### Video Transcript

By expressing the following
simultaneous equations as a matrix equation and finding the inverse of the
coefficient matrix, determine the values of π₯ and π¦ that solve the equations π₯
plus three π¦ equals negative four and two π₯ plus five π¦ equals 10.

In this question, we are given a
pair of simultaneous equations in two variables and asked to determine the solutions
of the system by finding the inverse of the coefficient matrix. To do this, we first recall that we
can write a pair of linear simultaneous equations as a matrix equation π΄π equals
π΅, where π΄ is the matrix of coefficients of the variables in the simultaneous
equations. In this case, itβs the two-by-two
matrix one, three, two, five.

π is the matrix of variables. It is a column matrix containing
the variables of the simultaneous equation. We have that π is the two-by-one
matrix π₯, π¦. Finally, matrix π΅ is a column
matrix containing the constants on the right-hand side of the simultaneous
equations. This is usually called the constant
matrix. We have that π΅ is the two-by-one
matrix negative four, 10.

If we were to evaluate the matrix
multiplication, we would see that the matrix equation only holds true when the
simultaneous equations are satisfied and vice versa. If π΄ is an invertible matrix, then
we can multiply on the left by the inverse of π΄ to obtain π equals π΄ inverse
times π΅. In other words, we can solve the
system by calculating the inverse of π΄ and then evaluating this expression. To find the inverse of the
coefficient matrix, we can recall that the inverse of a two-by-two matrix π, π,
π, π is given by one over ππ minus ππ times the two-by-two matrix π, negative
π, negative π, π, provided the determinant is nonzero.

To find the inverse of matrix π΄,
we first need to calculate its determinant, which we can do by finding the
difference in the product of its diagonals. We have the determinant of π΄ is
one times five minus three times two, which evaluates to give us negative one. We can now find the inverse of π΄
by using the formula for the inverse of a two-by-two matrix. We have that the inverse of π΄ is
equal to one over negative one times the two-by-two matrix five, negative three,
negative two, one. Evaluating this expression by
multiplying all of the elements of the matrix by negative one yields the two-by-two
matrix negative five, three, two, negative one.

We can now use the fact that π is
equal to π΄ inverse times π΅ to solve the system of equations. Substituting our matrices into this
equation gives us that π is equal to the two-by-two matrix negative five, three,
two, negative one times the column matrix negative four, 10.

We can now evaluate these products
by multiplying the elements in the rows of the first matrix with the corresponding
elements in the column of the second matrix and finding their sum. We get the following column matrix
for π. We can evaluate each entry
separately. This gives us the two-by-one matrix
50, negative 18. The first entry of this matrix is
the value of π₯, and the second entry is the value of π¦ that solves the
simultaneous equations. We can verify that these are the
solutions by substituting them back into the system. However, this is not necessary to
answer the question.

Therefore, we were able to show
that the solution to the given system of equations is π₯ equals 50 and π¦ equals
negative 18.