Page 400 - 35Linear Algebra

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of all zeros. Somehow the determinant is able to detect that there is only one
equation here. Even if we had a set of contradictory set of equations such as
x + y = 1
2x + 2y = 0,

where it is not possible for both of these equations to be true, the matrix M
is still the same, and still has a determinant zero.
Lets look at a three by three example, where the third equation is the sum
of the first two equations.

x + y + z = 1
y + z = 1
x + 2y + 2z = 2

and the matrix for this is
 
1 1 1
M = 0 1 1 

1 2 2
If we were trying to find the inverse to this matrix using elementary
matrices
   
1 1 1 1 0 0 1 1 1 1 0 0
 0 1 1 0 1 0  =  0 1 1 0 1 0 
1 2 2 0 0 1 0 0 0 −1 −1 1

And we would be stuck here. The last row of all zeros cannot be converted
into the bottom row of a 3 × 3 identity matrix. this matrix has no inverse,
and the row of all zeros ensures that the determinant will be zero. It can
be difficult to see when one of the rows of a matrix is a linear combination
of the others, and what makes the determinant a useful tool is that with this
reasonably simple computation we can find out if the matrix is invertible, and
if the system will have a solution of a single point or column vector.

Alternative Proof

Here we will prove more directly that the determinant of a product of matrices
is the product of their determinants. First we reference that for a matrix
0 0
M with rows r i , if M is the matrix with rows r = r j + λr i for j 6= i and
j
0 0 0
r = r i , then det(M) = det(M ) Essentially we have M as M multiplied by the
i
i
elementary row sum matrices S (λ). Hence we can create an upper-triangular
j
1
matrix U such that det(M) = det(U) by first using the first row to set m 7→ 0
i
for all i > 1, then iteratively (increasing k by 1 each time) for fixed k using
k
the k-th row to set m 7→ 0 for all i > k.
i
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