第12类RD Sharma解决方案–第6章行列式–练习6.1

问题1：写下以下矩阵第一列的每个元素的未成年人和辅因子，从而评估行列式。

$i) A= \begin{bmatrix} 5 &20\\ 0&-1\\ \end{bmatrix}$

解决方案：

i) Let M_ij and C_ij  represents minor and co-factor of element. They are placed in i^th  row and j^th column.

Here, a₁₁ = 5

Minor of a₁₁ = M₁₁ = -1

Note: In 2×2 matrix, minor is obtained for a particular element, by deleting that row and column were element is present.

Minor of a₁₂ = M₁₂ = 0

Minor of a₂₁ = M₂₁ = 20

Minor of a₂₂ = M₂₂ = 0

As M₁₂ and M₂₂ are zero so we don’t consider them. Hence we have got only two minors for this determinant.

M₁₁ = -1 & M₂₁  = 20

Now, co-factors for the determinants are

C₁₁ = (-1)¹⁺¹ x M₁₁                           {∵Cij =(-1)1+1 x Mij}

    = (+1)x(-1)

    = -1  

C₂₁ = (-1)²⁺¹ x M₂₁                            

    = (-1)³ x 20

    = -20

Evaluating the determinant,

|A| = a₁₁ x C₁₁ + a₂₁ x C₂₁

      =5 x (-1) + 0 x (-20)

      = -5

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

解决方案：

Let M_ij and C_ij  represents minor and co-factor of element. They are placed in i^th  row and j^th column.

Minor of a₁₁ = M₁₁ = 3

Note: In 2×2 matrix, minor is obtained for a particular element, by deleting that row and column were element is present.

Minor of a₂₁ = M₂₁ = 4

Now, co-factors for the determinants are

C₁₁ = (-1)¹⁺¹ x M₁₁                           {∵Cij =(-1)i+j x Mij}

     = (+1) x 3

     = 3

C₂₁ = (-1)²⁺¹ x M₂₁                           

     = (-1)³ x 4

     = -4

Evaluating the determinant,

|A| = a₁₁ x C₁₁ + a₂₁ x C₂₁

       =-1 x 3 + 2 x (-4)

       =-11

$iii) A= \begin{bmatrix} 1 & -3 &2 \\ 4 & -1 & 2\\ 3 & 5 & 2 \\ \end{bmatrix}$

解决方案：

Let M_ij and C_ij  represents minor and co-factor of element. They are placed in i^th row and j^th column.

C_ij = (-1)^i+j x M_ij

Given,

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

We have,

$M_{11}= \begin{bmatrix} -1 & 2 \\ 5 & 2 \\ \end{bmatrix}$

M₁₁ = -1×2 – 5×2

M₁₁ = -12

$M_{21}= \begin{bmatrix} -3 & 2 \\ 5 & 2 \\ \end{bmatrix}$

M₂₁ = -3×2 – 5×2

M₂₁ = -16

$M_{31}= \begin{bmatrix} -3 & 2 \\ -1 & 2 \\ \end{bmatrix}$

M₃₁ = -3×2 – (-1) x 2

M₃₁ = -4

Co-factors of the determinant are as follows,

C₁₁ = (-1)¹⁺¹ x M₁₁                            

     = 1x-12

     = -12

C₂₁ = (-1)²⁺¹ x M₂₁                            

     = (-1)³ x -16

     = 16

C₃₁ = (-1)³⁺¹ x M₃₁                            

     = (1)⁴ x (-4)

     = -4

To evaluate the determinant expand along first column,

|A| = a₁₁ x C₁₁ + a₂₁ x C₂₁+ a₃₁ x C₃₁

          =1x(-12) + 4×16 + 3x(-4)

       = -12 + 64 – 12

       = 40  

$iv) A= \begin{bmatrix} 1 & a &bc \\ 1 & b & ca\\ 1 & c & ab\\ \end{bmatrix}$

解决方案：

Let M_ij and C_ij  represents minor and co-factor of element. They are placed in ith  row and jth column.

