可逆矩阵

矩阵以矩形阵列的形式表示元素。矩阵由行和列组成。水平线称为行，垂直线称为列。矩阵的顺序定义为行数×列数。如果矩阵中的行数为“ m”且列数为“ n”，则矩阵的顺序表示为“ m×n”。

矩阵的逆

假设“ A”是一个正方形矩阵，那么现在，如果“ A”矩阵存在相同维数的另一个矩阵“ B”，则该A矩阵只有在一个条件下才是可逆的_{，从而AB = BA = I n} ，其中I _n是已知的作为相同阶的单位矩阵和矩阵“ B”被称为矩阵“ A”的逆。矩阵的逆可以表示为A ^-1 。也称为非奇异矩阵或非简并矩阵。

例如：

A = $\begin{bmatrix} 5 &6 \\ 4&5 \end{bmatrix}$

and B = $\begin{bmatrix} 5 &-6 \\ -4&5 \end{bmatrix}$

On multiplying A and B you get,

AB = $\begin{bmatrix} 5 &6 \\ 4&5 \end{bmatrix} \begin{bmatrix} 5&-6 \\ -4&5 \end{bmatrix}$

AB = $\begin{bmatrix} 25-24 &-30+30 \\ 20-20&-24+25 \end{bmatrix}$

AB = $\begin{bmatrix}1 &0 \\ 0&1 \end{bmatrix}$

AB = I ………. (1)

Similarly, you can get BA by multiplying matrix B and matrix A.

BA =$\begin{bmatrix} 5 &-6 \\ -4&5 \end{bmatrix} \begin{bmatrix} 5 &6 \\ 4&5 \end{bmatrix}$

BA = $\begin{bmatrix} 25-24 &30-30 \\ -20+20&-24+25 \end{bmatrix}$

BA =$\begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix}$

BA = I………… (2)

From (1) and (2), you can see that AB = BA = I_n

Hence, A is an invertible matrix and inverse of matrix A is matrix B. This can be written as A^-1 = B.

If B is inverse matrix for A then also, A is inverse matrix for B. So, you can write B^-1 = A.

注意：方矩阵A拥有逆矩阵的必要和充分条件是矩阵不应该是奇异的。如果矩阵的行列式为零，即| A |，则称该矩阵为奇异矩阵=0。所以| A |对于可逆矩阵A≠0。

可逆矩阵的应用：

可逆矩阵的应用是：

1. 最小二乘或回归
2. 模拟
3. MIMO无线通讯

矩阵求逆方法：

使用以下方法，您可以找到另一个矩阵，例如“ B”矩阵，它是矩阵“ A”的逆矩阵：

1. 高斯消除
2. 牛顿法
3. Cayley-Hamilton方法
4. 本征分解法

示例：检查矩阵A = $\begin{bmatrix} 0 &1 &2 \\ 1&2 &3 \\ 3& 1 & 1 \end{bmatrix}$是可逆的还是不可逆的。如果A是可逆的，则检查矩阵B是否= $\frac{-1}{2}\begin{bmatrix} -1 &1 &-1 \\ 8&-6 &2 \\ -5& 3 & -1 \end{bmatrix}$是否为矩阵A的逆。

解决方案：

First we check whether matrix A is invertible or not.

|A| = 0×(2-3) – 1×(1-2) + 3×(3-4)

|A| = 0+1-3

|A| = -2 ≠0

Hence. matrix A is invertible.

Now, check whether AB=BA=I_n or not

AB = $\begin{bmatrix} 0 &1 &2 \\ 1&2 &3 \\ 3& 1 & 1 \end{bmatrix}\frac{-1}{2}\begin{bmatrix} -1 &1 &-1 \\ 8&-6 &2 \\ -5& 3 & -1 \end{bmatrix}$

AB = $\frac{-1}{2} \begin{bmatrix} 0 &1 &2\\ 1&2 &3 \\ 3& 1& 1 \end{bmatrix} \begin{bmatrix} -1 &1 &-1 \\ 8&-6 &2 \\ -5& 3 & -1 \end{bmatrix}$

AB = $\frac{-1}{2}\begin{bmatrix} 0+8-10 &0-6+6 &0+2-2 \\ -1+16-15&1-12+9 &-1+4-3 \\ -3+8-5& 3-6+3 & -3+2-1 \end{bmatrix}$

AB = $\frac{-1}{2} \begin{bmatrix} -2&0 &0\\ 0&-2 &0 \\ 0& 0& -2 \end{bmatrix}$

AB = $\begin{bmatrix} 1&0 &0\\ 0&1 &0 \\ 0& 0& 1 \end{bmatrix}$

AB = I

BA =$\frac{-1}{2}\begin{bmatrix} -1 &1 &-1 \\ 8&-6 &2 \\ -5& 3 & -1 \end{bmatrix} \begin{bmatrix} 0 &1 &2 \\ 1&2 &3 \\ 3& 1 & 1 \end{bmatrix}$

