可逆矩阵的引入和介绍就很有意思，如果
P
P
P
是一系列初等矩阵的乘积，那么
B
=
P
A
B=PA
B
=
P
A
和
A
A
A
是row-equivalent的，所以
A
A
A
和
B
B
B
也是row-equivalent的，故有一个一系列初等矩阵的乘积
Q
Q
Q
使得
A
=
Q
B
A=QB
A
=
Q
B
，如果
A
=
I
m
A=I_m
A
=
I
m
，那么就会有
B
=
P
,
I
m
=
Q
B
=
Q
P
B=P,I_m=QB=QP
B
=
P
,
I
m
=
Q
B
=
Q
P
，所以通过
P
P
P
得到了一个
Q
Q
Q
，因而有inverse matrix的概念，inverse可以有左有右，不必同时存在，但如果同时存在，必相等。
Theorem 10
说明：矩阵的逆矩阵也可逆且逆逆得自身，两个（可推广到有限个）逆矩阵乘积也可逆。
Theorem 11
则证明初等矩阵是可逆的，其逆矩阵就是单位矩阵在该初等矩阵对应初等行变换的逆变换下的矩阵，计算也很简单。
Theorem 12
得出了重要的三个等价关系：可逆、与单位矩阵row-equivalent、是一系列初等矩阵的乘积，其两个重要推论，一是把
A
A
A
变成单位阵的行变换可以同时把
I
I
I
变成
A
−
1
A^{-1}
A
−
1
，二是两矩阵row-equivalent当且仅当一个矩阵等于某可逆阵左乘另一个矩阵（因为可逆阵就是一系列初等矩阵的乘积）。
Theorem 13
是另一组重要等价关系：
A
A
A
可逆、
A
X
=
0
AX=0
A
X
=
0
只有零解、
A
X
=
Y
AX=Y
A
X
=
Y
对每一个
Y
Y
Y
都有解。这两个定理说明了可逆阵的威力。推论一是如果是方阵，那么光有左逆或右逆就代表可逆。推论二是方阵的乘积可逆代表每个方阵均可逆，即Theorem 10在方阵约束下增强了。
这一节最后要说明的两个事，一个是如何用初等行变换或者初等矩阵乘积求逆，另一个是行的事情都可以复制到列上，只是初等矩阵会发生变化，且从左乘变成右乘。
Exercises
1. Let
A
=
[
1
2
1
0
−
1
0
3
5
1
−
2
1
1
]
A=\begin{bmatrix}1&2&1&0\\-1&0&3&5\\1&-2&1&1\end{bmatrix}
A
=
⎣
⎡
1
−
1
1
2
0
−
2
1
3
1
0
5
1
⎦
⎤
Find a row-reduced echelon matrix
R
R
R
which is row-equivalent to
A
A
A
and an invertible
3
×
3
3\times 3
3
×
3
matrix
P
P
P
such that
R
=
P
A
R=PA
R
=
P
A
.
solution
: We perform row operations on
A
′
=
[
A
Y
]
A’=\begin{bmatrix}A&Y\end{bmatrix}
A
′
=
[
A
Y
]
with
Y
=
[
y
1
,
y
2
,
y
3
]
T
Y=[y_1,y_2,y_3 ]^T
Y
=
[
y
1
,
y
2
,
y
3
]
T
, thus
A
′
=
[
1
2
1
0
y
1
−
1
0
3
5
y
2
1
−
2
1
1
y
3
]
→
[
1
2
1
0
y
1
0
2
4
5
y
2
+
y
1
0
−
4
0
1
y
3
−
y
1
]
→
[
1
2
1
0
y
1
0
2
4
5
y
2
+
y
1
0
0
8
11
2
y
2
+
y
3
+
y
1
]
→
[
1
2
1
0
y
1
0
1
2
5
2
1
2
(
y
2
+
y
1
)
0
0
1
11
8
1
8
(
2
y
2
+
y
3
+
y
1
)
]
→
[
1
2
0
−
11
8
7
8
y
1
−
1
4
y
2
−
1
8
y
3
0
1
0
−
1
4
1
4
(
y
1
−
y
3
)
0
0
1
11
8
1
8
(
2
y
2
+
y
3
+
y
1
)
]
→
[
1
0
0
−
7
8
3
8
y
1
−
1
4
y
2
+
3
8
y
3
0
1
0
−
1
4
1
4
(
y
1
−
y
3
)
0
0
1
11
8
1
8
(
2
y
2
+
y
3
+
y
1
)
]
\begin{aligned}A’&=\begin{bmatrix}1&2&1&0&y_1\\-1&0&3&5&y_2\\1&-2&1&1&y_3 \end{bmatrix}\rightarrow\begin{bmatrix}1&2&1&0&y_1\\0&2&4&5&y_2+y_1\\0&-4&0&1&y_3-y_1 \end{bmatrix}\\&\rightarrow\begin{bmatrix}1&2&1&0&y_1\\0&2&4&5&y_2+y_1\\0&0&8&11&2y_2+y_3+y_1 \end{bmatrix}\rightarrow\begin{bmatrix}1&2&1&0&y_1\\0&1&2&\frac{5}{2}&\frac{1}{2} (y_2+y_1 )\\0&0&1&\frac{11}{8}&\frac{1}{8} (2y_2+y_3+y_1 ) \end{bmatrix}\\&\rightarrow\begin{bmatrix}1&2&0&-\frac{11}{8}&\frac{7}{8} y_1-\frac{1}{4} y_2-\frac{1}{8} y_3\\0&1&0&-\frac{1}{4}&\frac{1}{4} (y_1-y_3 )\\0&0&1&\frac{11}{8}&\frac{1}{8} (2y_2+y_3+y_1 ) \end{bmatrix}\rightarrow\begin{bmatrix}1&0&0&-\frac{7}{8}&\frac{3}{8} y_1-\frac{1}{4} y_2+\frac{3}{8} y_3\\0&1&0&-\frac{1}{4}&\frac{1}{4} (y_1-y_3 )\\0&0&1&\frac{11}{8}&\frac{1}{8}(2y_2+y_3+y_1 )\end{bmatrix}\end{aligned}
A
′
=
⎣
⎡
1
−
1
1
2
0
−
2
1
3
1
0
5
1
y
1
y
2
y
3
⎦
⎤
→
⎣
⎡
1
0
0
2
2
−
4
1
4
0
0
5
1
y
1
y
2
+
y
1
y
3
−
y
1
⎦
⎤
→
⎣
⎡
1
0
0
2
2
0
1
4
8
0
5
1
1
y
1
y
2
+
y
1
2
y
2
+
y
3
+
y
1
⎦
⎤
→
⎣
⎡
1
0
0
2
1
0
1
2
1
0
2
5
8
1
1
y
1
2
1
(
y
2
+
y
1
)
8
1
(
2
y
2
+
y
3
+
y
1
)
⎦
⎤
→
⎣
⎡
1
0
0
2
1
0
0
0
1
−
8
1
1
−
4
1
8
1
1
8
7
y
1
−
4
1
y
2
−
8
1
y
3
4
1
(
y
1
−
y
3
)
8
1
(
2
y
2
+
y
3
+
y
1
)
⎦
⎤
→
⎣
⎡
1
0
0
0
1
0
0
0
1
−
8
7
−
4
1
8
1
1
8
3
y
1
−
4
1
y
2
+
8
3
y
3
4
1
(
y
1
−
y
3
)
8
1
(
2
y
2
+
y
3
+
y
1
)
⎦
⎤
thus
R
=
[
1
0
0
−
7
8
0
1
0
−
1
4
0
0
1
11
8
]
R=\begin{bmatrix}1&0&0&-\frac{7}{8}\\0&1&0&-\frac{1}{4}\\0&0&1&\frac{11}{8}\end{bmatrix}
R
=
⎣
⎡
1
0
0
0
1
0
0
0
1
−
8
7
−
4
1
8
1
1
⎦
⎤
, and the matrix
P
P
P
s.t.
R
=
P
A
R=PA
R
=
P
A
is
P
=
[
3
/
8
−
1
/
4
3
/
8
1
/
4
0
−
1
/
4
1
/
8
1
/
4
1
/
8
]
P=\begin{bmatrix}3/8&-1/4&3/8\\1/4&0&-1/4\\1/8&1/4&1/8\end{bmatrix}
P
=
⎣
⎡
3
/
8
1
/
4
1
/
8
−
1
/
4
0
1
/
4
3
/
8
−
1
/
4
1
/
8
⎦
⎤
2. Do Exercise 1, but with
A
=
[
2
0
i
1
−
3
−
i
i
1
1
]
A=\begin{bmatrix}2&0&i\\1&-3&-i\\i&1&1\end{bmatrix}
A
=
⎣
⎡
2
1
i
0
−
3
1
i
−
i
1
⎦
⎤
solution
: We perform row operations on
A
′
=
[
A
Y
]
A’=\begin{bmatrix}A&Y\end{bmatrix}
A
′
=
[
A
Y
]
with
Y
=
[
y
1
,
y
2
,
y
3
]
T
Y=[y_1,y_2,y_3 ]^T
Y
=
[
y
1
,
y
2
,
y
3
]
T
, thus
A
′
=
[
2
0
i
y
1
1
−
3
−
i
y
2
i
1
1
y
3
]
→
[
1
0
i
2
y
1
2
0
−
3
−
3
i
2
y
2
−
y
1
2
0
1
3
2
y
3
−
i
2
y
1
]
→
[
1
0
i
2
y
1
2
0
1
3
2
y
3
−
i
2
y
1
0
0
9
−
3
i
2
3
y
3
−
3
i
+
1
2
y
1
+
y
2
]
→
[
1
0
i
2
y
1
2
0
1
3
2
y
3
−
i
2
y
1
0
0
1
3
+
i
15
(
3
y
3
−
3
i
+
1
2
y
1
+
y
2
)
]
