How to solve augmented matrices work problem?

Question

 

The yearbook photographer sells sets of photos in three sizes (small, Medium, large). The price of each set includes a base price and the price for each size of print. The base price is twice the price of a large print. Find the base price and the price for each size of print.

SET A $19.75 : 1 large, 3 Medium, 5 small

SET B $ 32.75: 3 large, 5 Medium, 8 small

SET C $ 49.00: 5 large, 8 Medium, 15 small

Answer 1

Let the prices (in dollars) for each size of print be S, M, L for small, medium and large prints. 

Let B=base price in dollars. So for each set:

Set A: B+L+3M+5S=19.75

Set B: B+3L+5M+8S=32.75

Set C: B+5L+8M+15S=49

Also, B=2L, so the equations become:

3L+3M+5S=19.75,

5L+5M+8S=32.75,

7L+8M+15S=49.

To solve using matrices and Cramer's method, first evaluate the determinant, made up of the coefficients of the variables, Δ=

| 3 3   5 |

| 5 5   8 | = 3(75-64)-3(75-56)+5(40-35)=33-57+25=1.

| 7 8 15 |

Now evaluate this determinant, which replaces the L coefficients with the constants, Δ_L=

| 19.75 3   5 |

| 32.75 5   8 | = 19.75(75-64)-3(491.25-392)+5(262-245)=217.25-297.75+85=4.5.

| 49.00 8 15 |

From L=Δ_L/Δ=$4.50.

Similarly, Δ_M=

| 3 19.75   5 |

| 5 32.75   8 | = 3(491.25-392)-19.75(75-56)+5(245-229.25)=297.75-375.25+78.75=1.25.

| 7 49.00 15 |

M=Δ_M/Δ=$1.25.

And Δ_S=

| 3 3 19.75 |

| 5 5 32.75 | = 3(245-262)-3(245-229.25)+19.75(40-35)=-51-47.25+98.75=0.50.

| 7 8 49.00 |

S=Δ_S/Δ=$0.50.

B=2L=$9.

SOLUTION

Base price is $9; small photo=$0.50; medium photo=$1.25; large photo=$4.50.