hw912

8/10/2019 hw912

http://slidepdf.com/reader/full/hw912 1/3

§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. No: Let A =

a bc d

and B =

e f g h

.

Then,

det(

a bc d

) + det(

e f g h

) = ad − bc + eh − f g

and

det(a b

c d

+e f

g h

) = det(a + e b + f

c + g d + h

) = (a + e)(d + h)− (b + f )(c + g)

= 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).

b. Yes: Let A =

a bc d

and B =

e f g h

.

Then,

det(

a bc d

e f g h

) = det(

ae + bg af + bhce + dg cf + dh

) = (ae+bg)(cf +dh)−(af +bh)(ce+dg)

= adeh + bcfg − adfg − bceh

and

det(

a bc d

)det(

e f g h

) = (ad − bc)(eh − f g) = adeh + bcfg − adf g − bceh

It IS true that for all values of a, b, c, d, e, f , g, h that det(AB) = det(A)det(B).

c. Yes: Let A =

a bc d

and B =

e f g h

. Then, from part (b), we have

det(AB) = adeh + bcfg − adfg − bceh

and

det(

e f g h

a bc d

) = (eh − f g)(ad − bc) = adeh + bcfg − adfg − bceh

It IS true that det(AB) = det(BA).

§2.2 5. Let A be an n × n matrix and α a scalar. Show that

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8/10/2019 hw912


det(αA) = αndet(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, α A is not row equivalent to I because dividing each row in α A by1/α would get us back to A and A is singular. Thus, α A is singular anddet(αA) = 0. Also, αndet(A) = αn(0) = 0. Therefore, det(αA) = αndet(A)when A is singular.

In case 2 if A is nonsingular it can be reduced to strictly triangular form byusing only the first and third row operations. We will call the matrix to whichA is reduced T . Since we only used the first and third row operations to obtain

T we know:det(A) = ±det(T ) = ±(t11t22 · · · tnn) where tii are diagonal entries in T .Thus,αndet(A) = αn(±t11t22 · · · 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 wemultiply a row by α. Repeat n times.

αn det(A) = αn−1

αa11 αa12 ... αa1na21 a22 ... a2na31 a32 ... a3n... ... ... ...

an1 an2 ... ann

= αn−2

αa11 αa12 ... αa1nαa21 αa22 ... αa2na31 a32 ... a3n... ... ... ...

an1 an2 ... ann

= ... = α

αa11 αa12 ... αa1nαa21 αa22 ... αa2n

... ... ... ...αan−1,1 αan−1,2 ... αan−1,n

an1 an2 ... ann

=

αa11 αa12 ... αa1nαa21 αa22 ... αa2n

... ... ... ...αan1 αan2 ... αann

= det(αA)

§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

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8/10/2019 hw912


det(A−1) = 1

det(A)

as desired.

§2.2 14. Let A and B be n × n matricies. Prove that the product AB isnonsingular 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) = 0 by theorem 2.2.2. We also know

that det(AB) = det(A) · det(B) = 0 by theorem 2.2.3. Clearly the only waydet(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 thematricies is singular by theorem 2.2.2. Therefore, the determinants of A and Bmust both be nonzero in order to make the determinant of AB nonzero, andboth A and B are nonsingular when AB is nonsingular.

Now, I will prove that if A and B are both nonsingular then AB is nonsin-gular.

If A and B are both nonsingular then we have det(A) = 0 and det(B) = 0by 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) = 0 and AB is nonsingular when bothA and B are nonsingular.

We have clearly shown that the product AB is nonsingular if and only if Aand B are both nonsingular.

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