\begin{aligned}A’&=\begin{bmatrix}2&0&i&y_1\\1&-3&-i&y_2\\i&1&1&y_3 \end{bmatrix}\rightarrow\begin{bmatrix}1&0&\frac{i}{2}&\frac{y_1}{2}\\0&-3&-\frac{3i}{2}&y_2-\frac{y_1}{2}\\0&1&\frac{3}{2}&y_3-\frac{i}{2} y_1 \end{bmatrix}\\&\rightarrow\begin{bmatrix}1&0&\frac{i}{2}&\frac{y_1}{2}\\0&1&\frac{3}{2}&y_3-\frac{i}{2} y_1\\0&0&\frac{9-3i}{2}&3y_3-\frac{3i+1}{2} y_1+y_2 \end{bmatrix}\rightarrow\begin{bmatrix}1&0&\frac{i}{2}&\frac{y_1}{2}\\0&1&\frac{3}{2}&y_3-\frac{i}{2} y_1\\0&0&1&\frac{3+i}{15} \left(3y_3-\frac{3i+1}{2} y_1+y_2 \right)\end{bmatrix}\end{aligned}
A
′
=
⎣
⎡
2
1
i
0
−
3
1
i
−
i
1
y
1
y
2
y
3
⎦
⎤
→
⎣
⎡
1
0
0
0
−
3
1
2
i
−
2
3
i
2
3
2
y
1
y
2
−
2
y
1
y
3
−
2
i
y
1
⎦
⎤
→
⎣
⎡
1
0
0
0
1
0
2
i
2
3
2
9
−
3
i
2
y
1
y
3
−
2
i
y
1
3
y
3
−
2
3
i
+
1
y
1
+
y
2
⎦
⎤
→
⎣
⎡
1
0
0
0
1
0
2
i
2
3
1
2
y
1
y
3
−
2
i
y
1
1
5
3
+
i
(
3
y
3
−
2
3
i
+
1
y
1
+
y
2
)
⎦
⎤
It’s easy to see
R
=
I
R=I
R
=
I
and to calculate
P
P
P
, we see
A
′
→
[
I
Y
′
]
A’\rightarrow\begin{bmatrix}I&Y’\end{bmatrix}
A
′
→
[
I
Y
′
]
, in which
Y
′
=
[
y
1
2
−
2
(
3
+
i
)
15
i
(
3
y
3
−
3
i
+
1
2
y
1
+
y
2
)
y
3
−
i
2
y
1
−
2
(
3
+
i
)
45
(
3
y
3
−
3
i
+
1
2
y
1
+
y
2
)
3
+
i
15
(
3
y
3
−
3
i
+
1
2
y
1
+
y
2
)
]
=
[
1
2
+
2
(
3
+
i
)
(
3
i
+
1
)
30
i
−
2
(
3
+
i
)
15
i
−
2
(
3
+
i
)
5
i
−
i
2
+
(
3
+
i
)
(
3
i
+
1
)
45
−
2
(
3
+
i
)
45
1
−
2
(
3
+
i
)
15
−
(
3
+
i
)
(
3
i
+
1
)
30
3
+
i
15
3
+
i
5
]
[
y
1
y
2
y
3
]
\begin{aligned}Y’&=\begin{bmatrix}\dfrac{y_1}{2}-\dfrac{2(3+i)}{15i} \left(3y_3-\dfrac{3i+1}{2} y_1+y_2 \right)\\y_3-\dfrac{i}{2} y_1-\dfrac{2(3+i)}{45} \left(3y_3-\dfrac{3i+1}{2} y_1+y_2 \right)\\\dfrac{3+i}{15} \left(3y_3-\dfrac{3i+1}{2} y_1+y_2\right)\end{bmatrix}\\&=\begin{bmatrix}\dfrac{1}{2}+\dfrac{2(3+i)(3i+1)}{30i}&-\dfrac{2(3+i)}{15i}&-\dfrac{2(3+i)}{5i}\\-\dfrac{i}{2}+\dfrac{(3+i)(3i+1)}{45}&-\dfrac{2(3+i)}{45}&1-\dfrac{2(3+i)}{15}\\-\dfrac{(3+i)(3i+1)}{30}&\dfrac{3+i}{15}&\dfrac{3+i}{5}\end{bmatrix}\begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix}\end{aligned}
Y
′
=
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎡
2
y
1
−
1
5
i
2
(
3
+
i
)
(
3
y
3
−
2
3
i
+
1
y
1
+
y
2
)
y
3
−
2
i
y
1
−
4
5
2
(
3
+
i
)
(
3
y
3
−
2
3
i
+
1
y
1
+
y
2
)
1
5
3
+
i
(
3
y
3
−
2
3
i
+
1
y
1
+
y
2
)
⎦
⎥
⎥
⎥
⎥
⎥
⎥
⎤
=
⎣
⎢
⎢
⎢
⎢
⎡
2
1
+
3
0
i
2
(
3
+
i
)
(
3
i
+
1
)
−
2
i
+
4
5
(
3
+
i
)
(
3
i
+
1
)
−
3
0
(
3
+
i
)
(
3
i
+
1
)
−
1
5
i
2
(
3
+
i
)
−
4
5
2
(
3
+
i
)
1
5
3
+
i
−
5
i
2
(
3
+
i
)
1
−
1
5
2
(
3
+
i
)
5
3
+
i
⎦
⎥
⎥
⎥
⎥
⎤
⎣
⎡
y
1
y
2
y
3
⎦
⎤
thus
P
P
P
is the matrix on the left.