Also, C_ij = (-1)^i+j x M_ij

Given,

     $A= \begin{bmatrix} 1 & a &bc \\ 1 & b & ca\\ 1 & c & ab\\ \end{bmatrix}$

We have,

$M_{11}= \begin{bmatrix} b & ca \\ c & ab\\ \end{bmatrix}$

M₁₁ = b x ab – c x ca

M₁₁ = ab² – ac²

$M_{21}= \begin{bmatrix} a & bc \\ c & ab\\ \end{bmatrix}$

M₂₁ = a x ab – c x bc

M₂₁ = a²b – c²b

$M_{31}= \begin{bmatrix} a & bc \\ b & ca\\ \end{bmatrix}$

M₃₁ = a x ca – b x bc

M₃₁ = a²c – b²c

Co-factors of the determinant are as follows,

C₁₁ = (-1)¹⁺¹ x M₁₁                           

     = 1 x (ab² – ac²)

     = ab² – ac²

C₂₁ = (-1)²⁺¹ x M₂₁                           

     = (-1)³ x (a²b – c²b)

     = c²b – a²b  

C₃₁ = (-1)³⁺¹ x M₃₁                           

     = (1)⁴ x (a²c – b²c)

     = a²c – b²c

To evaluate the determinant expand along first column,

|A| = a₁₁ x C₁₁ + a₂₁ x C₂₁+ a₃₁ x C₃₁

     =1 x (ab² – ac²) + 1 x (c²b – a²b) + 1 x (a²c – b²c)

     = ab² – ac² + c²b – a²b + a²c – b²c

$v) A= \begin{bmatrix} 0 & 2 &6 \\ 1 & 5 & 0\\ 3 & 7 & 1 \\ \end{bmatrix}$

解决方案：

Let M_ij and C_ij  represents minor and co-factor of element. They are placed in i^th row and j^th column.

C_ij = (-1)^i+j x M_ij

Given,

     $A= \begin{bmatrix} 0 & 2 &6 \\ 1 & 5 & 0\\ 3 & 7 & 1 \\ \end{bmatrix}$

We have,

$M_{11}=\begin{bmatrix} 5 & 0 \\ 7 & 1\\ \end{bmatrix}$

M₁₁ = 5×1 – 7×0

M₁₁ = 5

$M_{21}=\begin{bmatrix} 2 & 6 \\ 7 & 1\\ \end{bmatrix}$

M₂₁ = 2×1 – 7×6

M₂₁ = -40

$M_{31}=\begin{bmatrix} 2 & 6 \\ 5 & 0\\ \end{bmatrix}$

M₃₁ = 2×0 – 5×6

M₃₁ = -30

Co-factors of the determinant are as follows,

C₁₁ = (-1)¹⁺¹ x M₁₁                            

     = 1×5

     = 5

C₂₁ = (-1)²⁺¹ x M₂₁                            

     = (-1)³ x -40

     = 40

C₃₁ = (-1)³⁺¹ x M₃₁                            

     = (1)⁴ x (-30)

     = -30

To evaluate the determinant expand along first column,

|A| = a₁₁ x C₁₁ + a₂₁ x C₂₁+ a₃₁ x C₃₁

          =0x5 + 1×40 + 3x(-20)

       = 0 + 40 – 90

       = 50

$vi) A= \begin{bmatrix} a & h & g \\ h & b & f\\ g & f & c \\ \end{bmatrix}$

解决方案：

Let M_ij and C_ij  represents minor and co-factor of element. They are placed in i^th  row and j^th column.

C_ij = (-1)^i+j x M_ij

Given,

     $A= \begin{bmatrix} a & h & g \\ h & b & f\\ g & f & c \\ \end{bmatrix}$

We have,

$M_{11}= \begin{bmatrix} b & f \\ f&c\\ \end{bmatrix}$

M₁₁ = b x c – f x f

M₁₁ = bc – f²

$M_{21}= \begin{bmatrix} h & g \\ f & c\\ \end{bmatrix}$

M₂₁ = h x c – f x g

M₂₁ = hc – fg

$M_{31}= \begin{bmatrix} h & g \\ b & f\\ \end{bmatrix}$

M₃₁ = h x f – b x g

M₃₁ = hf – bg

Co-factors of the determinant are as follows,

C₁₁ = (-1)¹⁺¹ x M₁₁                            

     = 1x (bc – f²)

     = bc – f²

C₂₁ = (-1)²⁺¹ x M₂₁                            

     = (-1)³ x (hc – fg)

