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## Linear algebra

### Course: Linear algebra > Unit 2

Lesson 6: More determinant depth- Determinant when row multiplied by scalar
- (correction) scalar multiplication of row
- Determinant when row is added
- Duplicate row determinant
- Determinant after row operations
- Upper triangular determinant
- Simpler 4x4 determinant
- Determinant and area of a parallelogram
- Determinant as scaling factor

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# Determinant when row multiplied by scalar

The determinant when a row is multiplied by a scalar. Created by Sal Khan.

## Want to join the conversation?

- At5:31Why is the cofactor determinant of f equal to (dh-eg) and not (ah-bg)? I might have misunderstood the pattern.(57 votes)
- You are correct; Sal made a mistake there.(43 votes)

- Do the same rules for the determinant when a row is multiplied by a scalar apply to when a column is multiplied by a scalar?(8 votes)
- Yes. If you transpose a matrix its determinant doesn't change so you can consider multiplying a column by a scalar as first transposing the matrix, then multiplying the equivalent row by the scalar.(7 votes)

- Do these rules apply for a m by n matrix?(4 votes)
- The determinant isn't defined for non-square matrices because non-square matrices aren't invertible in the traditional sense.(13 votes)

- at ~13:20, sal is writing out 3rd/final term of the 3x3 determinant |A|...but shouldn't it be "-f(ah-bg)" vs "-f(dh-eg)" ?(5 votes)
- Yes. Transcription error from a few minutes in. [I'm writing words to pass the comment screener.](1 vote)

- Can anyone tell me why in 3*3 matrix determinant is of order + - + -...? What is its proof? I am talking about the cofactor. Where does the cofactor emerge from?(2 votes)
- How would we use the Anxn formula in finding the determinant for example a 4x4 matrice? Can somebody tell me what values to place in i and j? And how will we find the determinant of Aij in this case?(2 votes)
- How to multiply 4m+ 3n = 1 by 8/x - 9/y = 7 ?(2 votes)
- Okay,

If we make that substitution we get the two equations 4m+3n=1 and 8m-9n=7. Lets solve it by solving the first equation for m to get m=(1-3n)/4 and then substitution that into the second to get 8((1-3n)/4)-9n=7. This simplifies to 2-15n=7. Solving for n gives us n=-1/3. Then we remember we have m=(1-3n)/4=1/2.(1 vote)

- At5:59, shouldn't he have written

| a b |

| g h |?(2 votes) - is there any general formula ?(1 vote)
- Sal's method of "minors and cofactors" is I guess a general formula. Based on two previous videos,

https://www.khanacademy.org/math/linear-algebra/matrix-transformations/inverse-of-matrices/v/linear-algebra-nxn-determinant, and https://www.khanacademy.org/math/linear-algebra/matrix-transformations/inverse-of-matrices/v/linear-algebra-determinants-along-other-rows-cols,

the (recursive) general formula for expanding along the ith row or jth column using minors and cofactors is:

|A| (expanding along row i) = sum(j=1,n) [(-1)^(i+j)*a_ij*A_ij], or

|A| (expanding along column j) = sum(i=1,n) [(-1)^(i+j)*a_ij*A_ij].

This gives you n products of elements ("a_ij") of A and their minors = (n-1)x(n-1) determinants "A_ij" (see the first video above) which can be solved by the same equations above and each yield n-1 products of elements and (n-2)x(n-2) minors, and so on to total n-2 recursions, until the minors are 2x2 and you get an answer.

There is another general formula using the definition of a determinant, which takes more time, and also explains more about what a determinant actually is and why "minors and cofactors" works.(2 votes)

- What does j mean? Is it he column?(1 vote)
- j is a column in
**A**just like i is a row in**A**, so yes.(1 vote)

