### Video Transcript

Consider the system of
equations. Express the system as a single
matrix equation.

Remember, a matrix equation can be
used to help us solve systems of linear equations. It will be of the form π΄π± equals
π, where π΄ is called the coefficient matrix, π± is a column vector containing the
variables of our equation, and π contains all of the constants in the equation. So letβs identify first the
coefficient matrix. Thatβs the matrix that weβre going
to call π΄. We have three equations with three
variables. And so matrix π΄ is going to be a
three-by-three matrix. The elements in the first row of
matrix π΄ are the coefficients of π, π, and π, respectively in our first
equation.

We see that the coefficients are
two, two, and four. And so that is the first row of our
matrix. The elements in the second row of
matrix π΄ are the coefficients of π, π, and π in our second equation. Those are negative one, negative
one, and negative one. And so those are the elements in
the second row of our matrix. Weβre going to repeat this process
for the third row of our matrix, looking for the coefficients of π, π, and π in
our third equation. We now see that those are two,
five, and six. And so weβve identified our
coefficient matrix. Itβs two, two, four; negative one,
negative one, negative one; and two, five, six.

Weβre then going to look at the
vector π±. Itβs a variable vector. And so we need to list the
variables in our equation. Those are π, π, and π. And it is in fact really important
that we write them in that order. We need it to be the case that when
we multiply the matrix π΄ by the column vector π±, we end up with the original
expressions two π plus two π plus four π, negative π negative π negative π,
and so on. If we were to mix these variables
around, weβre not going to end up with the exact same equation.

Finally, weβre interested in the
constant vector π that contains all of the constants from our equations. Once again, we need to give them in
the correct order. And so we find vector π is four,
14, 10. We write this as a single matrix
equation by putting it back into the form π΄π± equals π. And when we do, we see that the
matrix equation we need is two, two, four; negative one, negative one, negative one;
two, five, six times the vector π, π, π equals the vector four, 14, 10.