     = fg – hc

C₃₁ = (-1)³⁺¹ x M₃₁                            

     = (1)⁴ x (hf – bg)

     = hf – bg

To evaluate the determinant expand along first column,

|A| = a₁₁ x C₁₁ + a₂₁ x C₂₁+ a₃₁ x C₃₁

          =a x (bc – f²) + h x (fg – hc) + g x (hf – bg)

       = abc – af² + hgf – h²c + ghf –bg²

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

解决方案：

Let M_ij and C_ij  represents minor and co-factor of element. They are placed in i^th row and j^th column

Also, C_ij = (-1)^i+j x M_ij

Given,  

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

From the matrix we have,

$M_{11}= \begin{bmatrix} 0 & 1 & -2 \\ 1 & -1 & 1 \\ -1 & 5 & 0 \\ \end{bmatrix}$

M₁₁ = 0(-1 x 0 – 5 x 1) – 1(1 x 0 – (-1) x 1) + (-2)(1 x 5 – (-1) x (-1))

M₁₁ = -9

$M_{21}= \begin{bmatrix} -1 & 0 & 1 \\ 1 & -1 & 1 \\ -1 & 5 & 0 \\ \end{bmatrix}$

M₂₁ = -1(-1 x 0 – 5 x 1) – 0(1 x 0 – (-1) x 1) + (1 x 5 – (-1) x (-1))

M₂₁ = 9

$M_{31}= \begin{bmatrix} -1&0&1 \\ 0 & 1 & -2 \\ -1 & 5 & 0 \\ \end{bmatrix}$

M₃₁ = -1(1 x 0 – 5 x (-2)) – 0(0 x 0 – (-1) x (-2)) + 1(0 x 5 – (-1) x 1)

M₃₁ = -9

$M_{41}= \begin{bmatrix} -1& 0 & -1 \\ 0 & 1 & -2 \\ 1 & -1 &1 \\ \end{bmatrix}$

M₄₁ = -1(1 x 1 – (-1) x (-2)) – 0(0 x 1 – 1 x (-2)) + 1(0 x (-1) – 1 x 1)

M₄₁ = 0

Co-factors of the determinant are as follows,

C₁₁ = (-1)¹⁺¹ x M₁₁                            

     = 1x (-9)

     = -9

C₂₁ = (-1)²⁺¹ x M₂₁                           

     = (-1)³ x 9

     = -9

C₃₁ = (-1)³⁺¹ x M₃₁                            

     = (-1)⁴ x -9

     = -9

C₄₁ = (-1)⁴⁺¹ x M₄₁                            

     = (-1)⁵ x 0

     = 0

To evaluate the determinant expand along first column,

|A| = a₁₁ x C₁₁ + a₂₁ x C₂₁+ a₃₁ x C₃₁+ a₄₁ x C₄₁

          =2 x (-9) + (-3) x (-9) + 1 x (-9) + 2 x 0

       = -18 + 27 – 9

       = 0

问题2：评估以下决定因素

$i)A= \begin{vmatrix} x& -7\\ z&5x+1 \end{vmatrix}$

解决方案：

Given, $A= \begin{vmatrix} x& -7\\ z&5x+1 \end{vmatrix}$

Cross multiplying the values inside the determinant,

|A| = (5x + 1) – (-7)x

|A| = 5x² = 8x

$ii) A= \begin{vmatrix} cos\theta& -sin\theta \\ sin\theta&cos\theta \end{vmatrix}$

解决方案：

Given, $A= \begin{vmatrix} cos\theta& -sin\theta \\ sin\theta&cos\theta \end{vmatrix}$

$|A|=cos\theta \times sin\theta-(-sin\theta)\times sin\theta$                      {$\therefore cos^2\theta + sin^2\theta = 1$

$|A|=cos^2\theta+sin^2\theta \\ |A|=1$

$iii) A= \begin{vmatrix} cos15\degree & -sin15\degree\\ sin75\degree & cos75\degree \\ \end{vmatrix}$

解决方案：

Given, $A= \begin{vmatrix} cos15\degree & -sin15\degree\\ sin75\degree & cos75\degree \\ \end{vmatrix}$