## Video transcript

Let's explore what happens to
determinants when you multiply them by a scalar. So let's say we wanted to find
the determinant of this matrix, of a, b, c, d. By definition the determinant
here is going to be equal to a times d minus b times c, or
c times b, either way. ad minus bc. That's the determinant
right there. Now what if we were
to multiply one of these rows by a scalar. Let's say we multiply it by k. So we have the situation a, b. Let's multiply the second
row times k. So we kc and kd. Now what's the determinant
going to look like? We're going to have a times
kd, or we could just write that as kad, minus kc times b,
or we could right that as kbc. If we factor out the k,
we get that equals k times ad minus bc. Now you immediately see that
this thing is the same thing as this thing. So this is equal to k times the
determinant of a, b, c, d. So when you just multiply a
row by a scalar-- just one row, not the entire matrix--
when you just multiply a row by some scalar, the resulting
determinant will be the original determinant
times that scalar. Now you might say well what
happens if I multiply the whole matrix times
that scalar. Well that's equivalent to
multiplying by a scalar twice. Let's say I have the matrix A,
and the matrix A is equal to a, b, c, d. If I were to think about the
matrix kA, now I'm not just multiplying one row. I'm multiplying the whole
matrix by a scalar. This is going to be equal
to ka, kb, kc, and kd. When you figure out its
determinant, the determinant of k times A is going to be
equal to the determinant of ka, kb, kc, and kd. Here you're immediately going
to end up with k squared terms. You're going to have
k squared times ad. It's going to be equal to k
squared ad minus k squared bc, or k squared times ad minus
bc, or k squared times the determinant of just A. So you have to be
very careful. This is only for a
two by two case. You'll find out if this was an
n by n matrix that this would have been k to the n. So the take away is the only way
you can say that this is going to be some scalar multiple
times your original determinant, is only if you
multiply one row times that scalar multiple, not
the whole matrix. Let's see how this extends
to maybe a 3 by 3 case. You might say, hey Sal, you just
picked the second row. Does it work with
the first row? I'll leave that for you to
determine, but it does. It does work. It doesn't matter which row
I multiplied it by. Let's take the three
by three case. Let's say we have some matrix. Let's call this A again. I'm redefining A. It's going to be a, b,
c, d, e, f, g, h, i. Then if you take its
determinant, the determinant of A is going to be equal to--
we can do it a couple of different ways. But I'll just pick some
arbitrary row. Because that's the row that
we're going to multiply by some scalar. So let's just take that
row right there. Remember the plus
minus pattern. Remember plus, minus, plus,
minus, plus, minus, plus, minus, plus, that little
checkerboard pattern. So d is a minus right there. So it's going to be equal to
minus d times the determinant of its submatrix. So you cross out that
column in that row. It's b, c, h, i. It's going to be plus e times
its submatrix a, c, g, i. It's going to be minus f times--
you get rid of that row in that column--
d, e, g, h, the determinant of d, e, g, h. That's the determinant
of this matrix A. Now what if we define some
new matrix here? Let's call it A prime. Let me make this scroll
down a little bit. Let me define A prime
right here. A prime, I'm just going to
multiply this row by a scalar. So it's going to be equal to
a, b, c, kd, ke, and kf. I'm not multiplying the whole
matrix times the scalar. I can't say this is kA. I'm just multiplying
one of its rows. Then I have g, h, and i. So what's the determinant
of A prime going to be? I put that prime there. So it's different than A,
but it's derived from A. I just multiplied one row
of A times the scalar. Well I can go along that same
row that I did up here. I can go along that same row. The only difference is is that
instead of having a d, I now have a kd. Instead of an e, I
now have a ke. So instead of a d, I'm going
to have a kd there. Instead of an e, I'm going
to have an ke there. So it's going to be this exact
same thing, but I can replace this guy, this guy, and this guy
with them multiplied by k. So it's going to be equal to
minus kd times the determinant of the submatrix b, c, h, i--
I'm not even look over here because it's going to be the
same thing as that one up there-- plus ke times the
determinant of a, c, g, i, minus kf times the determinant
of d, e, g, h. What is this equal to? This is equal to, if you just
factor out the k, it's equal to k times this. So it's equal to k times
the determinant of A. So our result also worked
for three by three case. I just happened to pick
the middle row. But I encourage you
to pick other rows and to see what happens. So let's actually do it
for the general case. Because I've just been giving
you particular examples, and I like to show you the general
