1. The determinant of a zero matrix is 0. The transpose of a matrix P is equal to the transpose of PT. 2. Solving a system of linear equations by calculating the determinants of the coefficient matrices results in solutions for w=4, x=4, y=2, z=0. 3. The inverse of a non-singular matrix A is equal to the adjugate of A divided by the determinant of A.

1. 2
1. (5 ! )
# $%
1. (Zero matrices) 0 1 2 3 2 3 4 5 4 0 467 2 467 4 5
78 4
2. $ !4 % 0 78 : 7#
; !<! 78 3% :% =:% 24% %
:$ 8 7#
:24% A =
nna2na1na
n2a22a21a
000
0 :% 4 :@ !
@ A : (cofactor) ! %
det(A) = a11C11(A) + a12C12(A) + a13C13(A) + G + a1nC1n(A)
= 0 + 0 + 0 + G + 0
= 0
! % det(A) = 0 7 0 A : :% Q.E.D.
2. J 4 24% P =
33p32p31p
23p22p21p
13p12p11p
6! P 3% PT
@; $
P P = (PT
)T
(5 ! )
@ J 4 P 24% ! % PT
=
33p23p13p
32p22p12p
31p21p11p
; !<! 78 (PT
)T
=
33p32p31p
23p22p21p
13p12p11p
= P Q.E.D.

2. 3. 21% M$ % !## : 8 (10 ! )
w O x + y O z = O 4
4w O x + 3y + z = O 8
2w + x + y O z = 0
3w + 2x + y O 3z = 1
$ !## J 4 24% 2 7 8
3123
1112
1314
1111
z
y
x
w
=
1
0
8
4
det(A) =
3123
1112
1314
1111
= O 4, det(A1) =
3121
1110
1318
1114
= 8
det(A2) =
3113
1102
1384
1141
= O 12, det(A3) =
3123
1012
1814
1411
= O 4
det(A4) =
1123
0112
8314
4111
= 0
! % w = 4
8 = O 2, x = 4
12 = 3, y = 4
4 = 1, z = 0

3. 4. J 4 24% A =
33a32a31a
23a22a21a
13a12a11a
3% A
V4 AO 1
(10 ! )
: AO 1
= )Adet(
)A(adj
=
T
)A(33C)A(32C)A(31C
)A(23C)A(22C)A(21C
)A(13C)A(12C)A(11C
)Adet(
1
=
)A(33C)A(23C)A(13C
)A(32C)A(22C)A(12C
)A(31C)A(21C)A(11C
)Adet(
1
=
++
+
++
)12a21a22a11a()12a31a32a11a()22a31a32a21a(
)13a21a23a11a()13a31a33a11a()23a31a33a21a(
)13a22a23a12a()13a32a33a12a()23a32a33a22a(
)Adet(
1