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Linear Algebra: nxn Determinant

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    So far we've been able to define
    the determinant for a
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    2-by-2 matrix.
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    This was our definition right
    here: ad minus bc.
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    And then we were able to broaden
    that a bit by creating
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    a definition for the determinant
    of a 3-by-3
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    matrix, and we did that right
    here, where we essentially
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    said the determinant is equal
    to each of these terms-- you
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    could call these maybe the
    coefficient terms-- times the
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    determinant of the matrix-- you
    can kind of view it as the
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    submatrix produced-- when you
    get rid of each of these guys'
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    column and row.
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    So when you got rid of this
    guy's column and row, you're
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    left with this matrix.
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    So we said this guy times
    the determinant of this.
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    And we kept switching signs,
    minus this guy times the
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    determinant, if you move
    his column and his row.
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    So it was left with these terms
    right there to get that
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    determinant.
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    Then finally, you switched
    signs again.
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    So plus this guy times the
    determinant of the 2-by-2
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    matrix if you get rid of this
    row and this column.
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    So this thing right here,
    which was this matrix.
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    Now let's see if we can
    extend this to a
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    general n-by-n matrix.
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    So let's write out our n-by-n
    matrix right over here.
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    I'll do it in blue.
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    So let's say I have some matrix
    A that is an n-by-n
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    matrix, so it's going
    to look like this.
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    This would be a11, that would
    be a12, and we would go all
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    the way to-- you're going
    to have n columns, a1n.
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    And when you go down, this is
    going to be your second row:
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    a21, and it's going to go all
    the way down to an1, because
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    you have n rows as well.
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    And then if you go down the
    diagonal all the way, this
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    right here would be ann.
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    So there is my n-by-n matrix.
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    Now, before I define how to find
    the determinant of this,
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    let me make another
    definition.
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    Let me define-- so this
    is my matrix A.
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    Let me define a submatrix Aij to
    be equal to-- see this is n
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    by n, right?
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    So this is going to be an n
    minus 1 by n minus 1 matrix.
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    So if this is 7 by 7, the
    submatrix is going to be 6 by
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    6, one less in each direction.
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    So this is going to be the n
    minus 1 by n minus 1 matrix
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    you get if you essentially
    ignore or if you take away--
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    maybe I should say take away.
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    Let's say ignore, like
    the word ignore.
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    If you ignore the i-th row, this
    right here is the row,
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    the i-th row and the
    j-th column of A.
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    So, for example, let's go back
    to our 3 by 3 right here.
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    This thing could be denoted
    based on that definition I--
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    we could have called
    this, this was a11,
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    this term right here.
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    We could denote the matrix when
    you get rid of the first
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    column and the first row or the
    first row and the first
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    column, we could call this thing
    right here, we could
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    call that big matrix A11.
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    So this was big matrix A11.
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    This is big matrix A21, or
    actually, this matrix was
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    called C, so this would
    be C11 right there.
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    We could call this one, this
    would be matrix C12.
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    Why is that?
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    Because if you get rid of the
    first row, let me get rid of
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    the first row, right?
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    The first term is your row.
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    If you get rid of the first
    row and the second column,
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    this is the matrix that's
    left over: 2, 3, 4, 1.
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    So this is this guy
    and this guy.
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    2, 3, 4, 1.
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    So this is the submatrix C
    because this is the big matrix
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    C, But this one is C12.
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    I know it's a little
    bit messy there.
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    So that's all we mean
    by the submatrix.
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    Very similar to what we did
    in the 3 by 3 case.
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    You essentially get rid of--
    so if you want to find out
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    this guy's submatrix, you would
    call that A11, and you
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    would literally cross out the
    first row and the first
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    column, and everything
    left over here
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    would be that submatrix.
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    Now, with that out of the way,
    we can create a definition,
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    and it might seem a little
    circular to you at first, and
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    on some level it is.
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    We're going to define the
    determinant of A to be equal
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    to-- this is interesting.
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    It's actually a recursive
    definition.
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    I'll talk about that
    in a second.
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    It's equal to-- we start
    with a plus.
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    It's equal to a11 times the
    submatrix if you remove this
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    guy's row and column.
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    So that, by definition, is
    just A, big capital A11's
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    determinant.
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    So that's exactly what we did.
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    Let me write that a
    little bit neater.
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    So times the determinant of
    its submatrix, so the
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    determinant of A11.
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    So you take A11, you get rid of
    its column and its row or
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    its row and its column, and
    everything else, you find the
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    determinant of that.
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    Actually, let me write it
    in terms of-- let me
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    write it this way.
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    a11 times the determinant
    of the submatrix A11.
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    And then we switch signs.