proof when the general proof isn't too harry. So let's say I have
an n by n matrix. Let's say that I
have matrix A. Let's say that A is n by n. You can write it like this. This is the first row. This is the first column a11,
a12, all the way to a1n. I'm going to pick some arbitrary
row here that I'm going to end up multiplying
by a scalar. So we can go down here. Let's say row ai, so this is
ai1, ai2, all the way to ain. This is some row that I'm going
to use to determine the determinant. Remember we can go to any row
to get the determinant. Then finally you keep going. You get an1, an2, all
the way to ann. This is as general as you can
get for an n by n matrix. Now let's figure out its
determinant, so the determinant of A. I'm just going to go down
this row right there. So the determinant of
A is equal to what? Well we have to remember the
checkerboard pattern. We don't know where we are on
the checkerboard pattern because I just picked an
arbitrary general row here. But we can use the general
formula that the sign is going to be determined by negative 1
to the i-- I don't know if i is even or odd-- so it's
going to be i plus for this term 1 power. That's it's sign. This is what gives us the
checkerboard pattern. Let me make that clear. It looks complicated, but this
is just a checkerboard. Times this term right there,
so times ai, so the coefficient, ai1, and then times
this guy's submatrix. You remember the submatrix. You get rid of this row, and
this column is going to be everything that's left over. So times that submatrix
of ai1. Then it's going to be plus-- we
just keep doing it-- plus negative 1 to the i plus
2 times ai2 times its submatrix-- all the way, you
just keep going-- plus minus 1 to the i plus n times ai. You're in the nth column,
and then its submatrix. This is going to be an n minus
1 by n minus 1 matrix. All of these are going to be. Just like that, that's
the determinant of A. We could actually rewrite
it in sigma notation. That'll simplify things
a little bit. So the determinant of A, we can
rewrite it as the sum from j is equal to 1 to j-- I'll
write it explicitly here-- j is equal to n of negative 1 to
the i plus j times aij, and then each of the submatrices,
Aij. This thing right here is just
another way of writing this thing I wrote up there. I'm just saying the sum. You just take j equal 1,
put them in there. You get this term right there. You take j equal
2, you add it. You get this term right there. You keep doing it. You get j equal n. You get that term right there. So these two things
are equivalent. So what happens I have
some new matrix? Let me copy and paste
my current matrix. So let me copy and paste it. Actually let me copy and
paste everything. That'll make things
move quickly. I copied it. Now let me paste it
just like that. Let me define a new
matrix A prime. It's still an n by n matrix. But that row that I just
happened to use to determine my determinant, I'm going to
multiply it by a scalar k. So it's k ai1, k ai2 k
ain just like that. So what's the determinate
of A prime? Well we're just going to
go down this row again. But now instead of just an ai1,
we have a k ai1 Instead of an ai2, we have a
k ai2 Instead of an ain we have a k ain. So it's determinant is just
going to be this same thing, but instead of an aij
everywhere, we're going to have a k aij. So this is the determinant
of A prime. We could just take out this
constant right here. It has no i's or j's in it. I has no j's in it in
particular, so we can just take it out. So it's equal to k times the sum
from j is equal to 1 to j is equal to n of minus 1 to
the i plus j times aij. This is the coefficient. This is the submatrix for each
of those coefficients, aij. That's a matrix right
there, an n minus 1 by n minus 1 matrix. Then you immediately recognize--
I think you saw where this was going-- this
right here is just the determinant of A. So we get the result that the
determinant of A prime is equal to k times the
determinant of A. So we've just shown you in
general, if you have any n by n matrix, if you multiply only
one row, not the whole matrix, only one row by some scalar
multiple k, the resulting determinant will be your
original determinant times k. Now I touched on this in
the original video. What is the determinant
of k times A? So now we're multiplying
every row times k. Or another way to think about
is you're multiplying n rows times k. So you're doing this n times. So if you multiply
k times itself n times, what do you get? You get k to the n. So this is going to be equal
to k to the n times the determinant of A. If you just do it once,
you get k times the determinant of A. Now if you do a second row,
you're going to get k times k times the determinant of A. If you do a third row, you're
going to get k to the third times the determinant of A. The fourth row, k to the
fourth times the determinant of A. If you do them all, all n rows,
you're going to have k to the n times the
determinant of A. Anyway, hopefully you found
that interesting. I encourage you to experiment
with these other ways. Try going down a column and
seeing what happens.