### Video Transcript

Consider the simultaneous equations
four π₯ minus two π¦ is equal to zero and three π¦ plus five π₯ is negative 11. Express the given simultaneous
equations as a matrix equation. Write down the inverse of the
coefficient matrix. And multiply through by the inverse
on the left-hand side to solve the matrix equation.

There are three parts to this
question involving the set of simultaneous equations four π₯ minus two π¦ is equal
to zero and three π¦ plus five π₯ is negative 11. And these three parts lead us to
the solution of the equations via matrix methods. The first part is to express the
simultaneous equations as a matrix equation. We must then write down the inverse
of the coefficient matrix and use this by multiplying through on the left to solve
the matrix equation. So letβs begin with the first part,
which is to write the equations as a matrix equation.

The first thing we need to do is to
make sure that our π₯βs and π¦βs are vertically aligned on the left-hand side. In our second equation then, weβll
need to swap the three π¦ and the five π₯. And now our π₯βs and π¦βs are
vertically aligned. Letβs call our equations equations
one and two so that equation one is four π₯ minus two π¦ is equal to zero and
equation two is five π₯ plus three π¦ is negative 11. And this helps to read off our
coefficients so we can put these into a two-by-two matrix, which then multiplies a
column matrix of our variables π₯ and π¦. And we put this equal to the
constants on the right-hand side.

The first row of our coefficient
matrix contains the coefficients of π₯ and π¦ in equation one. That is four and negative two. Our associated constant on the
right-hand side is zero. The second row of our coefficient
matrix contains the constant coefficients of π₯ and π¦ in equation two. That is five and three. And our constant element on the
right-hand side is negative 11. So now we have our equations in
matrix equation form as required.

The second part of the question
asks us to write down the inverse of the coefficient matrix. And to do this, we recall that for
a nonsingular two-by-two matrix with elements π, π, π, π, the inverse of π΄ is
one over ππ minus ππ times the matrix with elements π, negative π, negative
π, and π. Recall that ππ minus ππ is the
determinant of π΄. And notice that weβve swapped the
elements π and π and taken the negative of π and π. In our case, our coefficient matrix
has elements four, negative two, five, and three. So its inverse, if π is four, π
is negative two, π is five, and π is three, is one over four times three minus
negative two times five, that is one over ππ minus ππ, times the matrix with
elements three, two, negative five, and four. The inverse of our coefficient
matrix is therefore one over 22 times the matrix with elements three, two, negative
five, and four.

Our final part is to multiply
through by the inverse on the left-hand side to solve the matrix equation. If we call our coefficient matrix
π΄, we have π΄ inverse times π΄ times the column matrix π₯ is equal to π΄ inverse
times the column matrix π, where π₯ is the matrix of variables and π is the matrix
of constants on the right-hand side. Remember, though, that for any
nonsingular matrix π΄, that is, a matrix with an inverse, π΄ inverse times π΄ is
equal to the identity matrix, which for a two-by-two matrix is the matrix with
elements one, zero, zero, one. So that in our left-hand side, we
have the identity matrix times π₯, π¦ and on our right, we have π΄ inverse times the
column matrix π.

Our left-hand side simplifies to
the column matrix π₯, π¦. And if we multiplied the right-hand
side, we have one over 22 times the matrix with elements three times zero plus two
times negative 11, negative five times zero plus four times negative 11. That is one over 22 times the
column matrix with elements negative 22, negative 44. Making some space and evaluating
this gives us the column matrix with elements negative one and negative two. By equality of matrices, this then
gives us π₯ is equal to negative one and π¦ is equal to negative two.

So for the simultaneous equations
four π₯ minus two π¦ is equal to zero and three π¦ plus five π₯ is equal to negative
11, we have the matrix equation where the coefficient matrix has elements four,
negative two, five, three multiplying the column matrix of variables equal to the
column matrix with elements zero, negative 11 of constants on the right-hand
side. Our matrix of coefficients has an
inverse, which is one over 22 times the matrix with elements three, two, negative
five, and four. And we use this to find our
solution: π₯ is negative one and π¦ is negative two.