BA = $\frac{-1}{2}\begin{bmatrix} -1 &1 &-1 \\ 8&-6 &2 \\ -5& 3 & -1 \end{bmatrix} \begin{bmatrix} 0 &1 &2\\ 1&2 &3 \\ 3& 1& 1 \end{bmatrix}$

BA = $\frac{-1}{2}\begin{bmatrix} 0+1-3 &-1+2-1 &-2+3-1 \\ 0-6+6&8-12+2 &16-18+2 \\ 0+3-3& -5+6-1 & -10+9-1 \end{bmatrix}$

BA = $\frac{-1}{2} \begin{bmatrix} -2&0 &0\\ 0&-2 &0 \\ 0& 0& -2 \end{bmatrix}$

BA = $\begin{bmatrix} 1&0 &0\\ 0&1 &0 \\ 0& 0& 1 \end{bmatrix}$

BA = I

You can see, AB = BA = I

Hence, matrix B is inverse of matrix A.

可逆矩阵定理

定理1：每个可逆矩阵都具有唯一的逆。

证明：

Let ‘A’ be an n×n invertible matrix.

Let us considered B and C be two inverse of A.

Then, AB = BA = I …….(1)

and AC = CA = I …….. (2)

From (1) you have

C(AB) = C(I_n) = C ……..(3)

From (2) you have

(CA)B = I_n(B) = B ……. (4)

Since, C(AB) = (CA)B [Associativity Law]

So, C = B

Hence, it is proved.

例子：

Let A = $\begin{bmatrix} 2 &9 \\ 1& 7 \end{bmatrix}$, B = $\begin{bmatrix} b_{1} &b_{2} \\ b_{3}& b_{4} \end{bmatrix}$, and C = $\begin{bmatrix} c_{1} &c_{2} \\ c_{3}& c_{4} \end{bmatrix}$

If both matrices B and C are inverse of matrix A then,

AB = BA = I and

AC = CA = I 

Taking AB = I

$\begin{bmatrix} 2 &9 \\ 1& 7 \end{bmatrix} \begin{bmatrix} b_{1} &b_{2} \\ b_{3}& b_{4} \end{bmatrix} = \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$

$\begin{bmatrix} 2b_{1}+9b_{3} &2b_{2}+9b_{4} \\ b_{1}+7b_{3}& b_{2}+7b_{4} \end{bmatrix} = \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$

From above, you get 4 equations 

2 b1 + 9 b3 = 1 ………. (1)

b1 + 7 b3 = 0 …………….(2)

2 b2 + 9 b4 = 0 …………..(3)

b2 + 7 b4 = 1 ……………….(4)

after solving these 4 equations, you will get 

b1 = 7/5

b2 = -9/5

b3 = -1/5

b4 = 2/5 

So matrix B = $\begin{bmatrix} \frac{7}{5} &\frac{-9}{5} \\ \frac{-1}{5}& \frac{2}{5} \end{bmatrix}$………… (5)

Now, consider AC = I 

$\begin{bmatrix} 2 &9 \\ 1& 7 \end{bmatrix} \begin{bmatrix} c_{1} &c_{2} \\ c_{3}& c_{4} \end{bmatrix} = \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$

$\begin{bmatrix} 2c_{1}+9c_{3} &2c_{2}+9c_{4} \\ c_{1}+7c_{3}& c_{2}+7c_{4} \end{bmatrix} = \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$

From above, you get 4 equations

2 b1 + 9 b3 = 1 ………. (6)

b1 + 7 b3 = 0 …………….(7)

2 b2 + 9 b4 = 0 …………..(8)

b2 + 7 b4 = 1 ……………….(9)

After solving these 4 equations, you will get

b1 = 7/5

b2 = -9/5

b3 = -1/5

b4 = 2/5

So matrix B =$\begin{bmatrix} \frac{7}{5} &\frac{-9}{5} \\ \frac{-1}{5}& \frac{2}{5} \end{bmatrix}$ ………..(10)

From (9) and (10) you can see that matrix B and C are equal.

Hence, it is proved that any invertible matrix posses unique inverse.

