第 12 课 NCERT 解决方案 - 数学第一部分 - 第 4 章行列式 - 练习 4.5

求练习 1 和 2 中每个矩阵的伴随矩阵。

问题 1。 $\begin{bmatrix} 1 &2 \\ 3 & 4 \end{bmatrix}$

解决方案：

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

A₁₁ = 4; A₁₂ = -3; A₂₁ = -2; A₂₂ = 1

adj A = $\begin{bmatrix} A_{11} &A_{21} \\ A_{12} & A_{22} \end{bmatrix}$

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

问题2。 $\begin{bmatrix} 1 & -1 &2 \\ 2&3 &5 \\ -2& 0 &1 \end{bmatrix}$

解决方案：

A = $\begin{bmatrix} 1 & -1 &2 \\ 2&3 &5 \\ -2& 0 &1 \end{bmatrix}$

A₁₁ = $\begin{vmatrix} 3 &5 \\ 0& 1 \end{vmatrix}$

A₁₁ = 3 – 0 = 3

A₁₂ = $-\begin{vmatrix} 2&5 \\ -2& 1 \end{vmatrix}$

A₁₂ = -(2 + 10) = -12

A₁₃ = $\begin{vmatrix} 2&3\\ -2& 0 \end{vmatrix}$

A₁₃ = 0 + 6 = 6

A₂₁ = $-\begin{vmatrix} -1&2\\ 0& 1 \end{vmatrix}$

A₂₁ = -(-1 – 0) = 1

A₂₂ = $\begin{vmatrix} 1&2\\ -2& 1 \end{vmatrix}$

A₂₂ = 1 + 4 = 5

A₂₃ = $-\begin{vmatrix} 1&-1\\ -2&0 \end{vmatrix}$

A₂₃ = -(0 – 2) = 2

A₃₁ = $\begin{vmatrix} -1&2\\ 3&5 \end{vmatrix}$

A₃₁ = -5 – 6 = -11

A₃₂ = $-\begin{vmatrix} 1&2\\ 2&5 \end{vmatrix}$

A₃₂ = -(5 – 4) = -1

A₃₃ = $\begin{vmatrix} 1&-1\\ 2&3 \end{vmatrix}$

A₃₃ = 3 + 2 = 5

adj A = $\begin{bmatrix} A_{11} & A_{21} &A_{31} \\ A_{12}&A_{22} &A_{32} \\ A_{13}& A_{23} &A_{33} \end{bmatrix}$

adj A = $\begin{bmatrix} 3& 1 &-11 \\ -12&5 &-1 \\ 6& 2 &5 \end{bmatrix}$

验证 A(adj A) = (adj A)A = |A|我在练习 3 和 4 中。

问题 3。 $\begin{bmatrix} 2 &3 \\ -4 & -6 \end{bmatrix}$

解决方案：

|A| = -12 -(-12) 

|A| = -12 + 12 = 0

so, |A|*I = 0 * $\begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}$

|A|*I =$\begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix}$ ……… (1)

Now, for adjoint of A

A₁₁ = -6

A₁₂ = 4

A₂₁ = -3

A₂₂ = 2

adj A = $\begin{bmatrix} A_{11} &A_{21} \\ A_{12} & A_{22} \end{bmatrix}$

adj A = $\begin{bmatrix} -6&-3 \\ 4&2 \end{bmatrix}$

Now, A(adj A) = $\begin{bmatrix} 2&3 \\ -4&-6 \end{bmatrix} \begin{bmatrix} -6&-3 \\ 4&2 \end{bmatrix}$

A(adj A) = $\begin{bmatrix} -12+12&-6+6 \\ 24-24&12-12 \end{bmatrix}$[Tex]\begin{bmatrix} 11& 0  &0 \\ 0 & 11 &0 \\ 0 & 0 & 11 \end{bmatrix}[/Tex]

A(adj A) = $\begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix}$…………(2)

Now, (adj A)A = $\begin{bmatrix} -6&-3 \\ 4&2 \end{bmatrix} \begin{bmatrix} 2&3 \\ -4&-6 \end{bmatrix}$

(adj A)A = $\begin{bmatrix} -12+12&-18+18 \\ 8-8&12-12 \end{bmatrix}$

(adj A)A = $\begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix}$ …………….(3)