3. For each of the two matrices
[
2
5
−
1
4
−
1
2
6
4
1
]
,
[
1
−
1
2
3
2
4
0
1
−
2
]
\begin{bmatrix}2&5&-1\\4&-1&2\\6&4&1\end{bmatrix},\quad\begin{bmatrix}1&-1&2\\3&2&4\\0&1&-2\end{bmatrix}
⎣
⎡
2
4
6
5
−
1
4
−
1
2
1
⎦
⎤
,
⎣
⎡
1
3
0
−
1
2
1
2
4
−
2
⎦
⎤
use elementary row operations to discover whether it is invertible, and to find the inverse in case it is.
solution
: As
[
2
5
−
1
1
0
0
4
−
1
2
0
1
0
6
4
1
0
0
1
]
→
[
2
5
−
1
1
0
0
0
−
11
4
−
2
1
0
0
−
11
4
−
3
0
1
]
→
[
2
5
−
1
1
0
0
0
−
11
4
−
2
1
0
0
0
0
−
1
−
1
1
]
\begin{aligned}\begin{bmatrix}2&5&-1&1&0&0\\4&-1&2&0&1&0\\6&4&1&0&0&1\end{bmatrix}&\rightarrow\begin{bmatrix}2&5&-1&1&0&0\\0&-11&4&-2&1&0\\0&-11&4&-3&0&1\end{bmatrix}\\&\rightarrow\begin{bmatrix}2&5&-1&1&0&0\\0&-11&4&-2&1&0\\0&0&0&-1&-1&1\end{bmatrix}\end{aligned}
⎣
⎡
2
4
6
5
−
1
4
−
1
2
1
1
0
0
0
1
0
0
0
1
⎦
⎤
→
⎣
⎡
2
0
0
5
−
1
1
−
1
1
−
1
4
4
1
−
2
−
3
0
1
0
0
0
1
⎦
⎤
→
⎣
⎡
2
0
0
5
−
1
1
0
−
1
4
0
1
−
2
−
1
0
1
−
1
0
0
1
⎦
⎤
the first matrix is not invertible.
As
[
1
−
1
2
1
0
0
3
2
4
0
1
0
0
1
−
2
0
0
1
]
→
[
1
−
1
2
1
0
0
0
5
−
2
−
3
1
0
0
1
−
2
0
0
1
]
→
[
1
−
1
2
1
0
0
0
1
−
2
0
0
1
0
0
8
−
3
1
−
5
]
→
[
1
−
1
0
7
/
4
−
1
/
4
5
/
4
0
1
0
−
3
/
4
1
/
4
−
1
/
4
0
0
1
−
3
/
8
1
/
8
−
5
/
8
]
→
[
1
0
0
1
0
1
0
1
0
−
3
/
4
1
/
4
−
1
/
4
0
0
1
−
3
/
8
1
/
8
−
5
/
8
]
\begin{aligned}&\begin{bmatrix}1&-1&2&1&0&0\\3&2&4&0&1&0\\0&1&-2&0&0&1\end{bmatrix}\rightarrow\begin{bmatrix}1&-1&2&1&0&0\\0&5&-2&-3&1&0\\0&1&-2&0&0&1\end{bmatrix}\\&\rightarrow\begin{bmatrix}1&-1&2&1&0&0\\0&1&-2&0&0&1\\0&0&8&-3&1&-5\end{bmatrix}\rightarrow\begin{bmatrix}1&-1&0&7/4&-1/4&5/4\\0&1&0&-3/4&1/4&-1/4\\0&0&1&-3/8&1/8&-5/8\end{bmatrix}\\&\rightarrow\begin{bmatrix}1&0&0&1&0&1\\0&1&0&-3/4&1/4&-1/4\\0&0&1&-3/8&1/8&-5/8\end{bmatrix}\end{aligned}
⎣
⎡
1
3
0
−
1
2
1
2
4
−
2
1
0
0
0
1
0
0
0
1
⎦
⎤
→
⎣
⎡
1
0
0
−
1
5
1
2
−
2
−
2
1
−
3
0
0
1
0
0
0
1
⎦
⎤
→
⎣
⎡
1
0
0
−
1
1
0
2
−
2
8
1
0
−
3
0
0
1
0
1
−
5
⎦
⎤
→
⎣
⎡
1
0
0
−
1
1
0
0
0
1
7
/
4
−
3
/
4
−
3
/
8
−
1
/
4
1
/
4
1
/
8
5
/
4
−
1
/
4
−
5
/
8
⎦
⎤
→
⎣
⎡
1
0
0
0
1
0
0
0
1
1
−
3
/
4
−
3
/
8
0
1
/
4
1
/
8
1
−
1
/
4
−
5
/
8
⎦
⎤
the second matrix is invertible and its inverse is
[
1
0
1
−
3
/
4
1
/
4
−
1
/
4
−
3
/
8
1
/
8
−
5
/
8
]
\begin{bmatrix}1&0&1\\-3/4&1/4&-1/4\\-3/8&1/8&-5/8\end{bmatrix}
⎣
⎡
1
−
3
/
4
−
3
/
8
0
1
/
4
1
/
8
1
−
1
/
4
−
5
/
8
⎦
⎤
4. Let
A
=
[
5
0
0
1
5
0
0
1
5
]
A=\begin{bmatrix}5&0&0\\1&5&0\\0&1&5\end{bmatrix}
A
=
⎣
⎡
5
1
0
0
5
1
0
0
5
⎦
⎤
For which
X
X
X
does there exist a scalar
c
c
c
such that
A
X
=
c
X
AX=cX
A
X
=
c
X
?
