Video Transcript
A motorist bought 83 liters of
gasoline and six liters of oil for 190 pounds. A motorcyclist bought 22 liters of
gasoline and 20 liters of oil for 124 pounds. Assuming both paid the same price
per liter, use matrices to find the price of one liter of gasoline and one liter of
oil.
We’re asked to find the price of
one liter of gasoline and one liter of oil using matrices and given some information
about the gas and oil purchases of a motorist and a motorcyclist. Let’s begin by writing down the
information we’ve been given as two simultaneous equations. In order to do this, let’s let 𝑔
equal the price of gas per liter and ℎ equal the price of oil per liter.
We’re told that our motorist bought
83 liters of gasoline and six liters of oil for 190 pounds. So for our motorist, we have 83𝑔
plus six ℎ is 190. Our motorcyclist bought 22 liters
of gasoline and 20 liters of oil for 124 pounds so that for our motorcyclist, we
have 22𝑔 plus 20ℎ is 124.
So we’re off to a good start. We have two simultaneous equations
and two unknowns 𝑔 and ℎ. Making sure that our variables are
vertically aligned, which they are in this case, and our constants are on the
right-hand side, this allows us to read off our variable coefficients into a
coefficient matrix. The elements in the first row of
our coefficient matrix are the coefficients of 𝑔 and ℎ in equation one. And the elements in the second row
of our coefficient matrix are the coefficients of 𝑔 and ℎ in equation two.
Our coefficient matrix therefore
has elements 83, six, 22, and 20. We multiply this matrix by a column
matrix of our variables 𝑔 and ℎ. And this equates to a column matrix
of our constants on the right-hand side so that if we multiplied out the left-hand
side, and by quality of matrices, we would regain our equations one and two.
So we have an equation of the form
𝐴𝑥 equals 𝑏, and we want to find 𝑥. Note that an 𝑚-by-one matrix can
be called a column vector, hence our use of vector notation. And to solve this, we’re going to
use the fact that for any nonsingular matrix 𝐴, 𝐴 inverse times 𝐴 is equal to 𝐴
times 𝐴 inverse. And that’s equal to the identity
matrix. And recall that the identity matrix
is the matrix with leading diagonal elements one and all the rest zero.
So if we multiply through our
matrix equation on the left with 𝐴 inverse and we use the fact that 𝐴 inverse 𝐴
is equal to 𝐼, then our left-hand side simplifies to 𝑥 and our right-hand side is
𝐴 inverse times 𝑏. So if our matrix of coefficients is
𝐴, all we need to do is to find the inverse of 𝐴 and multiply 𝑏 by this. And this gives us our solution
𝑥.
Now recall that for a two-by-two
matrix with elements 𝑎, 𝑏, 𝑐, 𝑑, the inverse of 𝐴 is one over 𝑎𝑑 minus 𝑏𝑐
times the matrix with elements 𝑑 negative 𝑏 negative 𝑐 and 𝑎. And remember that 𝑎𝑑 minus 𝑏𝑐
is the determinant of 𝐴, which must be nonzero. And in the matrix, we’ve simply
swapped the elements 𝑎 and 𝑑 and taken the negatives of 𝑏 and 𝑐. So in our case, our matrix 𝐴 has
elements 83, six, 22, 20. Then our inverse is one over 83
times 20, that’s 𝑎 times 𝑑, minus six times 22, that’s 𝑏 times 𝑐, times the
matrix with elements 20, negative six, negative 22, and 83.
Our fraction evaluates to one over
1528. So that the inverse of our
coefficient matrix is one over 1528 times the matrix with elements 20, negative six,
negative 22, and 83.
Referring back to our equation
then, on the left-hand side we have our column matrix with elements 𝑔 and ℎ, that’s
our 𝑥. On the right-hand side, we have our
inverse matrix multiplying our column matrix of constants. And multiplying out our right-hand
side, we have 20 times 190 plus negative six times 124 and negative 22 times 190
plus 83 times 124. And this gives us one over 1528
times the column matrix with elements 3056 and 6112.
And making some room, this gives us
the column matrix with elements two and four. And by equality of matrices, that
means 𝑔 is equal to two and ℎ is equal to four so that if a motorist buys 83 liters
of gasoline and six liters of oil for 190 pounds and a motorcyclist bought 22 liters
gasoline and 20 liters of oil for 124 pounds, gasoline is two pounds per liter and
oil is four pounds per liter.