From eq(1), (2), and (3), you can see that A(adj A) = (adj A)A = |A|I

问题 4。 $\begin{bmatrix} 1 &-1 &2 \\ 3 & 0 &-2 \\ 1 & 0 & 3 \end{bmatrix}$

解决方案：

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

|A| = 1(0 – 0) + 1(9 + 2) + 2(0 – 0)

|A| = 11

|A| * I = $11\begin{bmatrix} 1& 0 &0 \\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

|A| * I = $\begin{bmatrix} 11& 0 &0 \\ 0 & 11 &0 \\ 0 & 0 & 11 \end{bmatrix}$

Now, for adjoint of A

A₁₁ = 0

A₁₂ = -(9 + 2) = -11

A₁₃ = 0

A₂₁ = -(-3 – 0) = 3

A₂₂ = 3 – 2 = 1

A₂₃ = -(0 + 1) = -1

A₃₁ = 2 – 0 = 2

A₃₂ = -(-2 – 6) = 8

A₃₃ = 0 + 3 = 3

adj A = $\begin{bmatrix} 0& 3 &2 \\ -11 & 1 &8 \\ 0 & -1 & 3 \end{bmatrix}$

Now,

A(adjA) = $\begin{bmatrix} 1 &-1 &2 \\ 3 & 0 &-2 \\ 1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 0& 3 &2 \\ -11 & 1 &8 \\ 0 & -1 & 3 \end{bmatrix}$

A(adj A) = $\begin{bmatrix} 0+11+0& 3-1-2 &2-8+6 \\ 0+0+0 & 9+0+2 &6+0-6 \\ 0+0+0 & 3+0-3 & 2+0+9 \end{bmatrix}$

A(adj A) = $\begin{bmatrix} 11& 0 &0 \\ 0 & 11 &0 \\ 0 & 0 & 11 \end{bmatrix}$

Also,

(adj A).A = $\begin{bmatrix} 0& 3 &2 \\ -11 & 1 &8 \\ 0 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 &-1 &2 \\ 3 & 0 &-2 \\ 1 & 0 & 3 \end{bmatrix}$

(adj A).A = $\begin{bmatrix} 0+9+2& 0+0+0 &0-6+6 \\ -11+3+8 & 11+0+0 &-22-2+24 \\ 0-3+3 & 0+0+0 & 0+2+9 \end{bmatrix}$

(adj A).A = $\begin{bmatrix} 11& 0 &0 \\ 0 & 11 &0 \\ 0 & 0 & 11 \end{bmatrix}$

From above, you can see,

A(adj A) = (adj A)A = I

求练习 5 到 11 中给出的每个矩阵（如果存在）的逆矩阵。

问题 5。 $\begin{bmatrix} 2 & -2\\ 4 & 3 \end{bmatrix}$

解决方案：

|A| = 6 – (-8) = 14

|A| ≠ 0, So inverse exists.

A₁₁ = 3

A₁₂ = 2

A₂₁ = -4

A₂₂ = 2

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

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

A^-1 = $\frac{1}{14}\begin{bmatrix} 3 & 2\\ -4 & 2 \end{bmatrix}$

问题 6。 $\begin{bmatrix} -1 & 5\\ -3 & 2 \end{bmatrix}$

解决方案：

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

|A| = -2 + 15 = 13 ≠ 0

Hence, inverse exists.

A₁₁ = 2

A₁₂ = 3

A₂₁ = -5

A₂₂ = -1

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

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

A^-1 = $\frac{1}{13}\begin{bmatrix} 2 &-5 \\ 3& -1 \end{bmatrix}$

问题 7。 $\begin{bmatrix} 1& 2 &3 \\ 0 & 2 &4 \\ 0 & 0 & 5 \end{bmatrix}$

解决方案：

A = $\begin{bmatrix} 1& 2 &3 \\ 0 & 2 &4 \\ 0 & 0 & 5 \end{bmatrix}$

|A| = 1(10 – 0) – 2(0 – 0) + 3(0 – 0) = 10

For adj A

A₁₁ = 10 – 0 = 0

A₁₂ = -(0 – 0) = 0

A₁₃ = 0 – 0 = 0

A₂₁ = -(10 – 0) = -10

A₂₂ = 5 – 0 = 5

A₂₃ = -(0 – 0) = 0

A₃₁ = 8 – 6 = 2

A₃₂ = -(4 – 0) = -4

A₃₃ = 2 – 0 = 2

adj A = $\begin{bmatrix} 10& -10 &2 \\ 0 & 5 &-4 \\ 0 & 0 & 2 \end{bmatrix}$

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

A^-1 = $\frac{1}{10}\begin{bmatrix} 10& -10 &2 \\ 0 & 5 &-4 \\ 0 & 0 & 2 \end{bmatrix}$