solution
: If
X
=
0
X=0
X
=
0
, then
A
X
=
c
X
AX=cX
A
X
=
c
X
for any scalar
c
c
c
. If
X
≠
0
X\neq 0
X
=
0
, then the system
(
A
−
c
I
)
X
=
0
(A-cI)X=0
(
A
−
c
I
)
X
=
0
has non trivial solutions, if
c
≠
5
c\neq 5
c
=
5
we have
[
5
−
c
0
0
1
5
−
c
0
0
1
5
−
c
]
→
[
1
0
0
0
5
−
c
0
0
1
5
−
c
]
→
[
1
1
1
]
\begin{bmatrix}5-c&0&0\\1&5-c&0\\0&1&5-c \end{bmatrix} \rightarrow \begin{bmatrix}1&0&0\\0&5-c&0\\0&1&5-c \end{bmatrix}\rightarrow \begin{bmatrix}1&&\\&1&\\&&1 \end{bmatrix}
⎣
⎡
5
−
c
1
0
0
5
−
c
1
0
0
5
−
c
⎦
⎤
→
⎣
⎡
1
0
0
0
5
−
c
1
0
0
5
−
c
⎦
⎤
→
⎣
⎡
1
1
1
⎦
⎤
thus the system
(
A
−
c
I
)
X
=
0
(A-cI)X=0
(
A
−
c
I
)
X
=
0
has only trivial solutions, a contradiction. If
c
=
5
c=5
c
=
5
, then as long as
X
=
(
0
,
0
,
k
)
,
k
≠
0
X=(0,0,k),k\neq 0
X
=
(
0
,
0
,
k
)
,
k
=
0
, we have
A
X
=
5
X
AX=5X
A
X
=
5
X
. In conclusion we can say for
X
=
(
0
,
0
,
k
)
,
k
∈
R
X=(0,0,k),k\in R
X
=
(
0
,
0
,
k
)
,
k
∈
R
does there exists a scalar
c
c
c
s.t.
A
X
=
c
X
AX=cX
A
X
=
c
X
.
5. Discover whether
A
=
[
1
2
3
4
0
2
3
4
0
0
3
4
0
0
0
4
]
A=\begin{bmatrix}1&2&3&4\\0&2&3&4\\0&0&3&4\\0&0&0&4\end{bmatrix}
A
=
⎣
⎢
⎢
⎡
1
0
0
0
2
2
0
0
3
3
3
0
4
4
4
4
⎦
⎥
⎥
⎤
is invertible, and find
A
−
1
A^{-1}
A
−
1
if it exists.
