Determinant of an n × n matrix The determinant is a function from the set of n × n matrices to R. (Or from n-tuples of row vectors in Rn , to R). For us the determinant is defined by the first row expansion, explained later. The definition is recursive. Notation: det A or |A|. 

2 5 A= 3 4



 2 5 2 5 det = 3 4 3 4 

Definition For a 1 × 1 matrix [a] a = a

3 × 3 determinant, first row expansion The determinant of a 3 × 3 matrix is defined by the first row expansion. This is an sign-alternating sum of first row entries times 2 × 2 determinants. First row entries are in blue. Each blue entry is multiplied by a 2 × 2, obtained by crossing out the row and column of the blue entry.   a b c A = d e f  g h i  a b these matrices  d e are just pictures g h a b c d e f = a e h g h i

     c a b c a b c f  d e f  d e f  i g h i g h i d f d e f +c − b g h i g i

This defines a 3 × 3 determinant, using 2 × 2 determinants.

2 × 2 determinant, first row expansion

Blue gives first row entries. Signs alternate. To get the 1 × 1 determinant to multiply by a first row entry, cross out the row and the column of the entry.     a b a b just pictures c d c d a b c d = a|d| − b|c| = ad − bc.

3 × 3 determinant, first row expansion, continued

Now that we know 2 × 2 determinants, we finish the previous 3 × 3: a b c d e f = a e f − b d f + c d e g h g i h i g h i = a(ei − fh) − b(di − fg ) + c(dh − eg ) = aei − afh − bdi + bfg + cdh − ceg .

3 × 3 determinant, second column expansion Can expand along any row or column and the answer is the same.   a b c A = d e f  g h i 

  a b c a b d e f  d e just pictures g h i g h a b c d e f = −b d f + e a g i g g h i

  c a f  d i g a c − h d i

 b c e f h i c f

= −b(di − fg ) + e(ai − cg ) − h(af − cd) = −bdi + bfg + eai − ecg − haf + hcd Answer matches first row expansion.

General sign pattern



+ −  +  −  .. .

− + − + .. .

+ − + − .. .

− + − + .. .

 ··· · · ·  · · ·  · · ·  .. .

The assign associated with the (i, j)-th entry is (−1)i+j .

Scaling a row scales the determinant Why it works: multiply first row by m and expand along first row: ma mb mc d e f g h i d e d f e f − mb = ma g i + mc g h h i   e f d f d e = m a −b +c h i g i g h a b c = m d e f g h i

Determinant has been multiplied by m.

Swapping rows changes determinant by a sign: Alternating property Why it works for 2 × 2: check directly c d = cb − ad = −(ad − bc) = − a b c d a b Why it works for 3 × 3: swap rows 2 and 3. Use just-verified alternating property for 2 × 2: a b c g h i = a h i − b g i + c g h e f d f d e d e f   e f d f d e = − a −b +c h i g i g h a b c = − d e f g h i

The determinant of a matrix with two equal rows, is zero

If we swap the two identical rows, the determinant changes by a sign using the alternating property. However since the swapped rows are identical, the determinant did not change because the matrix did not change. The only number that doesn’t change when you negate it, is zero. Hence the determinant is zero.

Determinant is additive in any given row First row varies; other rows fixed. a b c a0 b 0 c 0 d e f + d e f g h i g h i d e d f e f +c −b = a g h g i h i 0 e f 0 d f 0 d e +a − b + c g i g h h i 0 e f 0 d f 0 d = (a + a ) − (b + b ) + c + c ) g i g h i a + a0 b + b 0 c + c 0 e f = d g h i

e h

Determinant is linear in its first row Let n = 3 and consider the determinant as a function of three row vectors u = [a, b, c], v = [d, e, f ], w = [g , h, i]. Let T : R3 → R be the function of a single vector u ∈ R3 defined by   u T (u) = det  v  with v, w fixed. w Claim: T is a linear function. Check additive property. Let u0 = [a0 , b 0 , c 0 ]. Then u + u0 = [a, b, c] + [a0 , b 0 , c 0 ] = [a + a0 , b + b 0 , c + c 0 ]. We have   a + a 0 b + b 0 c + c 0 u + u0 e f T (u + u0 ) = det  v  = d g w h i 0     0 a b c a b 0 c 0 u u = d e f + d e f = det  v  + det  v  = T (u) + T (u0 ) g h i g h i w w

Determinant is linear in its first row, continued n = 3: row vectors u = [a, b, c], v = [d, e, f ], w = [g , h, i]. Recall T : R3 → R is the function of a single vector u ∈ R3 defined by   u T (u) = det  v  with v, w fixed. w Check T has the scalar multiplication property. mu = m[a, b, c] = [ma, mb, mc].   ma mb mc a b c mu e f = m d e f = mT (u). T (mu) = det  v  = d g g h i w h i Therefore this function T is linear. “Determinant is linear in its first row”. This works for any row. “Determinant is multilinear”.

(i) Scaling a row of a matrix, scales its determinant. (ii) Viewing the determinant as a function of a sequence of n row vectors in Rn , if we fix all of those vectors but one, the resulting function of a single vector, is additive. Properties (i) and (ii) together say that the determinant is a multilinear function. (iii) Exchanging two rows of a matrix, changes its determinant by a sign (negation). The determinant is an alternating function (of n row vectors in Rn ).

Theorem The determinant is the unique multilinear alternating function of n row vectors in Rn to R, which sends In to 1.

Adding a multiple of one row to another, doesn’t change determinant Proof for 3 × 3: Let u, v, w be rows of a 3 × 3 matrix. Do the row operation R3 ← R3 + cR1 . Get the matrix with rows u, v, w + cu. Take the determinant:       u u u      v det = det v + det v  w + cu w cu     u u    = det v + c det v  w u     u u    = det v + c · 0 = det v  w w because det is additive in the 3rd row, det scales in the 3rd row, and if two rows are equal then the determinant is zero.