The matrix A is given
as well as the reduced form in row-echelon form.
Rest to find a basis for the null space of A.
To find a basis for the null space of A,
we solved A times vector X equals a 0 vector,
the vectors in the spanning set of these solution
form a basis for the null space.
So notice how we only have matrix A in row-echelon form.
If they weren't the augmented matrix for the equation,
A times vector X equals 0 in row-echelon form,
we would have a fifth column of 0s
I'm going to go ahead and include the fifth column of 0s
for the augmented matrix.
So again, now we have the augmented matrix
in row-echelon form for the equation,
A times vector X equals to 0 vector.
Now we could continue
and write the augmented matrix in reduced row-echelon form,
let's go ahead and leave it in this form.
Notice how we have pivots in row one column one,
row two column three,
and row three column four,
which indicates that x sub 1, x sub 3, and x sub 4
are the basic variables
and x sub 2 is a free variable.
And now, let's go ahead
and write the corresponding equations for each row.
The first one indicates a negative 4x sub 1
plus x sub 2 minus 3x sub 3
minus 4x sub 4 equals 0.
The second row indicates that negative x sub 3
minus 2x sub 4 equals 0.
The third row indicates a negative x sub 4 equals 0.
And because x sub 2 is a free variable,
we include x sub 2 equals x sub 2.
Because x sub 2 is a free variable,
we need to solve x sub 1, x sub 2, x sub 3
in terms of x sub 2,
but notice how we have negative x sub 4 equal 0,
so we know x sub 4 is equal to 0.
If x sub 4 is equal to 0 in the first equation,
minus 4x sub 4 is 0 in the second equation,
minus 2x sub 4 is 0,
and then simplifying the second equation,
notice how we have negative x sub 3 equals 0,
which indicates that x sub 3 equals 0.
And if x sub 3 is equal to 0 in the first equation,
minus 3x sub 3 is 0,
and the fourth equation remains x sub 2 equals x sub 2,
and now let's solved the first equation for x sub 1.
Let's show some work on this.
We have negative 4x sub 1 plus x sub 2 equals 0.
First step, subtract x sub 2 on both sides,
which gives us negative 4x sub 1 equals negative x sub 2.
To solve for x sub 1,
we divide both sides by negative 4
or multiply both sides by negative 1/4.
Simplifying, we have x sub 1 equals 1/4x sub 2.
Let's go ahead and prioritize the solution
by letting the free variable x sub 2 equal t.
So if x sub 2 is equal to t,
x sub 1 is equal to 1/4t,
x sub 2 is equal to t,
x sub 3 is equal to 0,
and x sub 4 is equal to 0.
Which means the solutions to the equation,
A times vector X equals the 0 vectors equal to vector X,
which has components x sub 1 through x sub 4.
Where one more time,
x sub one is equal to 1/4t,
x sub 2 is equal to t,
x sub 3 is equal to 0,
x sub 4 is equal to 0.
Let's go ahead and factor out the t,
giving us t times
the vector 1/4, 1, 0, 0.
Since the solution to A times vector X equal to 0 vector
is x equals t times the vector 1/4, 1, 0, 0,
and the vectors in the spanning set of these solution
form my basis for the null space
and basis for the null space is the vector 1/4, 1, 0, 0.
I hope you found this helpful.