solution
:
A
A
A
is invertible. we have
[
1
2
3
4
1
0
0
0
0
2
3
4
0
1
0
0
0
0
3
4
0
0
1
0
0
0
0
4
0
0
0
1
]
→
[
1
2
3
0
1
0
0
−
1
0
2
3
0
0
1
0
−
1
0
0
3
0
0
0
1
−
1
0
0
0
4
0
0
0
1
]
→
[
1
2
0
0
1
0
−
1
−
1
0
2
0
0
0
1
−
1
−
1
0
0
3
0
0
0
1
−
1
0
0
0
4
0
0
0
1
]
→
[
1
0
0
0
1
−
1
−
1
−
1
0
2
0
0
0
1
−
1
−
1
0
0
3
0
0
0
1
−
1
0
0
0
4
0
0
0
1
]
→
[
1
0
0
0
1
−
1
−
1
−
1
0
1
0
0
0
1
/
2
−
1
/
2
−
1
/
2
0
0
1
0
0
0
1
/
3
−
1
/
3
0
0
0
1
0
0
0
1
/
4
]
\begin{aligned}\begin{bmatrix}1&2&3&4&1&0&0&0\\0&2&3&4&0&1&0&0\\0&0&3&4&0&0&1&0\\0&0&0&4&0&0&0&1\end{bmatrix}&\rightarrow \begin{bmatrix}1&2&3&0&1&0&0&-1\\0&2&3&0&0&1&0&-1\\0&0&3&0&0&0&1&-1\\0&0&0&4&0&0&0&1\end{bmatrix}\\&\rightarrow \begin{bmatrix}1&2&0&0&1&0&-1&-1\\0&2&0&0&0&1&-1&-1\\0&0&3&0&0&0&1&-1\\0&0&0&4&0&0&0&1\end{bmatrix}\\&\rightarrow \begin{bmatrix}1&0&0&0&1&-1&-1&-1\\0&2&0&0&0&1&-1&-1\\0&0&3&0&0&0&1&-1\\0&0&0&4&0&0&0&1\end{bmatrix}\\&\rightarrow \begin{bmatrix}1&0&0&0&1&-1&-1&-1\\0&1&0&0&0&1/2&-1/2&-1/2\\0&0&1&0&0&0&1/3&-1/3\\0&0&0&1&0&0&0&1/4\end{bmatrix}\end{aligned}
⎣
⎢
⎢
⎡
1
0
0
0
2
2
0
0
3
3
3
0
4
4
4
4
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
⎦
⎥
⎥
⎤
→
⎣
⎢
⎢
⎡
1
0
0
0
2
2
0
0
3
3
3
0
0
0
0
4
1
0
0
0
0
1
0
0
0
0
1
0
−
1
−
1
−
1
1
⎦
⎥
⎥
⎤
→
⎣
⎢
⎢
⎡
1
0
0
0
2
2
0
0
0
0
3
0
0
0
0
4
1
0
0
0
0
1
0
0
−
1
−
1
1
0
−
1
−
1
−
1
1
⎦
⎥
⎥
⎤
→
⎣
⎢
⎢
⎡
1
0
0
0
0
2
0
0
0
0
3
0
0
0
0
4
1
0
0
0
−
1
1
0
0
−
1
−
1
1
0
−
1
−
1
−
1
1
⎦
⎥
⎥
⎤
→
⎣
⎢
⎢
⎡
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
1
0
0
0
−
1
1
/
2
0
0
−
1
−
1
/
2
1
/
3
0
−
1
−
1
/
2
−
1
/
3
1
/
4
⎦
⎥
⎥
⎤
thus
A
−
1
=
[
1
−
1
−
1
−
1
0
1
/
2
−
1
/
2
−
1
/
2
0
0
1
/
3
−
1
/
3
0
0
0
1
/
4
]
A^{-1}=\begin{bmatrix}1&-1&-1&-1\\0&1/2&-1/2&-1/2\\0&0&1/3&-1/3\\0&0&0&1/4\end{bmatrix}
A
−
1
=
⎣
⎢
⎢
⎡
1
0
0
0
−
1
1
/
2
0
0
−
1
−
1
/
2
1
/
3
0
−
1
−
1
/
2
−
1
/
3
1
/
4
⎦
⎥
⎥
⎤
6. Suppose
A
A
A
is a
2
×
1
2\times 1
2
×
1
matrix and that
B
B
B
is a
1
×
2
1\times 2
1
×
2
matrix. Prove that
C
=
A
B
C=AB
C
=
A
B
is not invertible.
solution
: Suppose
A
=
[
a
b
]
,
B
=
[
c
d
]
A=\begin{bmatrix}a\\b\end{bmatrix},B=\begin{bmatrix}c&d\end{bmatrix}
A
=
[
a
b
]
,
B
=
[
c
d
]
, then
C
=
A
B
=
[
a
c
a
d
b
c
b
d
]
C=AB=\begin{bmatrix}ac&ad\\bc&bd\end{bmatrix}
C
=
A
B
=
[
a
c
b
c
a
d
b
d
]
, use elementary row operation one can eliminate one of
C
C
C
’s rows if
a
≠
0
a\neq 0
a
=
0
, and the first row of
C
C
C
vanishes if
a
=
0
a=0
a
=
0
, thus
C
C
C
can’t be row equivalent to the identity matrix, and
C
C
C
is not invertible.
7. Let
A
A
A
be an
n
×
n
n\times n
n
×
n
(square) matrix. Prove the following two statements:
(a) If
A
A
A
is invertible and
A
B
=
0
AB=0
A
B
=
0
for some
n
×
n
n\times n
n
×
n
matrix
B
B
B
, then
B
=
0
B=0
B
=
0
.
(b) If
A
A
A
is not invertible, then there exists an
n
×
n
n\times n
n
×
n
matrix
B
B
B
such that
A
B
=
0
AB=0
A
B
=
0
but
B
≠
0
B\neq 0
B
=
0
.
solution
: (a) We have
P
A
=
I
PA=I
P
A
=
I
for some
P
P
P
, thus
B
=
I
B
=
P
A
B
=
P
0
=
0
B=IB=PAB=P0=0
B
=
I
B
=
P
A
B
=
P
0
=
0
.