∣A∣ = cos15°×cos75°+sin15°×sin75°

As per formula

cos(A−B)=cosAcosB+sinAsinB

Substitute this in |A| so we get,

∣A∣ = cos(75−15)°

∣A∣ = cos60°

∣A∣ = 0.5

$iv) A= \begin{vmatrix} a+ib & c+id \\ -c+id & a-ib\\ \end{vmatrix}$

解决方案：

∣A∣ = (a+ib)(a−ib)−(c+id)(−c+id)

Expanding the brackets we get,

∣A∣=(a+ib)(a−ib)+(c+id)(c−id)

|A| = a²-i²b²+c²-i²d²

We know i² = -1

|A| = a²-1b²+c²-(-1)d²

|A| = a²+b²+c²+d²

问题3：评估以下内容：

$\begin{vmatrix} 2&3&7\\ 13&17&5\\ 15&20&12\\ \end{vmatrix}^2$

解决方案：

In the given formula, ∣AB∣=∣A∣∣B∣

$\\ |A| = \begin{vmatrix} 2&3&7\\ 13&17&5\\ 15&20&12\\ \end{vmatrix}$

Cross multiplying the terms  in |A| 

$\\ |A| = 2 \begin{vmatrix} 17&5\\ 20&12\\ \end{vmatrix} -3 \begin{vmatrix} 13&5\\ 15&12\\ \end{vmatrix} +7 \begin{vmatrix} 13&17\\ 15&20\\ \end{vmatrix} \\$

∣A∣ = 2(17×12−5×20)−3(13×12−5×15)+7(13×20−15×17)

= 2(204−100)−3(156−75)+7(260−255)

= 2×104−3×81+7×5

= 208−243+45

= 0

Now ∣A∣²=∣A∣×∣A∣

∣A∣²=0

问题4：表明这一点，

$\begin{vmatrix} sin10\degree & -cos10\degree \\ sin80\degree & cos 80\degree \\ \end{vmatrix}$

解决方案：

Method 1:

Given,

$\\ \begin{vmatrix} sin10\degree & -cos10\degree \\ sin80\degree & cos 80\degree \\ \end{vmatrix}$

Let the given determinant as A,

Using sin(a+B) = sinA×cosB+cosA×sinB

∣A∣ = sin10°×cos80°+cos10°×sin80°

∣A∣ = sin(10+80)°

∣A∣ = sin90°

∣A∣ = 1

Method 2:

∣A∣ = sin10°×cos80°+cos10°×sin80°

[∴cosθ = sin(90−θ)]

∣A∣ = sin10°cos(90°−10°)+cos10°sin(90°−10°)

∣A∣ = sin10°sin10°+cos10°cos10°

∣A∣ = sin210°+cos210°

[∴sin2θ+cos2θ = 1]

∣A∣ = 1

问题5：通过两种方法评估以下行列式。

$\begin{vmatrix} 2 &3&-5 \\ 7&1&-2 \\ -3&4&1\\ \end{vmatrix}$

解决方案：

Method 1

Expanding along the first row

$\\ |A| = 2 \begin{vmatrix} 1&-2 \\ 4&1\\ \end{vmatrix} -3 \begin{vmatrix} 7&-2 \\ -3&1\\ \end{vmatrix} -5 \begin{vmatrix} 7&1 \\ -3&4\\ \end{vmatrix}$

∣A∣ = 2(1×1−4×−2)−3(7×1−(−2)×−3)−5(7×4−1×(−3))

∣A∣ = 2(1+8)−3(7−6)−5(28+3)

∣A∣ = 2×9−3×1−5×31

∣A∣ = 18−3−155

∣A∣ = −140

Method 2

Here it is Sarus Method, we adjoin the first two columns.

