How do you solve the system #3p+6q=-3,2p-3q=-9# using matrices?

Question

Sadie Alexander · Accepted Answer

<p>see below</p> <p>

Serenity Johnson · Answer

undefined To solve the system $3p + 6q = -3$ and $2p - 3q = -9$ using matrices, we can represent the system in matrix form $AX = B$, where $A$ is the coefficient matrix, $X$ is the matrix of variables, and $B$ is the matrix of constants.

$$ A = \begin{bmatrix} 3 & 6 \ 2 & -3 \end{bmatrix} $$
$$ X = \begin{bmatrix} p \ q \end{bmatrix} $$
$$ B = \begin{bmatrix} -3 \ -9 \end{bmatrix} $$

To find $X$, we'll use the formula $X = A^{-1}B$, where $A^{-1}$ is the inverse of matrix $A$.

First, we need to find the inverse of matrix $A$, denoted as $A^{-1}$.

$$ A^{-1} = \frac{1}{\text{det}(A)} \times \text{adj}(A) $$

Where $ \text{det}(A) $ is the determinant of $A$, and $ \text{adj}(A) $ is the adjugate of $A$.

For matrix $A$:
$$ \text{det}(A) = (3 \times -3) - (6 \times 2) = -9 - 12 = -21 $$

$$ \text{adj}(A) = \begin{bmatrix} -3 & -6 \ -2 & 3 \end{bmatrix} $$

Now, we can calculate $A^{-1}$:

$$ A^{-1} = \frac{1}{-21} \times \begin{bmatrix} -3 & -6 \ -2 & 3 \end{bmatrix} $$
$$ A^{-1} = \begin{bmatrix} \frac{1}{21} & \frac{2}{7} \ \frac{2}{21} & -\frac{1}{7} \end{bmatrix} $$

Next, we can find $X$ by multiplying $A^{-1}$ and $B$:

$$ X = A^{-1}B $$
$$ X = \begin{bmatrix} \frac{1}{21} & \frac{2}{7} \ \frac{2}{21} & -\frac{1}{7} \end{bmatrix} \begin{bmatrix} -3 \ -9 \end{bmatrix} $$
$$ X = \begin{bmatrix} \frac{1}{21} & \frac{2}{7} \ \frac{2}{21} & -\frac{1}{7} \end{bmatrix} \begin{bmatrix} -3 \ -9 \end{bmatrix} $$
$$ X = \begin{bmatrix} \frac{1}{21} \times -3 + \frac{2}{7} \times -9 \ \frac{2}{21} \times -3 + (-\frac{1}{7}) \times -9 \end{bmatrix} $$
$$ X = \begin{bmatrix} \frac{-3}{7} - \frac{18}{7} \ \frac{-6}{7} - \frac{-9}{7} \end{bmatrix} $$
$$ X = \begin{bmatrix} \frac{-21}{7} \ \frac{3}{7} \end{bmatrix} $$
$$ X = \begin{bmatrix} -3 \ \frac{3}{7} \end{bmatrix} $$

So, the solution to the system of equations is $p = -3$ and $q = \frac{3}{7}$.