For what value of k
will the linear system be consistent?
Remember, a system is consistent
if it has one solution or infinite solutions.
Looking at the system below,
notice k is the constant in equation two.
Because we have three equations with two unknowns,
if we write the augmented matrix
in reduced row echelon form,
we must have a row of zeros in row three
in order for the system to be consistent.
So, let's begin with the augmented matrix
for the given system, and then write the augmented matrix
in reduced row echelon form.
Notice equation one gives us the row one, three, four.
Equation two gives us the road nine, one, k.
Equation three gives us the row negative three, seven, two.
Let's begin by obtaining a zero in row two, column one,
as well as row three, column one.
Looking at row two and row three,
notice three times negative three plus nine is zero.
And therefore, we will replace row two
with three times row three plus row two.
And now, looking at row one and row three,
notice three times one plus negative three is zero.
Let's also replace row three
with three times row one plus row three.
So, we keep row one the same for now.
Now, replacing row two
with three times row three plus row two.
We begin with three times negative three,
which is negative nine, plus nine, which is zero.
Next, we have three times seven,
which is 21, plus one, which is 22.
And then, we have three times two, which is six,
plus k, giving us six plus k.
And now, we replace row three
with three times row one plus row three.
Three times one plus negative three is zero.
Three times three is nine, plus seven, is 16.
And three times four is 12, plus two, is 14.
For the next step, let's make the entry
in row two, column two a one,
by replacing row two with 1/22 times row two.
Row one and row three remain the same.
Multiplying row two by 1/22, we have zero, one.
And then, 1/22 times the quantity six plus k
is the quantity six plus k divided by 22.
And now, let's obtain a zero in row three, column two.
Looking at row two and row three,
notice negative 16 times one plus 16 is zero.
Let's replace row three with negative 16
times row two plus row three.
We will keep our a one and row two the same.
Replacing row three with negative 16 times row two
plus row three, we have negative 16 times zero,
plus zero, which is zero.
Negative 16 times one plus 16 is zero.
And then, we have negative 16
times the quantity six plus k divided by 22, plus 14.
Let's work this out below.
The element in row three, column three,
is equal to, again, we have negative 16,
or negative 16 over one,
times the quantity six plus k over 22,
and then plus 14, or plus 14 over one.
Notice we can simplify before multiplying here.
Negative 16 and 22 share a common factor of two.
22 simplifies to 11,
negative 16 simplifies to negative eight.
Notice now, we have a denominator of 11.
To write 14 as a fraction with the denominator of 11,
we multiply the numerator and denominator by 11.
So, the entire expression is going to be over 11.
In the numerator, we distribute negative eight.
Negative eight times six is negative 48,
and then we have negative eight times k,
giving us minus eight k.
And then, plus 14 times 11 is equal to 154.
So, simplifying one more time in the numerator,
negative 48 plus 154 is equal to 106.
The numerator is negative eight k plus 106.
And now, we almost have reduced row echelon form.
We just need a zero in row one, column two.
Looking at row one and row two,
notice negative three times one plus three is zero.
We will now replace row one
with negative three times row two plus row one.
To begin, we have negative three times zero plus one,
which is one.
And then, we have negative three times one,
plus three, which is zero.
And then, we have negative three
times the quantity six plus k over 22, plus four.
Again, let's work this out below.
The element in row one, column three, is equal to,
again, negative three, or negative three over one,
times the quantity six plus k over 22,
and then plus four, plus four over one.
For this first product, nothing simplifies.
So, the common denominator's going to be 22.
Let's write four over one as an equivalent fraction,
with a denominator of 22.
So, we have a common denominator of 22 now,
and the numerator is going to be,
distributing negative three,
we have negative 18 minus three k, and then plus 88.
Simplifying, negative 18 plus 88 is 70.
We have negative three k plus 70, all over 22.
And row two and row three remain the same.
Notice now we do have the augmented matrix
in reduced row echelon form.
And again, from here, because we have three equations
with two unknowns, in order for the system to be consistent,
the third row must all be zeros.
And therefore, to find the value of k,
we set the quantity negative eight k plus 106
all divided by 11 equal to zero, and solve for k.
If the fraction equals zero,
we should recognize the numerator has to be zero.
But if we don't include the fraction
by multiplying both sides by 11,
but again, notice how on the right side we still have zero.
So, we have negative eight k plus 106 equals zero.
Subtracting 106 on both sides,
we have negative eight k equals negative 106.
Dividing both sides by negative eight,
simplifying, we have k is equal to positive 106
divided by eight, but this does simplify.
Both have a common factor of two,
giving us 53/4.
So, now we know in order for the system to be consistent,
k must equal 53/4.
Let's take a look at what this actually means.
So, we just found the value of k.
So, we know the second equation must be nine x sub one
plus x sub two equals 53/4.
If we replace x sub one with x and x sub two with y,
we can graph the linear equations on the coordinate plane,
which I've done here on the right.
Notice how we have all three equations,
and all three lines intersect in one point,
which is the solution to the system,
verifying the system is consistent.
Without the graph, if we go back to the augmented matrix
in reduced row echelon form, which is this matrix here,
we know x sub one is equal to the quantity negative three k
plus 70, all over 22.
And x sub two is equal to six plus k divided by 22.
To find the values of x and y by hand,
we would now substitute 53/4 for k
in order to find x sub one and x sub two,
which I have shown below.
Notice how the solutions do match
what we found graphically on the right.
I hope you found this helpful.