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    We're just going to go along
    this row, and then you do
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    minus a12 times the determinant
    of its submatrix,
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    which we'll just call A12.
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    We would get rid of this row
    and this column, and
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    everything left would
    be this matrix A12.
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    We want to find its
    determinant.
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    And then we'll take the next
    guy over here would be a13.
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    So we switch signs with minus.
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    Now, you go plus,
    so a13 times the
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    determinant of its submatrix.
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    So if this is n by n, these each
    are going to be n minus 1
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    by n minus 1.
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    So the determinant of A13.
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    And you're just going to keep
    doing that, keep switching
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    signs, so it's going to be a
    minus and then a plus and you
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    keep going all the way--
    and then I don't know.
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    It depends on whether an,
    whether we're dealing with an
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    odd number or an even number.
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    If we're dealing with an even
    number, this is going to be a
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    minus sign.
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    If it's an odd number, it's
    going to be a plus sign, but
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    you get the idea.
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    It's either going to be a plus
    or a minus, not just-- if it's
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    odd, this is going
    to be a plus.
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    If it's an even n, it's going to
    be a minus, All the way to
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    a1n, the n-th column times
    its submatrix, A1n.
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    With that submatrix, you get rid
    of the first row and the
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    n-th column, and it's going
    to be everything
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    that's left in between.
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    And you immediately might
    say, Sal, what kind of a
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    definition is this?
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    You defined a determinant for an
    arbitrary n-by-n matrix in
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    terms of another definition
    of a determinant.
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    How does this work?
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    And the reason why this works
    is because the determinant
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    that you use in the definition
    are determinants
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    of a smaller matrix.
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    So this is a determinant of an n
    minus 1 by n minus 1 matrix.
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    And you're saying hey, Sal, that
    still doesn't make any
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    sense because we don't know how
    to find the determinant of
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    an n minus 1 by n
    minus 1 matrix.
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    Well, you apply this definition
    again, and then
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    it's going to be in terms of n
    minus 2 times n-- or n minus 2
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    by n minus 2 matrices.
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    And you're like how
    do you do that?
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    Well, you keep doing it, and
    you're going to get all the
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    way down to a 2-by-2 matrix.
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    And that one we defined well.
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    We defined the determinant of
    a 2-by-2 matrix not in terms
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    of a determinant.
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    We just defined it in terms of
    a times-- we defined it as--
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    let me write it up here.
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    It was a times d minus
    b times c.
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    And you can see.
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    I mean, we could just go down to
    the 3 by 3, but the 2 by 2
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    is really the most fundamental
    definition.
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    And you could see that the
    definition of a 3-by-3
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    determinant is a special
    case of the general
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    case for an n by n.
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    We take this guy and we
    multiply him times the
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    determinant of his submatrix
    right there.
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    Then we take this guy where
    we switch signs.
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    We have a minus.
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    And we multiply him times the
    determinant of his submatrix,
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    which is that right there.
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    Then you do a plus.
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    You switch signs and then you
    multiply this guy times the
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    determinant of his submatrix,
    which is that right there.
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    So this is a general case
    of what I just defined.
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    But we know it's never that
    satisfying to deal in the
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    abstract or the generalities.
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    We want to do a specific case.
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    And actually, before I do that,
    let me just introduce a
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    term to you.
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    This is called a recursive
    formula.
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    And if you become a computer
    science major,
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    you'll see this a lot.
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    But a recursive function or a
    recursive formula is defined
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    in terms of itself.
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    But the things that you use in
    the definition use a slightly
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    simpler version of it, and as
    you keep going through, or you
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    keep recursing through it, you
    get simpler and simpler
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    versions of it until you get
    to some type of base case.
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    In this case, our base case is
    the case of a 2-by-2 matrix.
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    You keep doing this, and
    eventually you'll get to a
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    determinant of a 2-by-2 matrix,
    and we know how to
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    find those.
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    So this is a recursive
    definition.
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    But let's actually apply it
    because I think that's what
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    actually makes things
    concrete.
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    So let's take-- this is going
    to be computationally
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    intensive, but I think if we
    focus, we can get there.
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    So I'm going to have a 4-by-4
    matrix: 1, 2, 3, 4.
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    1-- throw some zeroes in there
    to make the computation a
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    little bit simpler, 0, 1, 2,
    3, and then 2, 3, 0, 0.
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    So let's figure out this
    determinant right there.
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    This is the determinant
    of the matrix.
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    If I put some brackets
    there that would
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    have been the matrix.
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    But let's find the determinant
    of this matrix.
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    So this is going to be equal
    to-- by our definition, it's
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    going to be equal to 1 times the
    determinant of this matrix
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    right here if you get rid of
    this row and this column.