问题 8。 $\begin{bmatrix} 1& 0 &0 \\ 3 & 3 &0 \\ 5 & 2 & -1 \end{bmatrix}$

解决方案：

A = $\begin{bmatrix} 1& 0 &0 \\ 3 & 3 &0 \\ 5 & 2 & -1 \end{bmatrix}$

|A| = 1(-3 – 0) – 0 + 0 = -3 ≠ 0

Hence, inverse exists.

For adj A

A₁₁ = -3 – 0 = -3

A₁₂ = -(-3 – 0) = 3

A₁₃ = 6 – 15 = -9

A₂₁ = -(0 – 0) = 0

A₂₂ = -1 – 0 = -1

A₂₃ = -(2 – 0) = -2

A₃₁ = 0 – 0 = 0 

A₃₂ = -(0 – 0) = 0

A₃₃ = 3 – 0 = 3

adj A = $\begin{bmatrix} -3& 0 &0 \\ 3 & -1 &0 \\ -9 & -2 & 3 \end{bmatrix}$

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

A^-1 = $\frac{-1}{3}\begin{bmatrix} -3& 0 &0 \\ 3 & -1 &0 \\ -9 & -2 & 3 \end{bmatrix}$

问题 9。 $\begin{bmatrix} 2& 1 &3 \\ 4 & -1 &0 \\ -7 & 2 & 1 \end{bmatrix}$

解决方案：

A = $\begin{bmatrix} 2& 1 &3 \\ 4 & -1 &0 \\ -7 & 2 & 1 \end{bmatrix}$

|A| = 2(-1 – 0) – 1(4 – 0) + 3(8 – 7) = -3 ≠ 0

Hence, inverse exists.

For adj A

A₁₁ = -1 – 0 = -1 

A₁₂ = -(4 – 0) = -4

A₁₃ = 8 – 7 = 1

A₂₁ = -(1 – 6) = 5

A₂₂ = 2 + 21 = 23

A₂₃ = -(4 + 7) = -11

A₃₁ = 0 + 3 = 3

A₃₂ = -(0 – 12) = 12

A₃₃ = -2 – 4 = -6

adj A = $\begin{bmatrix} -1& 5 &3 \\ -4 & 23 &12 \\ 1 & -11 & -6 \end{bmatrix}$

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

A^-1 = $\frac{-1}{3}\begin{bmatrix} -1& 5 &3 \\ -4 & 23 &12 \\ 1 & -11 & -6 \end{bmatrix}$

问题 10。 $\begin{bmatrix} 1& -1 &2 \\ 0 & 2 &-3 \\ 3 & -2 & 4 \end{bmatrix}$

解决方案：

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

|A| = 1(8 – 6) – 0 + 3(3 – 4) = -1

Now for adj A

A₁₁ = 8 – 6 =2 

A₁₂ = -(0 + 9) = -9

A₁₃ = 0 – 6 = -6

A₂₁ = -(-4 + 4) =0

A₂₂ = 4 – 6 = -2

A₂₃ = -(-2 + 3) = -1

A₃₁ = 3 – 4 = -1

A₃₂ = -(-3 – 0) = 3

A₃₃ = 2 – 0 = 2

adj A = $\begin{bmatrix} 2& 0 &-1 \\ -9 & -2 &3 \\ -6 & -1 & 2 \end{bmatrix}$

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

A^-1 = $-\begin{bmatrix} 2& 0 &-1 \\ -9 & -2 &3 \\ -6 & -1 & 2 \end{bmatrix}$

A^-1 = $\begin{bmatrix} -2& 0 &1 \\ 9 & 2 &-3 \\ 6 & 1 & -2 \end{bmatrix}$

问题 11。 $\begin{bmatrix} 1& 0 &0 \\ 0 & cos\alpha &sin\alpha \\ 0 & sin\alpha & -cos\alpha \end{bmatrix}$