Expanding along second column,

$\\ |A|=2 \begin{vmatrix} 1&-2 \\ 4&1 \\ \end{vmatrix} -7 \begin{vmatrix} 3&-5 \\ 4&1 \\ \end{vmatrix} -3 \begin{vmatrix} 3&-5 \\ 1&-2\\ \end{vmatrix}$

∣A∣ = 2(1×1−4×(−2))−7(3×1−4×(−5))−3(3×(−2)−1×(−5))
∣A∣ = 2(1+8)−7(3+20)−3(−6+5)
∣A∣ = 2×9−7×23−3×(−1)
∣A∣ = 18−161+3
∣A∣ = −140

问题6：评估以下内容：

$A = \begin{vmatrix} 0&sin\alpha & -cos\alpha \\ -sin\alpha &0 & sin\beta \\ cos\alpha & -sin\beta & 0 \\ \end{vmatrix}$

解决方案：

$|A| =0 \begin{vmatrix} 0&sin\beta \\ -sin\beta &0 \\ \end{vmatrix} -sin\alpha \begin{vmatrix} -sin\alpha &sin\beta \\ cos\alpha & 0 \\ \end{vmatrix} -cos\alpha \begin{vmatrix} -sin\alpha &0 \\ cos\alpha&-sin\beta \\ \end{vmatrix}$

∣A∣ = 0(0−sinβ(−sinβ))−sinα(−sinα×0−sinβcosα)−cosα((−sinα)(−sinβ)−0×cosα)
∣A∣ = 0+sinαsinβcosα−cosαsinαsinβ
∣A∣ = 0

问题7：

$\begin{vmatrix} cos\alpha cos\beta&cos\alpha sin\beta & -sin\alpha \\ -sin\beta & cos\beta & 0 \\ sin\alpha cos\beta& sin\alpha sin\beta & cos\alpha \\ \end{vmatrix}$

解决方案：

Expand C3, we have
∣A∣ = sinα(−sinαsin²β − cos²βsinα) + cosα(cosαcos²β + cosαsin²β)
∣A∣ = sin2α(sin²β + cos²β) + cos²α(cos²β + sin²β)
∣A∣ = sin²α(1) + cos²α(1)
∣A∣ = 1

问题8：如果$A= \begin{bmatrix} 2&5 \\ 2&1 \\ \end{bmatrix}$$\ B= \begin{bmatrix} 4&-3 \\ 2&15\\ \end{bmatrix}$验证∣AB∣ = ∣A∣∣B∣

解决方案：

Let’s take LHS,

$AB = \begin{bmatrix} 2&5 \\ 2&1 \\ \end{bmatrix} \times \begin{bmatrix} 4&-3 \\ 2&5 \\ \end{bmatrix} \\ = \begin{bmatrix} 8+10 & -6+25 \\ 8+2 & -6+5 \\ \end{bmatrix} \\ =\begin{bmatrix} 18 & 19 \\ 10&-1 \\ \end{bmatrix}$
∣AB∣ = −18−190
∣AB∣ = −208

Now taking RHS and calculating,

∣A∣ = 2−10
∣A∣ = −8
∣B∣ = 20−(−6)
∣B∣ = 26
∣A∣∣B∣ = −8×26
∣A∣∣B∣ = −208
∴LHS = RHS
Hence, it is proved.

问题9：如果$A = \begin{bmatrix} 1&0&1 \\ 0&1&2 \\ 0&0&4\\ \end{bmatrix}$ ，则表明∣3A∣ = 27∣A∣。

解决方案：

Evaluate along the first column,

$|A|=1 \begin{vmatrix} 1&2 \\ 0&4\\ \end{vmatrix}-0 \begin{vmatrix} 0&1 \\ 0&4\\ \end{vmatrix}+0 \begin{vmatrix} 0&1 \\ 1&2\\ \end{vmatrix}$
Now every element with 3,
$\\ |3A|=3 \begin{vmatrix} 3&6 \\ 0&12\\ \end{vmatrix}-0 \begin{vmatrix} 0&3 \\ 0&12\\ \end{vmatrix}+0 \begin{vmatrix} 0&3 \\ 3&6\\ \end{vmatrix}$
= 3(36−0) − 0 + 0
= 108
Now, according to the question,
∣3A∣ = 27∣A∣
Substituting the values we get,
108 = 27(4)
108 = 108
Hence, proved.