定理2：如果A，B是两个n行非奇异矩阵，则AB也是非奇异的(AB) ^-1 = B ^-1 A ^-1

证明：

|A| ≠0, |B| ≠0

So, |AB| ≠0

Let a matrix C = B^-1A^-1

(AB)C = (AB)B^-1A^-1

(AB)C = A(BB^-1)A^-1

(AB)C = AI_nA^-1

(AB)C = AA^-1

(AB)C = I_n

C(AB) = B^-1A^-1(AB)

C(AB) = B^-1A^-1AB

C(AB) = B^-1B

C(AB) = I_n

Since (AB)C = C(AB) = I_n

Hence, C is inverse of (AB)

So (AB)^-1 = B^-1A^-1

例子：

Let A =$\begin{bmatrix} 1 &4 \\ 2& 0 \end{bmatrix}$ and B =$\begin{bmatrix} 3 &0 \\ 1& 2 \end{bmatrix}$

AB = $\begin{bmatrix} 1 &4 \\ 2& 0 \end{bmatrix} \begin{bmatrix} 3 &0 \\ 1& 2 \end{bmatrix}$

AB = $\begin{bmatrix} 3+4 &0+8 \\ 6+0& 0 \end{bmatrix}$

AB = $\begin{bmatrix} 7 &8 \\ 6& 0 \end{bmatrix}$

(AB)^-1 = \frac{-1}{48}\begin{bmatrix} 0 &-8 \\ -6& 7 \end{bmatrix} ………. (1)

Inverse of a matrix can be obtained by the given formula

A^-1 = adjoint of matrix A/ |A|

B = $\begin{bmatrix} 3 &0 \\ 1& 2 \end{bmatrix}$

B^-1 = $\frac{1}{6}\begin{bmatrix} 2 &0 \\ -1& 3 \end{bmatrix}$

A = $\begin{bmatrix} 1 &4 \\ 2& 0 \end{bmatrix}$

A^-1 = $\frac{1}{-8}\begin{bmatrix} 0 &-4 \\ -2& 1 \end{bmatrix}$

B^-1A^-1 = $\frac{1}{6}\begin{bmatrix} 2 &0 \\ -1& 3 \end{bmatrix} \frac{1}{-8}\begin{bmatrix} 0 &-4 \\ -2& 1 \end{bmatrix}$

B^-1A^-1 = $\frac{-1}{48}\begin{bmatrix} 0 &-8+0 \\ 0-6& 4+3 \end{bmatrix}$

B^-1A^-1 = $\frac{-1}{48}\begin{bmatrix} 0 &-8 \\ -6& 7 \end{bmatrix}$ …….. (2)

From (1) and (2), you can see that (AB)^-1 = B^-1A^-1

Hence, it is proved that (AB)^-1 = B^-1A^-1

方阵逆矩阵的性质

1.(A ^-1 ) ^-1 = A

证明：

If A is an invertible matrix then

AA^-1 = I 

Taking inverse on both sides

(AA^-1)^-1 = I^-1 

(A^-1)^-1A^-1 = I [from theorem 2 (AB)^-1 = B^-1A^-1]

Multiplying by A on both sides

(A^-1)^-1A^-1A = IA

(A^-1)^-1I = A

(A^-1)^-1 = A

Hence, it is proved that (A^-1)^-1 = A

例子：

Let A = $\begin{bmatrix} 3 &1 \\ 2 & 4 \end{bmatrix}$

|A| = 12 – 2 = 10

adj A = $\begin{bmatrix} 4 &-1 \\ -2 & 3 \end{bmatrix}$

A^-1 = adj A /|A|

A^-1 = $\frac{1}{10}\begin{bmatrix} 4 &-1 \\ -2 & 3 \end{bmatrix}$

(A^-1)^-1 = adj (A^-1) / |A^-1|

adj(A^-1) = $\frac{1}{10}\begin{bmatrix} 3 &1 \\ 2 & 4 \end{bmatrix}$

|A^-1| = (12 – 10)/100

|A^-1| = 1/10

(A^-1)^-1 = $10*\frac{1}{10}\begin{bmatrix} 3 &1 \\ 2 & 4 \end{bmatrix}$

(A^-1)^-1 = $\begin{bmatrix} 3 &1 \\ 2 & 4 \end{bmatrix}$

From above you can see that (A^-1)^-1 = A

2. (A ₁ A ₂ A ₃ ………..A _n ) ^-1 = A _n ^-1 A _n-1 ^-1 ……….A ₂ ^-1 A ₁ ^-1

您还可以编写它：

(AB) ^-1 = A ^-1 B ^-1

(ABC) ^-1 = A ^-1 B ^-1 C ^-1

证明：

This can be proved by mathematical induction

for n = 2 

(A₁A₂)^-1 = A₂^-1A₁^-1 ……….(1)

This statement is true. [by theorem 2]

Let this is true for n = k

(A₁A₂A₃……….A_k)^-1 = A_k^-1…………A₂^-1A₁^-1……..(2)

For n = k+1, you have to prove this.

