If $$ A=\left[\begin{array}{cc} 3 & 6 \ 2 & -5 \end{array}\right], $$ then $$ A^{-1}=\left[\begin{array}{cc|cc} \frac{5}{27} & & \frac{6}{27} & \ \frac{2}{27} & \frac{1}{9} & \ \hline \end{array}\right] $$ Given $$ \vec{b}=\left[\begin{array}{l}-3 \ -4\end{array}\right] $$, solve $$ A \vec{x}=\vec{b} $$. $$ \vec{x}=\left[\begin{array}{r} \sigma_{0} \end{array}\right] $$

Question

UpStudy AI · Accepted Answer

To solve the linear system $$ A \vec{x} = \vec{b} $$ where

$$
A = \begin{bmatrix} 3 & 6 \ 2 & -5 \end{bmatrix}, \quad \vec{b} = \begin{bmatrix} -3 \ -4 \end{bmatrix},
$$

we can use the inverse of matrix $$ A $$ to find the solution vector $$ \vec{x} $$.

### Step 1: Compute the Inverse of $$ A $$

First, calculate the determinant of $$ A $$:

$$
\text{det}(A) = (3)(-5) - (6)(2) = -15 - 12 = -27
$$

The inverse of $$ A $$ is then:

$$
A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} -5 & -6 \ -2 & 3 \end{bmatrix} = \frac{1}{-27} \begin{bmatrix} -5 & -6 \ -2 & 3 \end{bmatrix} = \begin{bmatrix} \frac{5}{27} & \frac{6}{27} \ \frac{2}{27} & -\frac{1}{9} \end{bmatrix}
$$

### Step 2: Multiply $$ A^{-1} $$ by $$ \vec{b} $$

Now, compute $$ \vec{x} = A^{-1} \vec{b} $$:

$$
\vec{x} = \begin{bmatrix} \frac{5}{27} & \frac{6}{27} \ \frac{2}{27} & -\frac{1}{9} \end{bmatrix} \begin{bmatrix} -3 \ -4 \end{bmatrix}
$$

Calculate each component of $$ \vec{x} $$:

$$
x_1 = \left(\frac{5}{27}\right)(-3) + \left(\frac{6}{27}\right)(-4) = -\frac{15}{27} - \frac{24}{27} = -\frac{39}{27} = -\frac{13}{9}
$$

$$
x_2 = \left(\frac{2}{27}\right)(-3) + \left(-\frac{1}{9}\right)(-4) = -\frac{6}{27} + \frac{4}{9} = -\frac{2}{9} + \frac{4}{9} = \frac{2}{9}
$$

### Final Solution

$$
\vec{x} = \begin{bmatrix} -\dfrac{13}{9} \ \dfrac{2}{9} \end{bmatrix}
$$
$$ \vec{x} = \begin{bmatrix} -\dfrac{13}{9} \ \dfrac{2}{9} \end{bmatrix} $$

If

then

Given , solve .

Upstudy AI Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Algebra Questions

If then Given , solve .