The supply function for a product is given by $p=0.1 q^2+50 q+1027.50$ and the demand function for this product is p=6000-20q, where p is the price in dollars and q is the number of hundreds of units. Find the price that gives market equilibrium and the equilibrium quantity.

Suppose

$$A=\left[\begin{array}{rrr}1 & -\frac{1}{2} & -\frac{1}{4} \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ 0 & \frac{1}{2} & -\frac{1}{4}\end{array}\right]$$

and

$$B=\left[\begin{array}{lll}2 & 2 & 2 \\ 1 & 2 & 3 \\ 2 & 4 & 2\end{array}\right]$$

Compute AB and BA, if possible

A satellite company and a cable company compete for customers in a city. The satellite company retains $\frac{2}{3}$ of its customers and loses $\frac{1}{3}$ of its customers to the cable company, while the cable company retains $\frac{3}{5}$ of its customers and loses $\frac{2}{5}$ of its customers to the satellite company. If we represent the fraction of the market held last year by

$$\left[\begin{array}{l} a \\ b \end{array}\right]$$

where a is the number of customers that the satellite company had last year and b is the number that the cable company had last year, then the number that each company will have this year can be found by

$$\left[\begin{array}{l} A \\ B \end{array}\right]=\left[\begin{array}{ll} \frac{2}{3} & \frac{2}{5} \\ \frac{1}{3} & \frac{3}{5} \end{array}\right]\left[\begin{array}{l} a \\ b \end{array}\right]$$

If the satellite company has A = 90,000 customers and the cable company has B = 85,000 customers this year, how many customers did each have last year?

The supply function for a product is given by $p=q^2+500$ and the demand function for this product is p=1124-40q, where p is the price in dollars and q is the number of hundreds of units. Find the price that gives market equilibrium and the equilibrium quantity.

Solve for x.

$$\left|\begin{array}{rr}5 & x \\ -1 & -3\end{array}\right|=5$$

The supply function for a product is given by $p=q^2+2 q+122$ and the demand function for this product is p=650-30q, where p is the price in dollars and q is the number of hundreds of units. Find the price that gives market equilibrium and the equilibrium quantity.

If

$$A=\left[\begin{array}{rrr}1 & -1 & 2 \\ 3 & 4 & 4\end{array}\right]$$

and

$$B=\left[\begin{array}{rr}3 & 1 \\ 1 & 3 \\ -2 & 1\end{array}\right]$$

, compute A B and B A, if possible.

Solve for x.

$$\left|\begin{array}{rr}-2 & -11 \\ x & 6\end{array}\right|=32$$

Two phone companies compete for customers in the southeastern region of a state. Each year, company X retains $\frac{3}{5}$ of its customers and loses $\frac{2}{5}$ of its customers to company Y, while company Y retains $\frac{2}{3}$ of its customers and loses $\frac{1}{3}$ of its customers to company X. If we represent the fraction of the market held last year by

$$\left[\begin{array}{l} a \\ b \end{array}\right]$$

where a is the number that company X had last year and b is the number that company Y had last year, then the number that each company will have this year can be found by

$$\left[\begin{array}{l} A \\ B \end{array}\right]=\left[\begin{array}{ll} \frac{3}{5} & \frac{1}{3} \\ \frac{2}{5} & \frac{2}{3} \end{array}\right]\left[\begin{array}{l} a \\ b \end{array}\right]$$

If company A has A=150,000 customers and company Y has B=120,000 customers this year, how many customers did each have last year?

If

$$A=\left[\begin{array}{rr}1 & 4 \\ 3 & -1 \\ -2 & 2\end{array}\right]$$

and

$$B=\left[\begin{array}{rrr}4 & 2 & 2 \\ -1 & 3 & 1\end{array}\right]$$

compute A B and B A, if possible.

Use Cramer’s Rule to solve the system, if possible. If it is not possible, state whether the system is inconsistent or has infinitely many solutions.

$$\left\{\begin{array}{l}x-2 y+z=0 \\ 2 x+y-2 z=2 \\ 3 x+2 y-3 z=2\end{array}\right.$$

The system do not have unique solution. Solve the system, if possible.

$$\left\{\begin{aligned} 3 x-9 y+4 z & =24 \\ 2 x-6 y+2 z & =6 \\ x-3 y+2 z & =20\end{aligned}\right.$$

The system do not have unique solution. Solve the system, if possible.

$$\left\{\begin{array}{rr}3 x-4 y-6 z= & 10 \\ 2 x-4 y-5 z= & -14 \\ x-z= & 0\end{array}\right.$$

If

$$A=\left[\begin{array}{ll}1 & 5 \\ 3 & 2\end{array}\right]$$

and

$$B=\left[\begin{array}{rr}2 & 3 \\ -1 & -2\end{array}\right]$$

, compute A B and B A, if possible.