(b) The system
A
X
=
0
AX=0
A
X
=
0
has non trivial solutions, let
x
≠
0
x\neq 0
x
=
0
be a solution and let
B
=
[
x
0
…
0
]
B=\begin{bmatrix}x&0&\dots&0\end{bmatrix}
B
=
[
x
0
…
0
]
, then
A
B
=
0
AB=0
A
B
=
0
.
8. Let
A
=
[
a
b
c
d
]
A=\begin{bmatrix}a&b\\c&d\end{bmatrix}
A
=
[
a
c
b
d
]
Prove, using elementary row operations, that
A
A
A
is invertible if and only if
(
a
d
−
b
c
)
≠
0
(ad-bc)\neq0
(
a
d
−
b
c
)
=
0
.
solution
: If
c
=
0
c=0
c
=
0
, then to make
A
A
A
invertible if and only if
(
a
≠
0
)
∧
(
d
≠
0
)
(a\neq 0)\wedge (d\neq 0)
(
a
=
0
)
∧
(
d
=
0
)
, which is equivalent to
a
d
=
a
d
−
b
c
≠
0
ad=ad-bc\neq 0
a
d
=
a
d
−
b
c
=
0
.
If
c
≠
0
c\neq 0
c
=
0
, then use elementary row operations, we have
A
=
[
a
b
c
d
]
→
[
a
c
b
c
a
c
a
d
]
→
[
a
c
b
c
0
a
d
−
b
c
]
→
[
a
b
0
a
d
−
b
c
]
,
a
≠
0
A
=
[
a
b
c
d
]
→
[
c
d
a
b
]
=
[
c
d
0
b
]
,
a
=
0
A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\rightarrow\begin{bmatrix}ac&bc\\ac&ad\end{bmatrix}\rightarrow\begin{bmatrix}ac&bc\\0&ad-bc\end{bmatrix}\rightarrow\begin{bmatrix}a&b\\0&ad-bc\end{bmatrix},\quad a\neq 0 \\ A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\rightarrow\begin{bmatrix}c&d\\a&b\end{bmatrix}=\begin{bmatrix}c&d\\0&b\end{bmatrix},\quad a=0
A
=
[
a
c
b
d
]
→
[
a
c
a
c
b
c
a
d
]
→
[
a
c
0
b
c
a
d
−
b
c
]
→
[
a
0
b
a
d
−
b
c
]
,
a
=
0
A
=
[
a
c
b
d
]
→
[
c
a
d
b
]
=
[
c
0
d
b
]
,
a
=
0
thus if
A
A
A
is invertible, we shall have
a
d
−
b
c
≠
0
ad-bc\neq 0
a
d
−
b
c
=
0
. Conversely if
a
d
−
b
c
≠
0
ad-bc\neq 0
a
d
−
b
c
=
0
, then in the case
a
≠
0
a\neq 0
a
=
0
we directly have
A
A
A
is invertible since
[
a
b
0
a
d
−
b
c
]
\begin{bmatrix}a&b\\0&ad-bc\end{bmatrix}
[
a
0
b
a
d
−
b
c
]
is row equivalent to
I
2
I_2
I
2
, in the case
a
=
0
a=0
a
=
0
, notice then
b
c
≠
0
bc\neq 0
b
c
=
0
and thus
b
≠
0
b\neq 0
b
=
0
, so
[
c
d
0
b
]
\begin{bmatrix}c&d\\0&b\end{bmatrix}
[
c
0
d
b
]
is row equivalent to
I
2
I_2
I
2
, in either case,
A
A
A
is invertible.
9. An
n
×
n
n\times n
n
×
n
matrix
A
A
A
is called upper-triangular if
A
i
j
=
0
,
i
>
j
A_{ij}=0,i>j
A
i
j
=
0
,
i
>
j
, that is, if every entry below the main diagonal is 0. Prove that an upper-triangular (square) matrix is invertible if and only if every entry on its main diagonal is different from 0.
solution
: If every entry on its main diagonal is not
0
0
0
, then we use row
n
n
n
to eliminate the last column to
e
n
e_n
e
n
, and next use row
n
−
1
n-1
n
−
1
to eliminate column
n
−
1
n-1
n
−
1
to
e
n
−
1
e_{n-1}
e
n
−
1
, continuing we make
A
A
A
to
I
n
I_n
I
n
, thus
A
A
A
is invertible.
Conversely, suppose
A
A
A
is invertible and assume some entry on its main diagonal is
0
0
0
, let
k
k
k
be the largest integer s.t.
A
k
k
=
0
A_{kk}=0
A
k
k
=
0
, then if
k
=
n
k=n
k
=
n
then row
n
n
n
is all
0
0
0
, if not then repeating the same steps in the previous paragragh
n
−
k
n-k
n
−
k
times, we can get row
k
k
k
to be all
0
0
0
, then
A
A
A
is not invertible, as the matrix can’t be row-equivalent to
I
n
I_n
I
n
.
