Probability density function: f(x) =  λ.e^{- λx} , x>0

Formula: Var(X) = E(X²) – [E(X)]²

E(X) = ${\int }_0^{\mathrm{\infty }}x.f\left(x\right).dx$

E(X) = ${\int }_0^{\mathrm{\infty }}x\lambda .e^{-\lambda \mathrm{x}}.dx$

E(X) = $\frac{1}{\lambda }$

E(X²) = ${\int }_0^{\mathrm{\infty }}{x}^2.f\left(x\right).dx$

E(X²) = ${\int }_0^{\mathrm{\infty }}{x}^2\lambda .e^{-\lambda \mathrm{x}}.dx$

E(X²) = $\frac{2}{{\lambda }^2}$

Now, put both the values in the variance formula

Var(X) = E(X²) – [E(X)]²

Var(X) = $\frac{2}{{\lambda }^2}$  - ${\left(\frac{1}{\lambda }\right)}^2$ = $\frac{2}{{\lambda }^2}$ - $\frac{1}{{\lambda }^2}$

Var(X) = $\frac{1}{{\lambda }^2}$

$\left|A\right|=\left|\begin{array}{ccccc}3& 1& 1& 1& 1\\ 1& 3& 1& 1& 1\\ 1& 1& 3& 1& 1\\ 1& 1& 1& 3& 1\\ 1& 1& 1& 1& 3\end{array}\right|$

R₁ ↔ R₅

$\left|A\right|=-\left|\begin{array}{ccccc}1& 1& 1& 1& 3\\ 1& 3& 1& 1& 1\\ 1& 1& 3& 1& 1\\ 1& 1& 1& 3& 1\\ 3& 1& 1& 1& 1\end{array}\right|$

R₂ → R₂ – R₁

R₃ → R₃ – R₁

R₄ → R₄ – R₁

R₅ → R₅ – 3R₁

$\left|A\right|=-\left|\begin{array}{ccccc}1& 1& 1& 1& 3\\ 0& 2& 0& 0& -2\\ 0& 0& 2& 0& -2\\ 0& 0& 0& 2& -2\\ 0& -2& -2& -2& -8\end{array}\right|$

R₅ → R₅ + R₂ + R₃ + R₄

$\left|A\right|=-\left|\begin{array}{ccccc}1& 1& 1& 1& 3\\ 0& 2& 0& 0& -2\\ 0& 0& 2& 0& -2\\ 0& 0& 0& 2& -2\\ 0& 0& 0& 0& -14\end{array}\right|$

Since it is a diagonal matrix it’s determinant is the product of diagonal element

$\left|A\right|=-\left(1\times 2\times 2\times 2\times -14\right)$

$\left|A\right|=112$

x + y = 1

y + z = 1

x + z = 1

$A=\left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 1& 0& 1\end{array}\right]B=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$

$\left[A:B\right]=\left[\begin{array}{cccc}1& 1& 0& 1\\ 0& 1& 1& 1\\ 1& 0& 1& 1\end{array}\right]$

R3 → R3 – R1

$=\left[\begin{array}{cccc}1& 1& 0& 1\\ 0& 1& 1& 1\\ 0& -1& 1& 0\end{array}\right]$

R3 → R3 + R2

$=\left[\begin{array}{cccc}1& 1& 0& 1\\ 0& 1& 1& 1\\ 0& 0& 2& 1\end{array}\right]$

Rank (A) = Rank (A : B) = n

Concepts:

1. Switching two rows or columns causes the determinant to switch sign
2. Adding a multiple of one row/column to another row/column causes the determinant to remain the same

Calculations:

Now we can use the above concepts to determine the answer.

For options 1.

$\left|\begin{array}{ccc}1& x\left(x+1\right)& x+1\\ 1& y\left(y+1\right)& y+1\\ 1& z\left(z+1\right)& z+1\end{array}\right|=\left|\begin{array}{ccc}1& x^2+x& x+1\\ 1& y^2+y& y+1\\ 1& z^2+z& z+1\end{array}\right|$

$c3\to c3-c1.$

$\left|\begin{array}{ccc}1& x^2+x& x\\ 1& y^2+y& y\\ 1& z^2+z& z\end{array}\right|$

$\mathrm{c}2\to \mathrm{c}2-\mathrm{c}3$

$\left|\begin{array}{ccc}1& x^2& x\\ 1& y^2& y\\ 1& z^2& z\end{array}\right|$

Since,

$\left|\begin{array}{ccc}1& x^2& x\\ 1& y^2& y\\ 1& z^2& z\end{array}\right|=-\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|$

