2x+3y= -17
5x+2y= -4
You must cancel either the x or y out to solve so i'm going to cancel the x.
-5(2x+3y= -17)
2(5x+2y= -4)
This simplifies to:
-15x -15y=85
15x+4y= -8
The -15x and 15x cancel out leaving you:
-15y=85
4y= -8
You add them top numbers by the bottom and get:
-11y=77
Then divide by -11 to get y by its self.
y= -7
Now that you know what y is just plug it in to either original problem. I will use the second one.
5x + 2(-7) = -4
5x -14 = -4
5x = 10
x = 2
So the answer is (2,-7) or x=2 and y=-7

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3 years ago

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A) The equilibrium prices P1 and P2 for two goods satisfy the equations:

Allisa [31]

The value of P1 and P2 using the inverse matrix is 5 and 6 respectively.

<h3 />

The equilibrium prices of the three independent commodities using the inverse matrix are P1 = 4, P2 = 7, P3 = 8 respectively.

<h3>What is a matrix?</h3>

A matrix can be defined as a collection of integers(numbers that are either positive or negative) that are organized in rows and columns to construct a rectangular array. The numbers in this matrix system are referred to as elements.

To determine the values of P1 and P2 for the system of equations given by using an inverse matrix, we have:

9P1 + P2 = 51

3P1 +4P2 = 39

Representing the above data in matrix form, we have:

$\left[\begin{array}{cc}9&1\\3&4\\ \end{array}\right] \left[\begin{array}{c} \mathbf{P_1} \\ \mathbf{P_2}\\ \end{array}\right] = \left[\begin{array}{c} \mathbf{51} \\ \mathbf{39} \\ \end{array}\right]$ which is in the form AX = B

In order for us to determine the values of P1 and P2, Let take the inverse of A⁻¹ on both sides of the AX= B, we have:

$\mathbf{{A^{-1} AX} = A^{-1} B}$

X = A⁻¹ B

Let's start by finding A⁻¹;

$\mathbf{A = \left[\begin{array}{cc}9&1\\3&4\end{array}\right] }$

$\mathbf{A^{-1} = \dfrac{1}{36-3}\left[\begin{array}{cc}4&-1\\3&9\end{array}\right] }$

$\mathbf{A^{-1} = \dfrac{1}{33}\left[\begin{array}{cc}4&-1\\3&9\end{array}\right] }$

Now, Let's Find A⁻¹B;

$\mathbf{A^{-1}B = \dfrac{1}{33}\left[\begin{array}{cc}4&-1\\3&9\end{array}\right] \left[\begin{array}{c}51\\39\\ \end{array}\right] }$

$\mathbf{\implies \dfrac{1}{33}\left[\begin{array}{cc}204&-39\\-153&+351\end{array}\right] }$

$\mathbf{\implies \left[\begin{array}{c}\dfrac{165}{33}\\ \\ \dfrac{198}{33}\end{array}\right] }$

$\mathbf{\implies \left[\begin{array}{c}5\\ \\ 6\end{array}\right] }$

$\left[\begin{array}{c}\mathbf{P_1}\\ \mathbf{P_2}\end{array}\right]= \left[\begin{array}{c}5\\ 6 \end{array}\right] }$

Therefore, we can conclude that the value of P1 and P2 using the inverse matrix is 5 and 6 respectively.

To determine the equilibrium prices of the three independent commodities using the inverse matrix, we have:

P₁ + 2P₂ + 3P₃ = 42

2P₁ + P₂ + 4P₃ = 47

P₁ + 3P₂ + 4P₃ = 57

The matrix in AX = B form is computed as:

$\implies\left[ \begin{array}{ccc}1&2&3\\2&1&4\\1&3&4\end{array}\right] \left[\begin{array}{c}P_1\\P_2\\P_3 \end{array}\right] = \left[\begin{array}{c}42\\47\\ 57\end{array}\right]$