解决方案：

A = $\begin{bmatrix} 1& 0 &0 \\ 0 & cos\alpha &sin\alpha \\ 0 & sin\alpha & -cos\alpha \end{bmatrix}$

|A| = 1(-cos2α – sin2α) = -1

Now,

A₁₁ = -cos2α – sin2α = -1

A₁₂ = 0

A₁₃ = 0

A₂₁ = 0

A₂₂ = -cosα

A₂₃ = -sinα

A₃₁ = 0

A₃₂ = -sinα

A₃₃ = cosα

adj A = $\begin{bmatrix} -1& 0 &0 \\ 0 & -cos\alpha &-sin\alpha \\ 0 & -sin\alpha & cos\alpha \end{bmatrix}$

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

A^-1 = $-\begin{bmatrix} -1& 0 &0 \\ 0 & -cos\alpha &-sin\alpha \\ 0 & -sin\alpha & cos\alpha \end{bmatrix}$

A^-1 = $\begin{bmatrix} 1& 0 &0 \\ 0 & cos\alpha &sin\alpha \\ 0 & sin\alpha & -cos\alpha \end{bmatrix}$

问题 12. 让 A = $\begin{bmatrix} 3 &7 \\ 2& 5 \end{bmatrix}$和 B = $\begin{bmatrix} 6 &8 \\ 7& 9 \end{bmatrix}$ , 验证 (AB) ^{– 1} = B ^{– 1} A ^{– 1}

解决方案：

A = $\begin{bmatrix} 3 &7 \\ 2& 5 \end{bmatrix}$

|A| = 15 – 14 = 1

A₁₁ = 5

A₁₂ = -2

A₂₁ = -7

A₂₂ = 3

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

A^-1 = $\begin{bmatrix} 5 &-7 \\ -2& 3 \end{bmatrix}$

B = $\begin{bmatrix} 6 &8 \\ 7& 9 \end{bmatrix}$

|B| = 54 – 56 = -2

adj B = $\begin{bmatrix} 9 &-8 \\ -7& 6 \end{bmatrix}$

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

B^-1 = $\frac{-1}{2}\begin{bmatrix} 9 &-8 \\ -7& 6 \end{bmatrix}$

B^-1 = $\begin{bmatrix} \frac{-9}{2} &4 \\ \frac{7}{2}& -3 \end{bmatrix}$

Now,

B^-1A^-1 = $\begin{bmatrix} \frac{-9}{2} &4 \\ \frac{7}{2}& -3 \end{bmatrix} \begin{bmatrix} 5 &-7 \\ -2& 3 \end{bmatrix}$

B^-1A^-1 = $\begin{bmatrix} \frac{-45}{2}-8 &\frac{63}{2}+12 \\ \frac{35}{2}+6& \frac{-49}{2}-9 \end{bmatrix}$

B^-1A^-1 = $\begin{bmatrix} \frac{-61}{2} &\frac{87}{2} \\ \frac{47}{2}& \frac{-67}{2} \end{bmatrix}$

Now, AB = $\begin{bmatrix} 3 &7 \\ 2& 5 \end{bmatrix} \begin{bmatrix} 6 &8 \\ 7& 9 \end{bmatrix}$

AB = $\begin{bmatrix} 18+49 &24+63 \\ 12+35& 16+45 \end{bmatrix}$

AB = $\begin{bmatrix} 67 &87 \\ 47& 61 \end{bmatrix}$

|AB| = 67 * 61 – 87 * 47 = -2

adj (AB) = $\begin{bmatrix} 61 &-87 \\ -47& 67 \end{bmatrix}$

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

(AB)^-1 = $\begin{bmatrix} 61 &-87 \\ -47& 67 \end{bmatrix}$

(AB)^-1= $\begin{bmatrix} \frac{-61}{2} &\frac{87}{2} \\ \frac{47}{2}& \frac{-67}{2} \end{bmatrix}$

From above, you can see that (AB)^-1 = B^-1A^-1.

Hence, it is proved.

