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Ch1 矩陣與線性方程組
第 15 題／共 20 題
◀
LA 15/20
錯題回報
00:00
15. Elementary Row Operations、Column Relations、Cross Product
#LA-01-015
難
Elementary Row Operations
Column Relations
Cross Product
📝
☆
[
a
1
b
1
c
1
d
1
a
2
b
2
c
2
d
2
a
3
b
3
c
3
d
3
]
\begin{bmatrix} a_1 & b_1 & c_1 & d_1 \\ a_2 & b_2 & c_2 & d_2 \\ a_3 & b_3 & c_3 & d_3 \end{bmatrix}
a
1
a
2
a
3
b
1
b
2
b
3
c
1
c
2
c
3
d
1
d
2
d
3
is processed by row operations to form
[
1
2
3
5
2
5
7
13
0
0
1
0
]
\begin{bmatrix} 1 & 2 & 3 & 5 \\ 2 & 5 & 7 & 13 \\ 0 & 0 & 1 & 0 \end{bmatrix}
1
2
0
2
5
0
3
7
1
5
13
0
Which of the following are true?
A
∣
d
1
d
2
d
3
b
1
b
2
b
3
c
1
c
2
c
3
∣
=
∣
b
1
b
2
b
3
a
1
a
2
a
3
c
1
c
2
c
3
∣
\begin{vmatrix} d_1 & d_2 & d_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = \begin{vmatrix} b_1 & b_2 & b_3 \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3 \end{vmatrix}
d
1
b
1
c
1
d
2
b
2
c
2
d
3
b
3
c
3
=
b
1
a
1
c
1
b
2
a
2
c
2
b
3
a
3
c
3
B
After row operations,
[
a
1
b
1
c
1
d
1
a
2
b
2
c
2
d
2
a
3
b
3
c
3
d
3
]
\begin{bmatrix} a_1 & b_1 & c_1 & d_1 \\ a_2 & b_2 & c_2 & d_2 \\ a_3 & b_3 & c_3 & d_3 \end{bmatrix}
a
1
a
2
a
3
b
1
b
2
b
3
c
1
c
2
c
3
d
1
d
2
d
3
can become
[
2
−
1
0
−
5
3
0
−
1
−
3
0
3
−
2
9
]
\begin{bmatrix} 2 & -1 & 0 & -5 \\ 3 & 0 & -1 & -3 \\ 0 & 3 & -2 & 9 \end{bmatrix}
2
3
0
−
1
0
3
0
−
1
−
2
−
5
−
3
9
.
C
If
a
1
2
+
a
2
2
+
a
3
2
=
2
a_1^2 + a_2^2 + a_3^2 = 2
a
1
2
+
a
2
2
+
a
3
2
=
2
,
b
1
2
+
b
2
2
+
b
3
2
=
2
b_1^2 + b_2^2 + b_3^2 = 2
b
1
2
+
b
2
2
+
b
3
2
=
2
,
d
1
2
+
d
2
2
+
d
3
2
=
16
d_1^2 + d_2^2 + d_3^2 = 16
d
1
2
+
d
2
2
+
d
3
2
=
16
, then
∣
(
a
1
,
a
2
,
a
3
)
×
(
d
1
,
d
2
,
d
3
)
∣
=
4
2
|(a_1, a_2, a_3) \times (d_1, d_2, d_3)| = 4\sqrt{2}
∣
(
a
1
,
a
2
,
a
3
)
×
(
d
1
,
d
2
,
d
3
)
∣
=
4
2
, where
×
\times
×
is the cross product.
D
{
a
1
x
+
c
1
y
+
3
b
1
=
d
1
a
2
x
+
c
2
y
+
3
b
2
=
d
2
\begin{cases} a_1 x + c_1 y + 3b_1 = d_1 \\ a_2 x + c_2 y + 3b_2 = d_2 \end{cases}
{
a
1
x
+
c
1
y
+
3
b
1
=
d
1
a
2
x
+
c
2
y
+
3
b
2
=
d
2
has a unique solution.
E
{
(
3
b
1
+
c
1
)
x
−
2
a
1
y
+
d
1
z
+
2
c
1
=
0
(
3
b
2
+
c
2
)
x
−
2
a
2
y
+
d
2
z
+
2
c
2
=
0
(
3
b
3
+
c
3
)
x
−
2
a
3
y
+
d
3
z
+
2
c
3
=
0
\begin{cases} (3b_1 + c_1)x - 2a_1 y + d_1 z + 2c_1 = 0 \\ (3b_2 + c_2)x - 2a_2 y + d_2 z + 2c_2 = 0 \\ (3b_3 + c_3)x - 2a_3 y + d_3 z + 2c_3 = 0 \end{cases}
⎩
⎨
⎧
(
3
b
1
+
c
1
)
x
−
2
a
1
y
+
d
1
z
+
2
c
1
=
0
(
3
b
2
+
c
2
)
x
−
2
a
2
y
+
d
2
z
+
2
c
2
=
0
(
3
b
3
+
c
3
)
x
−
2
a
3
y
+
d
3
z
+
2
c
3
=
0
has a unique solution.
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