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    So it's going to be 1 times the
    determinant of 0, 2, 0; 1,
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    2, 3; 3, 0, 0.
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    That's just this guy right here,
    this matrix right there.
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    Then I'm going to have a 2, but
    I'm going to switch signs.
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    So it's minus 2 times the
    determinant if I get rid of
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    that row and this column,
    so it's 1, 2, 0.
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    I'm ignoring the zero because
    it's in the same column as the
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    2: 1, 2, 0; 0, 2, 3,
    and then 2, 0, 0.
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    And then I switch signs again.
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    It was a minus, so now
    I go back to plus.
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    So I do that guy, so
    plus 3 times the
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    determinant of his submatrix.
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    Get rid of that row and get
    rid of that columm,
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    I get a 1, 0, 0.
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    I get a 0, 1, 3.
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    I skip this column every time.
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    Then I get a 2, 3, 0,
    just like that.
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    We're almost done.
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    One more in this column.
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    Let me switch to
    another color.
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    I haven't used the
    blue in this yet.
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    So then I'm going
    to do a minus 4.
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    Remember, it's plus, minus,
    plus, minus 4 times the
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    determinant of its submatrix.
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    That's going to be
    that right there.
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    So it's 1, 0, 2; 0, 1, 2;
    2, 3, 0, just like that.
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    And now we're down to
    the 3-by-3 case.
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    We could use the definition of
    the 3 by 3, but we could just
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    keep applying this recursive
    definition.
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    So this is going to be equal
    to-- let me write it here.
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    It's 1 times-- what's
    this determinant?
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    This determinant's going to be
    0 times the determinant of
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    that submatrix, 2, 3, 0, 0.
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    That was this one right here.
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    And then we have minus 2, minus
    this 2-- remember, we
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    switched signs-- plus, minus,
    plus, so minus 2 times its
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    submatrix, so it's 1, 3, 3, 0.
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    And then finally plus 0 times
    its submatrix, which is this
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    thing right here: 1, 2,
    3, 0, just like that.
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    And then we have this
    next guy right here.
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    As you can see, this can get a
    little bit tedious, but we'll
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    keep our spirits up.
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    So minus 2 times 1 times its
    submatrix, so that's this guy
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    right here-- times the
    determinant of its submatrix
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    2, 3, 0, 0.
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    Then minus 2 times-- get
    rid of that row, that
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    column-- 0, 3, 2, 0.
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    And then plus 0 times
    0, 2, 2, 0.
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    That's this one right there.
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    Halfway there, at
    least for now.
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    And then we get this next one,
    so we have a plus 3.
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    Bring out our parentheses.
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    And then we're going to have 1
    times its sub-- I guess call
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    it sub-determinant.
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    So 1 times the determinant
    1, 3, 3, 0, right?
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    You get rid of this guy's row
    and column, you get this guy
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    right there.
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    And then minus 0-- get rid
    of this row and column--
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    times 0, 3, 2, 0.
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    Then you have plus 0 times its
    sub-determinant 0, 1, 2, 3.
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    Three-fourths of
    the way there.
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    One last term.
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    Let's hope we haven't made
    any careless mistakes.
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    Minus 4 times 1 times 1,
    2, 3, 0 right there.
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    Minus 0 times-- get rid of those
    two guys-- 0, 2, 2, 0.
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    And then plus 2 times
    0, 1, 2, 3, right?
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    Plus 2-- get rid of these
    guys-- 0, 1, 2, 3.
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    Now, we've defined or we've
    calculated or we've defined
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    our determinant of this matrix
    in terms of just a bunch of
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    2-by-2 matrices.
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    So hopefully, you saw in this
    example that the recursion
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    worked out.
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    So let's actually find what
    this number is equal to.
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    A determinant is always just
    going to be a number.
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    So let me get a nice
    vibrant color.
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    This is 0 times--
    I don't care.
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    0 times anything's
    going to be 0.
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    0 times anything is
    going to be 0.
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    0 times anything's
    going to be 0.
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    0 times anything's
    going to be 0.
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    Just simplifying it.
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    These guys are 0 because
    it's 0 times that.
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    0 times this is going
    to be equal to 0.
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    So what are we left with?
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    This is going to be equal to 1
    times-- this is all we have
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    left here is a minus 2 times--
    and what is this determinant?
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    It's 1 times 0, which is 0.
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    It's 0-- let me write this.
  • 15:30 - 15:35
    This is going to be 1 times 0
    is 0, minus 3 times 3 is 0
  • 15:35 - 15:38
    minus 9, so minus 9.
  • 15:38 - 15:40
    This right here is
    just minus 9.
  • 15:40 - 15:42
    So minus 2 times minus 9.
