§2.1 11. Let A and B be 2 × 2 matrices.
a. Does det(A + B) = det(A) + det(B)?
b. Does det(AB) = det(A)det(B)?
c. Does det(AB) = det(BA)?
Justify your answers.
Solution:(Joe) a b
e f
a. No: Let A =
and B =
.
c d
g h
Then,
a b
e f
det(
) + det(
) = ad − bc + eh − f g
c d
g h
and
a b
e
det(
+
c d
g
f
a+e b+f
) = det(
) = (a + e)(d + h) − (b + f )(c + g)
h
c+g d+h
= ad + eh + ah + ed − bc − f g − bg − f c
It is NOT true that for all values of a, b, c, d, e, f , g, h that det(A) + det(B)
equals det(A + B). a b
e f
b. Yes: Let A =
and B =
.
c d
g h
Then,
a b
e f
ae + bg af + bh
det(
) = det(
) = (ae+bg)(cf +dh)−(af +bh)(ce+dg)
c d
g h
ce + dg cf + dh
= adeh + bcf g − adf g − bceh
and
a b
e
det(
)det(
c d
g
f
) = (ad − bc)(eh − f g) = adeh + bcf g − adf g − bceh
h
It IS true that for allvaluesof a, b, c, d,e, f , g,
h that det(AB) = det(A)det(B).
a b
e f
c. Yes: Let A =
and B =
. Then, from part (b), we have
c d
g h
det(AB) = adeh + bcf g − adf g − bceh
and
e
det(
g
f
h
a b
) = (eh − f g)(ad − bc) = adeh + bcf g − adf g − bceh
c d
It IS true that det(AB) = det(BA).
§2.2 5.
Let A be an n × n matrix and α a scalar. Show that
1
det(αA) = αn det(A)
Solution: (Jeff)
For this particular problem there are two case. Specifically:
Case 1: A is singular.
Case 2: A is nonsingular.
In case 1 if A is singular then it is not row equivalent to I and det(A) = 0.
Clearly, αȦ is not row equivalent to I because dividing each row in αȦ by
1/α would get us back to A and A is singular. Thus, αȦ is singular and
det(αA) = 0. Also, αn det(A) = αn (0) = 0. Therefore, det(αA) = αn det(A)
when A is singular.
In case 2 if A is nonsingular it can be reduced to strictly triangular form by
using only the first and third row operations. We will call the matrix to which
A is reduced T . Since we only used the first and third row operations to obtain
T we know:
det(A) = ±det(T ) = ±(t11 t22 · · · tnn ) where tii are diagonal entries in T .
Thus,
αn det(A) = αn (±t11 t22 · · · tnn )
= ±((αt11 )(αt22 ) · · · (αtnn ))
= ±det(αT )
= det(αA)
Now we have shown what was desired.
Remark:(Li) It can proved without dividing into two cases. Each time we
multiply a row by α. Repeat n times.
αa11 αa12 ... αa1n αa11 αa12 ... αa1n αa21 αa22 ... αa2n a21
a22 ... a2n a32 ... a3n a32 ... a3n = αn−2 a31
αn det(A) = αn−1 a31
...
...
...
...
... ...
...
... an1
an2 ... ann an2 ... ann
an1
αa11
αa12
...
αa1n αa11 αa12 ... αa1n αa21
αa
...
αa
22
2n
αa21 αa22 ... αa2n = det(αA)
=
...
...
...
...
= ... = α ...
...
... αan−1,1 αan−1,2 ... αan−1,n ...
αan1 αan2 ... αann an1
an2
...
ann
§2.2 6. Let A be a nonsingular matrix. Show that
det(A−1 ) =
1
det(A)
Solution:(Joe)
We know that AA−1 = I. If we take the determinant of each side, we get:
det(AA−1 ) = det(I)
det(A)det(A−1 ) = 1
2
det(A−1 ) =
1
det(A)
as desired.
§2.2 14. Let A and B be n × n matricies. Prove that the product AB is
nonsingular if and only if A and B are both nonsingular.
Solution:(Jeff)
I will first prove that if AB is nonsingular then A and B are both nonsingular.
If AB is nonsingular then det(AB) 6= 0 by theorem 2.2.2. We also know
that det(AB) = det(A) · det(B) 6= 0 by theorem 2.2.3. Clearly the only way
det(A) · det(B) = 0 is if one or both of the determinants of A and B are zero. If
one or both of the determinats of A and B equal zero then at least one of the
matricies is singular by theorem 2.2.2. Therefore, the determinants of A and B
must both be nonzero in order to make the determinant of AB nonzero, and
both A and B are nonsingular when AB is nonsingular.
Now, I will prove that if A and B are both nonsingular then AB is nonsingular.
If A and B are both nonsingular then we have det(A) 6= 0 and det(B) 6= 0
by theorem 2.2.2. We know from theorem 2.2.3 that det(A) · det(B) = det(AB).
Thus, it must be the case that det(AB) 6= 0 and AB is nonsingular when both
A and B are nonsingular.
We have clearly shown that the product AB is nonsingular if and only if A
and B are both nonsingular.
3