Therefore, switching two rows or columns causes the determinant to switch sign

$\left|\begin{array}{ccc}1& x^2& x\\ 1& y^2& y\\ 1& z^2& z\end{array}\right|\ne \left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|$ 

Therefore option 1 correct

Concept:

Determinant matrix M = product of its eigenvalues

Calculation:

Since it is 3 × 3 matrix, so roots will be 3 means 3 eigenvalues. Two eigenvalues are given as 12 and 6 + √-1)

As 3 × 3 is a real matrix, in real matrix roots can never be imaginary. We have to take the complex roots in pair. So,

Three eigenvalues are:

12, (6 + √-1) and (6 - √-1), that is, 12, (6 + i) and (6 - i)

Determinant matrix M = 12 × (6 + i) ×  (6 - i) = 444

Important point: 

√-1 = i → iota (imaginary number)

Uniform density function:

If the probability density function (pdf) is constant, the random variable is said to be uniformly distributed and the function is called ‘Uniform density function’

Calculation:

 From the given pdf,

f(x) = ¼

Mean is calculated as:

E[X] = $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}xf\left(x\right)dx$ 

$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\times \frac{1}{4}dx$

$=\frac{1}{4}{\left[\frac{x^2}{2}\right]}_{-1}^3$

$=\frac{1}{4}\left[\frac{9-1}{2}\right]$

$=\frac{8}{8}=1$

Var[X] = E[X²] – {E[X]}²

E[X²] = ​$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^{2}f\left(x\right)dx$ 

$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2\frac{1}{4}dx$

$=\frac{1}{4}{\left[\frac{x^3}{3}\right]}_{-1}^3$

$=\frac{1}{4}\left[\frac{27-\left(-1\right)}{3}\right]$

$=\frac{28}{12}=\frac{7}{3}$

Var [X] = E[X²] - {E[X]}²

= 7/3 – (1)²

= 4/3

= 1.33

Direct Formula:

From the Uniform Density function we know two limits ‘a’ and ‘b’

Mean is defined as:

$m=\frac{a+b}{2}$

$=\frac{-1+3}{2}$

= 1

Variance is defined as:

 $=\frac{(b-a{)}^2}{12}$ 

$=\frac{(3-(-1){)}^2}{12}$

$=\frac{16}{12}=\frac{4}{3}$

= 1.33

Extra concept:

Statistical averages are defined below:

1. Mean or average value

$\overline{X}=E\left[x\right]=m_1=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}xf\left(x\right)dx$

2. Mean square value

$\overline{X^2}=E\left[X^2\right]=m_2=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^{2}f\left(x\right)dx$

3. Variance

${\sigma }^2=E\left[{\left(X-\overline{X}\right)}^2\right]=m_2-m_1^2$

4. Standard deviation

$SD=\sqrt{variance}=\sigma$

$M=\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right]$

Determinant = 0 – 1 (-1) + 1 (1) = 2 ≠ 0

So, the rank of M is 3

Alternate method:

$M=\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right]$

R₃ ↔ R₁

$\sim \left[\begin{array}{ccc}1& 1& 0\\ 1& 0& 1\\ 0& 1& 1\end{array}\right]$

R₂ → R₂ – R₁

$\sim \left[\begin{array}{ccc}1& 1& 0\\ 0& -1& 1\\ 0& 1& 1\end{array}\right]$

R₃ → R₃ + R₂

$\sim \left[\begin{array}{ccc}1& 1& 0\\ 0& -1& 1\\ 0& 0& 2\end{array}\right]$

Now it is in the Echelon form

Rank = number of non-zero rows = 3

Explanation:

The augmented matrix of the system is

$\left[A/B\right]=\left[\begin{array}{cccc}1& 2& 1& 6\\ 1& 4& 3& 10\\ 1& 4& a& d\end{array}\right]$ 

R₂ – R₁. R₃ – R₁

$\sim \left[\begin{array}{cccc}1& 2& 1& 6\\ 0& 2& 2& 4\\ 0& 2& a-1& b-6\end{array}\right]$ 

R₃ – R₂

$\sim \left[\begin{array}{cccc}1& 2& 1& 6\\ 0& 2& 2& 4\\ 0& 0& a-3& b-10\end{array}\right]$ 