  • 15:42 - 15:45
    That's our first thing, I'll
    simplify it in a second.
  • 15:45 - 15:47
    Now let's do this
    term right here.
  • 15:47 - 15:51
    So it's minus 2 times-- now
    what's this determinant?
  • 15:51 - 15:54
    2 times 0 minus 0 times 3.
  • 15:54 - 15:55
    That's 0 minus 0.
  • 15:55 - 15:57
    So this is 0.
  • 15:57 - 16:00
    That guy became 0, so we
    can ignore that term.
  • 16:00 - 16:03
    This term right here is
    0 times 0, which is 0,
  • 16:03 - 16:04
    minus 2 times 3.
  • 16:04 - 16:06
    So it's minus 6.
  • 16:06 - 16:11
    So it's minus 2 times-- so this
    is a minus 6 right here.
  • 16:11 - 16:15
    You have a minus 2 times a minus
    6, so that's a plus 12.
  • 16:15 - 16:17
    So I'll just write
    a plus 12 here.
  • 16:17 - 16:20
    This minus 2 is that minus
    2 right there.
  • 16:20 - 16:23
    And then we have a plus 3.
  • 16:23 - 16:27
    And then this first term is 1
    times 0, which is 0, minus--
  • 16:27 - 16:30
    let me make the parentheses
    here-- 1 times 0, which is 0,
  • 16:30 - 16:33
    minus 3 times 3, which
    is minus 9 times 1.
  • 16:33 - 16:35
    So it's minus 9.
  • 16:35 - 16:37
    Everything else was a 0.
  • 16:37 - 16:39
    We're in the home stretch.
  • 16:39 - 16:42
    We have a minus 4.
  • 16:42 - 16:48
    Let's see, this is 1 times 0,
    which is 0, minus 3 times 2,
  • 16:48 - 16:49
    so minus 6.
  • 16:49 - 16:52
    So this is minus 6 right here.
  • 16:52 - 16:55
    Minus 6, this is 0, and then we
    have this guy right here.
  • 16:55 - 16:59
    So we have 0 times 3, which
    is 0, minus 2 times 1.
  • 16:59 - 17:03
    So that's minus 2, and then
    you have a minus 2 times a
  • 17:03 - 17:06
    plus 2 is minus 4.
  • 17:06 - 17:08
    So now we just have
    to make sure we do
  • 17:08 - 17:11
    our arithmetic properly.
  • 17:11 - 17:15
    This is 1 times plus 18,
    so this is 18, right?
  • 17:15 - 17:17
    Minus 2 times minus 9.
  • 17:17 - 17:22
    This right here is minus 24.
  • 17:22 - 17:27
    This right here is minus 27.
  • 17:27 - 17:30
    And then this right here, let's
    see, this is minus 10
  • 17:30 - 17:31
    right here.
  • 17:31 - 17:33
    That is minus 10.
  • 17:33 - 17:40
    Minus 4 times minus
    10 is plus 40.
  • 17:40 - 17:43
    Let's see if we can simplify
    this a little bit.
  • 17:43 - 17:45
    If we simplify this a little
    bit-- I don't want to make a
  • 17:45 - 17:48
    careless mistake right
    at the end.
  • 17:48 - 17:55
    So 18 minus 24, 24 minus 18 is
    6, so this is going to be
  • 17:55 - 17:58
    equal to minus 6, right?
  • 17:58 - 18:00
    18 minus 24 is minus 6.
  • 18:00 - 18:03
    And then-- let me do it in
    green-- now what's the
  • 18:03 - 18:04
    difference?
  • 18:04 - 18:10
    If we have minus 27 plus
    40, that's 13, right?
  • 18:10 - 18:13
    It's positive 13.
  • 18:13 - 18:18
    So minus 6 plus positive
    13 is equal to 7.
  • 18:18 - 18:19
    And so we are done!
  • 18:19 - 18:22
    After all of that computation,
    hopefully we haven't made a
  • 18:22 - 18:23
    careless mistake.
  • 18:23 - 18:25
    The determinant of this
    character right
  • 18:25 - 18:28
    here is equal to 7.
  • 18:28 - 18:31
    The determinant is equal to 7.
  • 18:31 - 18:33
    And so the one useful takeaway,
    we know that this is
  • 18:33 - 18:37
    invertible because it has
    a non-zero determinant.
  • 18:37 - 18:39
    Hopefully, you found
    that useful.
  • 18:39 - 18:39
Title:
Linear Algebra: nxn Determinant
Description:

Defining the determinant for nxn matrices. An exampled of a 4x4 determinant.

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Video Language:
English
Duration:
18:40

English subtitles

Incomplete

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