问题 13. A = $\begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix}$ ，证明 A ² – 5A + 7I = O。因此找到 A ^-1 。

解决方案：

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

A² = $\begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix} \begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix}$

A² = $\begin{bmatrix} 9-1 &3+2\\ -3-2& -1+4 \end{bmatrix}$

A² = $\begin{bmatrix} 8 &5\\ -5&3 \end{bmatrix}$

So, A² – 5A + 7I

= $\begin{bmatrix} 8 &5\\ -5&3 \end{bmatrix}$– 5$\begin{bmatrix} 8 &5\\ -5&3 \end{bmatrix}$+ 7$\begin{bmatrix} 1 &0\\ 0&1 \end{bmatrix}$

= $\begin{bmatrix} 8 &5\\ -5&3 \end{bmatrix}$ – $\begin{bmatrix} 15 &5 \\ -5& 10 \end{bmatrix}$+ $\begin{bmatrix} 7 &0 \\ 0& 7 \end{bmatrix}$

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

= O

Hence, A² – 5A + 7I = O

It can be written as 

A.A – 5A = -7I

Multiplying by A^-1 in both sides

A.A(A^-1) – 5AA^-1 = 7IA^-1

A(AA^-1) – 5I = -7A^-1

AI – 5I = -7A^-1

A^-1 = -(A – 5I)/7

A^-1 =1/7($\begin{bmatrix} 5 &0\\ 0&5 \end{bmatrix}$ – $\begin{bmatrix} 3 &1\\ -1&2 \end{bmatrix}$)

A^-1 = $\frac{1}{7}\begin{bmatrix} 2 &-1 \\ 1& 3 \end{bmatrix}$

问题 14. 对于矩阵 A = $\begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix}$ ，找到数字 a 和 b 使得 A2 + aA + bI = O。

解决方案：

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

A² = $\begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix} \begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix}$

A² = $\begin{bmatrix} 9+2 &6+2 \\ 3+1& 2+1 \end{bmatrix}$

A2 = $\begin{bmatrix} 11 &8 \\ 4& 3 \end{bmatrix}$

Now,

A² – aA + bI = O

Multiplying by A^-1 in both sides

(AA)A^-1 + aAA^-1 + bIA^-1 = O

A(AA^-1) + aI + b(IA^-1) = O

AI + aI + bA^-1 = O

A + aI = -bA^-1

A^-1 = -(A + aI)/b

Now,

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

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

Now, 

$\begin{bmatrix} 1 &-2 \\ -1& 3 \end{bmatrix} = -1/b[\begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix} + \begin{bmatrix} a &0 \\ 0& a \end{bmatrix}]$

= -1/b$\begin{bmatrix} 3+a &2 \\ 1& 1+a \end{bmatrix}$

= $\begin{bmatrix} \frac{-3-a}{b} &\frac{-2}{b} \\ \frac{-1}{b}& \frac{-1-a}{b} \end{bmatrix}$

On comparing elements you will get

-1/b = -1

b = 1

(-3 – a)/b = 1

-3 – a = 1

a = -4

Hence, a = -4 and b = 1

问题 15. A = $\begin{bmatrix} 1& 1 &1 \\ 1 & 2 &-3 \\ 2 & -1 & 3 \end{bmatrix}$ , 证明 A ³ – 6A ² + 5A + 11I = O. 因此找到 A ^-1

解决方案：

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

A² = $\begin{bmatrix} 1& 1 &1 \\ 1 & 2 &-3 \\ 2 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1& 1 &1 \\ 1 & 2 &-3 \\ 2 & -1 & 3 \end{bmatrix}$

A² = $\begin{bmatrix} 1+1+2& 1+2-1 &1-3+3 \\ 1+2-6 & 1+4+3 &1-6-9 \\ 2-1+6 & 2-2-3 & 2+3+9 \end{bmatrix}$

A² = $\begin{bmatrix} 4& 2 &1 \\ -3 & 8 &-14 \\ 7 & -3 & 14 \end{bmatrix}$

A³ = A².A

A³ = $\begin{bmatrix} 4& 2 &1 \\ -3 & 8 &-14 \\ 7 & -3 & 14 \end{bmatrix} \begin{bmatrix} 1& 1 &1 \\ 1 &2 &-3 \\ 2 &-1 & 3 \end{bmatrix}$

A³ = $\begin{bmatrix} 4+2+2& 4+4-1 &4-6+3 \\ -3+8-28 &-3+16+14 &-3-24-42 \\ 7-3+28 &7-6-14 & 7+9+42 \end{bmatrix}$

A³ = $\begin{bmatrix} 8& 7 &1 \\ -23 &27 &-69 \\ 32&-13 & 58 \end{bmatrix}$

A³ – 6A² + 5A + 11I

$\begin{bmatrix} 8& 7 &1 \\ -23 &27 &-69 \\ 32&-13 & 58 \end{bmatrix}$– 6$\begin{bmatrix} 4& 2 &1 \\ -3 & 8 &-14 \\ 7 & -3 & 14 \end{bmatrix}$+ 5$\begin{bmatrix} 1& 1 &1 \\ 1 & 2 &-3 \\ 2 & -1 & 3 \end{bmatrix}$ + 11$\begin{bmatrix} 1& 0 &0 \\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