(A₁A₂A₃……….A_kA_k+1)^-1

=((A₁A₂A₃………A_k)A_k+1)^-1

=((A_k^-1…………A₂^-1A₁^-1)A_k+1)^-1

=(A_k+1)^-1 (A_k^-1…………A₂^-1A₁^-1)    [using theorem 2]

= A_k+1^-1A_k^-1………….A₂^-1A₁^-1 

Hence, it is proved.

例子：

Suppose there are two matrices A and B, 

Let A = $\begin{bmatrix} 1 &4 \\ 2& 0 \end{bmatrix}$

and B =$\begin{bmatrix} 3 &0 \\ 1& 2 \end{bmatrix}$

AB = $\begin{bmatrix} 1 &4 \\ 2& 0 \end{bmatrix} \begin{bmatrix} 3 &0 \\ 1& 2 \end{bmatrix}$

AB = $\begin{bmatrix} 3+4 &0+8 \\ 6+0& 0 \end{bmatrix}$

AB = $\begin{bmatrix} 7 &8 \\ 6& 0 \end{bmatrix}$

(AB)^-1 = $\frac{-1}{48}\begin{bmatrix} 0 &-8 \\ -6& 7 \end{bmatrix}$ ………. (1)

Inverse of a matrix can be obtained by the given formula

A^-1 = adjoint of matrix A/ |A|

B = $\begin{bmatrix} 3 &0 \\ 1& 2 \end{bmatrix}$

B^-1 = $\frac{1}{6}\begin{bmatrix} 2 &0 \\ -1& 3 \end{bmatrix}$

A = $\begin{bmatrix} 1 &4 \\ 2& 0 \end{bmatrix}$

A^-1 = $\frac{1}{-8}\begin{bmatrix} 0 &-4 \\ -2& 1 \end{bmatrix}$

B^-1A^-1 = $\frac{1}{6}\begin{bmatrix} 2 &0 \\ -1& 3 \end{bmatrix} \frac{1}{-8}\begin{bmatrix} 0 &-4 \\ -2& 1 \end{bmatrix}$

B^-1A^-1 = $\frac{-1}{48}\begin{bmatrix} 0 &-8+0 \\ 0-6& 4+3 \end{bmatrix}$

B^-1A^-1 = $\frac{-1}{48}\begin{bmatrix} 0 &-8 \\ -6& 7 \end{bmatrix}$…….. (2)

From (1) and (2), you can see that (AB)^-1 = B^-1A^-1

Hence, it is proved that (AB)^-1 = B^-1A^-1

3. AA ^-1 = A ^-1 A = I _n

证明：

A matrix is invertible if AA^-1 = I

Multiply by A on both sides

AAA^-1 = AI

AI = A

Multiplying by A^-1 on both sides

A^-1AI = A^-1A

I = A^-1A

Hence, it is proved that AA^-1 = I = A^-1A

例子：

Let A = $\begin{bmatrix} 1 &2 \\ -1 &3 \end{bmatrix}$

|A| = 3 + 2 = 5

adj A = $\begin{bmatrix} 3 &-2 \\ 1 &1 \end{bmatrix}$

A^-1= adj A \|A|

A^-1 = $\frac{1}{5}\begin{bmatrix} 3 &-2 \\ 1 &1 \end{bmatrix}$

Now to prove AA^-1 = A^-1A = I

Taking left hand side

AA^-1

$\begin{bmatrix} 1 &2 \\ -1 &3 \end{bmatrix}\frac{1}{5}\begin{bmatrix} 3 &-2 \\ 1 &1 \end{bmatrix}$

$\frac{1}{5}\begin{bmatrix} 3 +2 &-2 +2\\ -3+3 &2+3 \end{bmatrix}$

$\frac{1}{5}\begin{bmatrix} 5 & 0\\ 0&5 \end{bmatrix}$

$\begin{bmatrix} 1 & 0\\ 0&1 \end{bmatrix}$

The above matrix is equal to the identity matrix.

Now taking right side

A^-1A

$\frac{1}{5} \begin{bmatrix} 3 &-2 \\ 1 &1 \end{bmatrix} \begin{bmatrix} 1 &2 \\ -1 &3 \end{bmatrix}$

$\frac{1}{5}\begin{bmatrix} 3+2 & 6-6\\ 1-1&2+3 \end{bmatrix}$

$\frac{1}{5}\begin{bmatrix} 5 & 0\\ 0&5 \end{bmatrix}$

$\begin{bmatrix} 1 & 0\\ 0&1 \end{bmatrix}$

The above matrix is equal to the identity matrix.

Hence, it is proved that AA^-1 = A^-1A = I

更多属性：

• (A ^T ) ^-1 =(A ^-1 ) ^T
• (kA) ^-1 =(1 / k)A ^-1
• AB = I _n ，其中A和B彼此相反。
• 如果A是其中n> 0的方阵，则(A ^-1 ) ⁿ = A ^-n