We're given a three by three matrix,
and asked to write the matrix in reduced row echelon form.
For a quick review, these four conditions must be met
for a matrix to be in reduced row echelon form.
Number one, the first non-zero element in each row,
called the leading entry, must be a one.
Two, each leading entry, or one, is in a column to the right
of the leading entry, or one, in the previous row.
Three, rows with all zero elements, if any,
are below rows having a non-zero element.
Four, in each column that contains a leading entry,
or a one, all other elements of the column are zeros.
So here are two examples of three by three matrices
in reduced row echelon form.
Notice this three by three matrix is the identity matrix,
and this three by three matrix is not.
The reason I mention this now is because
if an N by N matrix A is invertible or nonsingular,
meaning it has an inverse,
then matrix A in reduced row echelon form
would be equal to the identity matrix,
and if the N by N matrix is non invertible, or singular,
meaning it does not have an inverse,
then matrix A in reduced row echelon form
would not be equal to the identity matrix.
Let's begin by performing row operations,
just write this matrix in reduced row echelon form.
Let's first obtain a zero in this position here.
Notice two times negative two plus four would be zero,
so let's replace row three with
two times row three plus row two.
Notice how all the elements in row two
have a common factor of two,
let's replace row two with 1/2 times row two.
Row one will stay the same.
Multiplying row two by one half would give us
two, five, negative five.
Now we're going to replace row three
with two times row three plus row two,
so two times negative two plus four, that's zero.
Two times negative three plus 10 is positive four,
and two times seven plus negative 10 is positive four.
Let's work on obtaining a zero in this position,
and this position.
Notice how there's a four here in row three,
and three times four is 12.
12 plus negative 12 would be zero.
So let's replace row one with
three times row three plus row one.
Looking at this two and this negative 10,
notice that five times two plus negative 10 would be zero.
Let's replace row two with five times row two plus row one.
Notice here we have two fours.
Let's replace row three with 1/4th times row three.
So again, row one is replaced with
three times row three plus row one,
so three times zero plus negative 10 is negative 10.
Three times four is 12 plus negative 12 is zero.
Three times four is 12 plus 38, that's 50.
Now we'll replace row two with
five times row two plus row one.
So five times two plus negative 10 is zero.
Five times five plus negative 12 is positive 13.
Five times negative five plus 38 is also 13.
Multiplying row three by 1/4th,
we will have zero, one, one.
I think we almost have it now.
Notice row one has a common factor of 10,
let's replace row one with negative 1/10th times row one,
so this first element is positive one.
Let's replace row two with 1/13th times row two,
and to obtain a zero in this position here,
let's use row two and row three, let's replace row three
with negative 13 times row three plus row two.
Notice negative 13 times one plus 13 is equal to zero.
Negative 1/10 times row one is one, zero, negative five.
1/13th times row two would be zero, one, one.
And now we'll replace row three with
negative 13 times row three plus row two,
so we'd have negative 13 times zero plus zero is zero.
Negative 13 times one plus 13 is zero.
And again, negative 13 times one plus 13 is zero.
Notice how this matrix is now in reduced row echelon form.
We have the row of zeros at the bottom,
the leading entry in row one and row two are ones,
the remaining elements in those columns are zero,
and the one in row two is
to the right of the one in row one.
Notice how this matrix, in reduced row echelon form,
does not equal the identity matrix,
which means the given matrix is not invertible, or singular,
meaning it does not have an inverse.
I hope you found this helpful.