The system has no solution, if a = 3 and b ≠ 10

Given:  f(x) = x² for -1 ≤ x ≤ 1, and

= 0 for any other value of x

P $\left(-\frac{1}{3}\le x\le \frac{1}{3}\right)$ = ${\int }_{-\frac{1}{3}}^{\frac{1}{3}}f\left(x\right)dx$

= ${\int }_{-\frac{1}{3}}^{\frac{1}{3}}{x}^{2}dx$ 

= ${\left|\frac{x^3}{3}\right|}_{-1/3}^{+1/3}$

= $\frac{2}{81}$

The probability expressed in percentage,

P = $\frac{2}{81}\times 100$ = 2.469 %

Hence, P = 2.47 %

Using L Hospital’s

$f\left(x\right)=\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{4x^2}{\left(\cos 8x-cos4x\right)}$

$=4\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{2x}{\left(-8sin8x+4sin4x\right)}?$

$=4\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{2}{\left(-64\times cos8x+16cos4x\right)}$

$=\frac{8}{-64+16}=\frac{8}{-48}=-\frac{1}{6}$

18× f(x) = 18 × -1/6 = -3

Concept:

If A is any square matrix of order n, we can form the matrix [A – λI], where I is the nth order unit matrix. The determinant of this matrix equated to zero i.e. |A – λI| = 0 is called the characteristic equation of A.

The roots of the characteristic equation are called Eigenvalues or latent roots or characteristic roots of matrix A.

Eigenvector (X) that corresponding to Eigenvalue (λ) satisfies the equation AX = λX.

Properties of Eigenvalues:

The sum of Eigenvalues of a matrix A is equal to the trace of that matrix A

The product of Eigenvalues of a matrix A is equal to the determinant of that matrix A

Calculation:

$A=\left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)$

The given matrix is triangular.

Therefore, Eigenvalues = diagonal elements

= 1, 1

As the Eigenvalues are not unique, it is not diagonalizable.

Determinant of A = 1 ≠ 0

Therefore A is non-singular

To get Eigenvector,

Ax = λx

$\Rightarrow \left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=1\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}{x}_1-x_2\\ x_2\end{array}\right]=\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]$

⇒ x₂ = 0 and x₁ can be any value.

The Eigenvector is $\left[\begin{array}{c}k\\ 0\end{array}\right]$ 

Therefore, the given vector is Eigenvector.

Probability of (X or Y) = P(X ∪ Y)  = $\frac{3}{4}$

Formula: 

P(X ∪ Y) = P(X) + P(Y) – P(X ∩ Y)

Two events are independent

P(X ∩ Y) = P(X).P(Y)

The two events are equally likely

P(X) = P(Y) = p

P(X ∪ Y) = p + p – p.p

∴ $\frac{3}{4}$ = 2p – p2 

3 = 8p  – 4p2

4p2 – 8p + 3 = 0

4p2 – 6p - 2p + 3 = 0

2p (2p - 3) - (2p - 3) = 0

(2p - 3)(2p - 1)

$\therefore p=\frac{1}{2}or\frac{3}{2}$

Since p  is always ≤ 1

$\therefore p=\frac{1}{2}=0.5$

the probability of Y is 0.5

Given $LU=\left[\begin{array}{cc}2& 2\\ 4& 9\end{array}\right]$

$\left[\begin{array}{cc}x& 0\\ z& y\end{array}\right]\left[\begin{array}{cc}1& p\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}2& 3\\ 4& 9\end{array}\right]$

$\left[\begin{array}{cc}x& xp\\ z& zp+y\end{array}\right]=\left[\begin{array}{cc}2& 2\\ 4& 9\end{array}\right]$

z = 4

x = 2,

xp = 2 

∴ p = 1

z = 4

zp + y = 9

(4)(1) + y = 9

y = 9 - 4

∴ y = 5

Here, we have assume some function by checking options to solve these kind of questions

Consider the function φ(x) = f(x) – 2 g(x)

φ(0) = φ(1) = 2,

So, f(x) satisfies the conditions of Roll’s theorem in [0, 1]

Hence, φ’(x) = f’(x) – 2g’(x) has at least one 0 at in (0,1)

i.e, φ’(c) = 0

f’(c) – 2g’(c) = 0

f’(c) = 2g’(c)

Hence, There must exist a constant c in (0, 1), such that f’(c) = 2g’(c)

The area under the curve must be 1, i.e.