= $\begin{bmatrix} 8& 7 &1 \\ -23 &27 &-69 \\ 32&-13 & 58 \end{bmatrix}$– $\begin{bmatrix} 24& 12 &6 \\ -18&48 &-84 \\ 42&-18 & 84 \end{bmatrix}$ + $\begin{bmatrix} 5& 5 &5 \\ 5&10 &-15 \\ 10&-5 & 15 \end{bmatrix}$ + $\begin{bmatrix} 11& 0 &0 \\ 0&11 &0 \\ 0&0 & 11 \end{bmatrix}$

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

= O

Hence, A³ – 6A² + 5A + 11I = O

Now,

A³ – 6A² + 5A + 11I = O

(AAA)A^-1 – 6(AA)A^-1 + 5(AA^-1) + 11IA^-1 = O

AA(AA^-1) – 6A(AA^-1) + 5(AA^-1) = -11(IA^-1)

A² – 6A + 5I = -11 A^-1

A^-1 = -1/11(A2 – 6A + 5I) ………….(1)

Now, A² – 6A + 5I

=$\begin{bmatrix} 4& 2 &1 \\ -3 & 8 &-14 \\ 7 & -3 & 14 \end{bmatrix}$ – 6$\begin{bmatrix} 1& 1 &1 \\ 1 & 2 &-3 \\ 2 & -1 & 3 \end{bmatrix}$ + 5$\begin{bmatrix} 1& 0 &0 \\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

=$\begin{bmatrix} 4& 2 &1 \\ -3 & 8 &-14 \\ 7 & -3 & 14 \end{bmatrix}$ – $\begin{bmatrix} 6& 6 &6 \\ 6&12 &-18 \\ 12&-6 & 18 \end{bmatrix}$ + $\begin{bmatrix} 5& 0 &0 \\ 0&5 &0 \\ 0&0 &5 \end{bmatrix}$

= $\begin{bmatrix} 3& -4 &-5 \\ -9&1 &4 \\ -5&3 &1 \end{bmatrix}$

From eq(1) you have 

A^-1 = -1/11 $\begin{bmatrix} 3& -4 &-5 \\ -9&1 &4 \\ -5&3 &1 \end{bmatrix}$

问题 16. A = $\begin{bmatrix} 2& -1 &1 \\ -1&2 &-1 \\ 1&-1 &2 \end{bmatrix}$ ，验证 A ³ – 6A ² + 9A – 4I = O 和因此鳍 A ^-1 。

解决方案：

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

A² = $\begin{bmatrix} 2& -1 &1 \\ -1&2 &-1 \\ 1&-1 &2 \end{bmatrix} \begin{bmatrix} 2& -1 &1 \\ -1&2 &-1 \\ 1&-1 &2 \end{bmatrix}$

A² = $\begin{bmatrix} 4+1+1& -2-2-1 &2+1+2 \\ -2-2-1&1+4+1 &-1-2-2 \\ 2+1+2&-1-2-2 &1+1+4 \end{bmatrix}$

A²= $\begin{bmatrix} 6& -5 &5 \\ -5&6 &-5 \\ 5&-5 &6 \end{bmatrix}$

A³ = A².A

A³ = $\begin{bmatrix} 6& -5 &5 \\ -5&6 &-5 \\ 5&-5 &6 \end{bmatrix} \begin{bmatrix} 2& -1 &1 \\ -1&2 &-1 \\ 1&-1 &2 \end{bmatrix}$

A³ = $\begin{bmatrix} 12+5+5& -6-10-5 &6+5+10 \\ -10-6-5&5+12+5 &-5-6-10 \\ 10+5+6&-5-10-6 &5+5+12 \end{bmatrix}$