10. Prove the following generalization of Exercise 6. If
A
A
A
is an
m
×
n
m\times n
m
×
n
matrix,
B
B
B
is an
n
×
m
n\times m
n
×
m
matrix and
n
<
m
n<m
n
<
m
, then
A
B
AB
A
B
is not invertible.
solution
: Since
n
<
m
n<m
n
<
m
, if row-reduce
A
A
A
to an row-deduced echelon form, the last row must be all
0
0
0
, i.e. there’s an
m
×
m
m\times m
m
×
m
invertible matrix
P
P
P
s.t.
P
A
=
C
PA=C
P
A
=
C
, where the last row of
C
C
C
is all
0
0
0
, now assume
A
B
AB
A
B
is invertible, as it’s square we can have
D
D
D
be it’s right inverse, so
A
B
D
=
I
ABD=I
A
B
D
=
I
, thus
P
A
B
D
=
C
B
D
=
P
PABD=CBD=P
P
A
B
D
=
C
B
D
=
P
, now the last row of
C
C
C
is
0
0
0
, means the last row of
C
B
D
CBD
C
B
D
,or
P
P
P
, is
0
0
0
, this contradicts
P
P
P
being invertible.
11. Let
A
A
A
be an
m
×
n
m\times n
m
×
n
matrix. Show that by means of a finite number of elementary row and/or column operations one can pass from
A
A
A
to a matrix
R
R
R
which is both ‘row-reduced echelon’ and ‘column-reduced echelon’, i.e.
R
i
j
=
0
R_{ij}=0
R
i
j
=
0
if
i
≠
j
i\neq j
i
=
j
,
R
i
i
=
1
,
1
≤
i
≤
r
,
R
i
i
=
0
R_{ii}=1,1\leq i\leq r, R_{ii}=0
R
i
i
=
1
,
1
≤
i
≤
r
,
R
i
i
=
0
if
i
>
r
i>r
i
>
r
. Show that
R
=
P
A
Q
R=PAQ
R
=
P
A
Q
, where
P
P
P
is an invertible
m
×
m
m\times m
m
×
m
matrix and
Q
Q
Q
is an invertible
n
×
n
n\times n
n
×
n
matrix.
solution
: We’ve proved use elementary row operations we can get a row-reduced echelon matrix
R
′
R’
R
′
such that
R
′
=
P
A
R’=PA
R
′
=
P
A
, where
P
P
P
is an invertible
m
×
m
m\times m
m
×
m
matrix. Starting from
R
′
R’
R
′
we can use elementary column operations to get a column-reduced echelon matrix
R
R
R
, and an invertible
n
×
n
n\times n
n
×
n
matrix
Q
Q
Q
s.t.
R
=
R
′
Q
R=R’Q
R
=
R
′
Q
, thus
R
=
P
A
Q
R=PAQ
R
=
P
A
Q
.
12. The result of Example 16 suggests that perhaps the matrix
[
1
1
2
⋯
1
n
1
2
1
3
⋯
1
n
+
1
⋮
⋮
⋮
1
n
1
n
+
1
⋯
1
2
n
−
1
]
\begin{bmatrix}1&\dfrac{1}{2}&\cdots&\dfrac{1}{n}\\ \dfrac{1}{2}&\dfrac{1}{3}&\cdots&\dfrac{1}{n+1}\\\vdots&\vdots&&\vdots\\\dfrac{1}{n}&\dfrac{1}{n+1}&\cdots&\dfrac{1}{2n-1}\end{bmatrix}
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎡
1
2
1
⋮
n
1
2
1
3
1
⋮
n
+
1
1
⋯
⋯
⋯
n
1
n
+
1
1
⋮
2
n
−
1
1
⎦
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎤
is invertible and
A
−
1
A^{-1}
A
−
1
has integer entries. Can you prove that?
solution
: This matrix is called the Hilbert matrix, it’s invertible and its inverse is given by
B
=
(
B
i
j
)
B=(B_{ij})
B
=
(
B
i
j
)
, where
B
i
j
=
(
−
1
)
i
+
j
(
i
+
j
+
1
)
(
i
+
j
i
)
(
i
+
j
j
)
(
n
+
i
+
1
n
−
j
)
(
n
+
j
+
1
n
−
i
)
B_ij=(-1)^{i+j} (i+j+1)\binom{i+j}{i}\binom{i+j}{j}\binom{n+i+1}{n-j}\binom{n+j+1}{n-i}
B
i
j
=
(
−
1
)
i
+
j
(
i
+
j
+
1
)
(
i
i
+
j
)
(
j
i
+
j
)
(
n
−
j
n
+
i
+
1
)
(
n
−
i
n
+
j
+
1
)
One can direct compute the product
A
B
AB
A
B
and verify that
A
B
=
I
AB=I
A
B
=
I
.