We want to write the given
three by three matrix in reduced row echelon form.
So for review, these four conditions must be met
for a matrix to be in reduced row echelon form.
If the first three conditions are met,
the matrix would be in row echelon form.
So number one, the first non-zero element in each row,
called the leading entry, must be a one.
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.
And then finally four, in each column that contains
a leading entry or one, all other elements
of the column are zeros.
And here are two examples of three by three matrices
in reduced row echelon form.
Now let's begin performing row operations
to write this matrix in reduced row echelon form.
Let's begin by obtaining a zero in this position
as well as this position.
Now if we take a lot at row one and row three,
notice six times negative two
plus 12 would be zero.
So let's replace row one with six times row three
plus row one.
Now looking at row two and three,
notice how three times one plus negative three
would be zero, let's replace row three with
three times row three plus row two.
So replacing row one with six times row three
plus row one, we would have six times one
plus negative five, that's one.
Six times negative two plus 12 is zero.
And six times three plus negative 18 is zero.
Row two stays the same.
Now we'll replace row three with three times row three
plus row two.
So three times one plus negative three is zero.
Three times negative two plus seven is positive one.
Three times three plus negative 10 is negative one.
Now let's obtain a zero in this position here.
Notice three times one plus negative three
would be zero.
So let's replace row two with three times row one
plus row two.
We'll keep the other two rows the same for right now.
So row one and row three stay the same.
Now for row two, we'll replace row two with
three times row one plus row two.
So three times one plus negative three is zero.
Three times zero plus seven is seven.
Three times zero plus negative 10 is negative 10.
Now let's focus on row two and row three.
Let's obtain a zero here and a zero here.
Looking at row two and row three notice
how negative 10 times negative one plus negative 10
would be zero.
So let's replace row two with
negative 10 times row three plus row two.
And notice how negative seven times one plus seven
is zero.
So let's replace row three with negative seven times
row three plus row two.
So row one stays the same.
And we'll replace row two with negative ten times row three
plus row two, so negative ten times zero plus zero is zero.
Negative ten times one plus seven is negative three.
Negative ten times negative one is ten,
plus negative ten, that's zero.
Now we'll replace row three with
negative seven times row three plus row two.
So negative seven times zero plus zero is zero.
Negative seven times one plus seven is zero.
Then we have negative seven times negative one
which is seven, plus negative ten,
which is negative three.
Now we almost have this in reduced row echelon form,
we just need this element here and this element here
both to be positive one.
So let's replace row two with
negative 1/3 times row two,
and we'll replace row three with negative 1/3
times row three.
In reduced row echelon form,
the first row is one zero zero,
the second row would be zero one zero,
the third row would be zero zero one.
So notice how now the matrix is in
reduced row echelon form, the first entry in any
non-zero row is one.
The remaining elements in the same column are zero,
and each one is to the right of the one above it.
And notice in this case,
the given matrix in reduced row echelon form
is the three by three identity matrix
which tells us the given matrix,
this matrix here,
is invertible or nonsingular,
meaning this matrix here does have an inverse,
again because in reduced row echelon form,
we did get the identity matrix.
I hope you found this helpful.