A³ = $\begin{bmatrix} 22& -21 &21 \\ -21&22 &-21 \\ 21&-21 &22 \end{bmatrix}$

Now,

A³ – 6A² + 9A – 4I

$\begin{bmatrix} 22& -21 &21 \\ -21&22 &-21 \\ 21&-21 &22 \end{bmatrix}$ – 6$\begin{bmatrix} 6& -5 &5 \\ -5&6 &-5 \\ 5&-5 &6 \end{bmatrix}$ + 9$\begin{bmatrix} 2& -1 &1 \\ -1&2 &-1 \\ 1&-1 &2 \end{bmatrix}$– 4$\begin{bmatrix} 1& 0 &0 \\ 0&1 &0 \\ 0&0 &1 \end{bmatrix}$

=$\begin{bmatrix} 22& -21 &21 \\ -21&22 &-21 \\ 21&-21 &22 \end{bmatrix}$ – $\begin{bmatrix} 36& -30 &30 \\ -30&36 &-30 \\ 30&-30 &36 \end{bmatrix}$ + $\begin{bmatrix} 18& -9 &9 \\ -9&18 &-9 \\ 9&-9 &18 \end{bmatrix}$– $\begin{bmatrix} 4& 0 &0 \\ 0&4 &0 \\ 0&0 &4 \end{bmatrix}$

=$\begin{bmatrix} 40& -30 &30 \\ -30&40 &-30 \\ 30&-30 &40 \end{bmatrix}$ – $\begin{bmatrix} 40& -30 &30 \\ -30&40 &-30 \\ 30&-30 &40 \end{bmatrix}$

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

= O

So, A³ – 6A² + 9A – 4I = O

Now,

A³ – 6A² + 9A – 4I = O

Multiplying by A^-1 in both sides

(AAA)A^-1 – 6(AA)A^-1 + 9AA^-1 – 4IA^-1 = O

AA(AA^-1) – 6A(AA^-1) + 9(AA^-1) = 4(IA^-1)

AAI – 6AI +9I = 4A^-1

A² – 6A + 9I = 4A^-1

A^-1 = 1/4(A2 – 6A + 9I) ……….(1)

A² – 6A + 9I

= $\begin{bmatrix} 6& -5 &5 \\ -5&6 &-5 \\ 5&-5 &6 \end{bmatrix}$ – 6$\begin{bmatrix} 2& -1 &1 \\ -1&2 &-1 \\ 1&-1 &2 \end{bmatrix}$ + 9$\begin{bmatrix} 1& 0 &0 \\ 0&1 &0 \\ 0&0 &1 \end{bmatrix}$

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

From eq(1), you have 

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

问题 17. 令 A 为 3 * 3 阶的非奇异矩阵。则 |adj A|等于

(一) |一| (B) |A| ² (C) |A| ³ (D) 3|A|

解决方案：

You know,

(adj A)A = |A|I

(adj A)A = $\begin{bmatrix} |A| &0 &0 \\ 0 & |A| &0 \\ 0 &0 &|A| \end{bmatrix}$

|(adj A)A| = $\begin{vmatrix} |A| &0 &0 \\ 0 & |A| &0\\ 0 &0 &|A| \end{vmatrix}$

|(adj A)A| = |A|³$\begin{vmatrix} 1 &0 &0 \\ 0 & 1 &0\\ 0 &0 &1 \end{vmatrix}$

|(adj A)A| = |A|³ I

|adj A| = |A|³

Hence, option B is correct.

问题 18. 如果 A 是 2 阶可逆矩阵，则 det(A ^-1 ) 等于

(A) 检测 (A) (B) 1/(检测 A) (C) 1 (D) 0

解决方案：

Since A is an invertible matrix then A^-1 exists.

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

Suppose a 2 order matrix is A = $\begin{bmatrix} a &b \\ c &d \end{bmatrix}$

Then |A| = ad – bc

and adj A = $\begin{bmatrix} d &-b \\ -c &a \end{bmatrix}$

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

A^-1 = $\frac{1}{|A|}\begin{bmatrix} d &-b \\ -c &a \end{bmatrix}$

A^-1 = $\begin{bmatrix} \frac{d}{|A|} &\frac{-b}{|A|} \\ \frac{c}{|A|}&\frac{a}{|A|} \end{bmatrix}$

|A^-1| = $\begin{vmatrix} \frac{d}{|A|} &\frac{-b}{|A|} \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{vmatrix}$

|A^-1| = $\frac{1}{|A|^{2}}\begin{vmatrix} d &-b \\ -c &a \end{vmatrix}$

|A^-1| = (ad – bc)/|A|²

|A^-1| = |A|/|A|²

|A^-1| = 1/|A|

det(A^-1) = 1/(det A)

Hence, option B is correct.