{2} = -3
\end{cases} }
$$
The solution to the system is $$ x_{1} = 8 $$ and $$ x_{2} = -3 $$.

Solve the system by inverting the coefficient matrix and using the following theorem:
If is an invertible matrix, then for each matrix , the system of equations has exactly one solution, namely,
.

Upstudy AI Solution

Answer

Solution

Step 1: Represent the System in Matrix Form

Step 2: Calculate the Inverse of Matrix

Step 3: Multiply by to Find

Step 4: Verify the Solution

Final Solution

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Solve the system by inverting the coefficient matrix and using the following theorem: If is an invertible matrix, then for each matrix , the system of equations has exactly one solution, namely, .