问题10：在以下情况下，找到x的值：

$i)\begin{vmatrix} 2&4 \\ 5&1\\ \end{vmatrix}= \begin{vmatrix} 2x&4 \\ 6&x\\ \end{vmatrix} \\$

解决方案：

2−20 = 2x²−24
−18 = 2x²−24
2x² = 6
Taking the square root,
x² = 3
x = ±√3

$ii)\begin{vmatrix} 2&3 \\ 4&5\\ \end{vmatrix}= \begin{vmatrix} x&3 \\ 2x&5\\ \end{vmatrix} \\$

解决方案：

2 × 5 − 3 × 4 = 5 × x − 3 × 2x
10 − 12 = 5x − 6x
−2 = −x
x = 2

$iii)\begin{vmatrix} 3&x \\ x&1\\ \end{vmatrix}= \begin{vmatrix} 3&2 \\ 4&1\\ \end{vmatrix} \\$

解决方案：

3(1)−x(x) = 3(1)−2(4)
3−x² = 3−8
−x² = −8
x² = 8
x = ±2√2
​

$iv)\begin{vmatrix} 3x&7 \\ 2&4\\ \end{vmatrix}=10$

解决方案：

3x(4)−7(2) = 10
12x−14 = 10
12x = 24
x = 24/12
​x = 2

$v)\begin{vmatrix} x+1&x-1 \\ x-3&x+2\\ \end{vmatrix}= \begin{vmatrix} 4&-1 \\ 1&3\\ \end{vmatrix} \\$

解决方案：

Cross multiplying elements from LHS,
(x+1)(x+2)−(x−3)(x−1) = 12+1
x² + 3x + 2 − x²+4x − 3 = 13
7x−1 = 13
7x = 14
x = 2

$vi)\begin{vmatrix} 2x&5 \\ 8&x\\ \end{vmatrix}= \begin{vmatrix} 6&5 \\ 8&3\\ \end{vmatrix} \\$

解决方案：

2x(x)−5(8) = 6(3)−5(8)
2x²−40 = 18−40
2x² = 18
x² = 9
x = ±3

问题11：求x的积分值，如果

$\begin{vmatrix} x^2 & x& 1 \\ 0&2&1\\ 3&1&4\\ \end{vmatrix}=28$

解决方案：

Here we have to take the determinant of the 3×3 matrix
x²(8−1)−x(0−3)+1(0−6)
8x²−x²+3x−6 = 28
7x²+3x−6 = 28
7x²+3x−34 = 0
Factorization of the above equation we get,
(7x+17)(x−2) = 0
x = 2
Integral value of x is 2. Thus, x = −17/7 is not an integer.

问题12：矩阵A对于x的哪个值是奇异的?

$i)A=\begin{vmatrix} 1+x&7 \\ 3-x&8\\ \end{vmatrix}=0$

解决方案：

Matrix A is singular if,
∣A∣ = 0
$\\ |A| =\begin{vmatrix} 1+x&7 \\ 3-x&8\\ \end{vmatrix}=0$
Cross−multiply the elements in the determinant,
8 + 8x − 21 + 7x = 0
15x − 13 = 0
15x = 13
x = 13/15

$ii)A=\begin{vmatrix} x-1&1&1 \\ 1&x-1&1 \\ 1&1&x-1\\ \end{vmatrix}$

解决方案：

Matrix A is singular if ∣A∣=0
Expanding along first row,

$|A|=(x-1)\begin{vmatrix} x-1&1 \\ 1&x-1\\ \end{vmatrix} -1 \begin{vmatrix} 1&1 \\ 1&x-1\\ \end{vmatrix} +1 \begin{vmatrix} 1&1 \\ x-1&1\\ \end{vmatrix} \\$

∣A∣ = (x−1)[(x−1)²−1] − 1[x−1−1] + 1[1−x+1]
∣A∣ = (x−1)(x²+1−2x−1) − 1(x−2) + 1(2−x)

Expanding the brackets to factorize
|A| = (x−1)(x²−2x) − x + 2 + 2 − x
|A| = (x-1) × x × (x-2) + (4-2x)
|A| = (x−1)× x ×(x−2) + 2(2−x)
|A| = (x−1)× x ×(x−2) − 2(x−2)
[∴ Take (x−2) as common]
|A| = (x−2)[x(x−1)−2]
Since A is a singular matrix, so ∣A∣ = 0
(x−2)(x²−x−2) = 0
There are two cases,
Case1:
(x−2) = 0
x = 2
Case2:
x²−x−2 = 0
x²−2x + x−2 = 0
x(x−2) + 1(x−2) = 0
(x−2)(x+1) = 0
x = 2,−1
∴ x = 2 or −1