(10 + 2) × a = 1

$a=\frac{1}{12}$

$f_x\left(x\right)=\left\{\begin{array}{c}\frac{1}{12}-2\le x\le 10\\ 0otherwise\end{array}\right\}$

Now, y = 2x – 6

∴ f_y(y) can be defined as:

$f_y\left(y\right)=\{\begin{array}{c}\frac{1}{24}-10\le x\le 14\\ 0otherwise\end{array}$

If x ≥ 5, then y ≥ 4

So, P(y ≤ 7|x ≥ 5) = P(y ≤ 7|y ≥ 4)

$P(y\le 7|x\ge 5)=\frac{P(4\le y\le 7)}{P(4\le y\le 14)}$

$P(y\le 7|x\ge 5)=\frac{P(4\le y\le 7)}{P(4\le y\le 14)}$

$=\frac{3}{10}$

= 0.3

Using characteristic equation | A - λI| = 0, we get

(1-λ) (λ²-2) + (2-λ) – λ = -λ³ + λ² = 0

Hence, λ = 0, λ = 1

The largest eigen value is 1, so we calculate the eigen vector corresponding to eigen value 1.

A – I = $\left[\begin{array}{ccc}0& -1& 1\\ 1& -2& 1\\ -1& 1& 0\end{array}\right]$ , R₁↔ R₂

= $\left[\begin{array}{ccc}1& -2& 1\\ 0& -1& 1\\ -1& 1& 0\end{array}\right]$ , R₃ ← R₂ + R₁

= $\left[\begin{array}{ccc}1& -2& 1\\ 0& -1& 1\\ 0& -1& 1\end{array}\right]$ , R₃ ← R₃ + R₂

= $\left[\begin{array}{ccc}1& -2& 1\\ 0& -1& 1\\ 0& 0& 0\end{array}\right]$ , R₁ ← R₁ + 2R₂

= $\left[\begin{array}{ccc}1& 0& 1\\ 0& -1& 1\\ 0& 0& 0\end{array}\right]$ 

[ A – I] $\overrightarrow{x}$ = 0

x₁ – x₃ = 0, x₁ = x₃

-x₂ + x₃ = 0, x₂ = x₃

= $\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}{x}_3\\ x_3\\ x_3\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]x^3$

Vector corresponding to eigen value 1 is $\overrightarrow{x}$ = $\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$

Data:  

$p\left(A\right)=\frac{3}{5}$

$P\left(\frac{A}{B}\right)=\frac{4}{5}$ , $P\left(\frac{B}{A}\right)=\frac{2}{3}$

Formula

$P\left(\frac{B}{A}\right)=\frac{P\left(A\cap B\right)}{P\left(A\right)}$

Calculation:

$\frac{2}{3}=\frac{P\left(A\cap B\right)}{\frac{3}{5}}$  , 

$P\left(A\cap B\right)=\frac{2}{5}$

Also, $P\left(\frac{A}{B}\right)=\frac{A\left(A\cap B\right)}{P\left(B\right)},$

$\frac{4}{5}=\frac{\frac{2}{5}}{P\left(B\right)}$ ,

$P\left(B\right)=\frac{1}{2}$

Required probability, $P\left(\frac{A^{\prime }}{B^{\prime }}\right)=\frac{P\left(A^{\prime }\cap B^{\prime }\right)}{P\left(B^{\prime }\right)}$

$=\frac{P{\left(A\cup B\right)}^{\prime }}{1-P\left(B\right)}=\frac{1-P\left(A\cup B\right)}{1-P\left(B\right)}$

$=\frac{1-\left(P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)\right)}{1-P\left(B\right)}$

$=\frac{1-\left(\frac{3}{5}+\frac{1}{2}-\frac{2}{5}\right)}{1-\frac{1}{2}}$

Given:

f(-1) = 0

|f’(x)| ≤ 2

-2 ≤ f ’(x) ≤ 2

Using Lagrange mean value theorem:

$f^{\prime }\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Since value of f (-1) = 0 is given, we take the interval [-1, 2], i.e.

-2 ≤ f’(x) ≤ 2

$-2\le \frac{f\left(2\right)-f\left(-1\right)}{2-\left(-1\right)}\le 2$

$-2\le \frac{f\left(2\right)}{3}\le 2$

-6 ≤ f(2) ≤ 6

We observe that option (a) satisfies this condition.

Concept:

Properties of Rank:

Rank of a matrix is the number of independent rows in the given matrix. Given two square matrices A and B of order n × n, we have following properties:

1. Rank of product of A and B i.e. Rank (AB) ≥ Rank (A) + Rank (B) – order of square matrix

2. Rank of sum of A and B i.e. Rank (A + B) ≤ Rank (A) + Rank (B).

Properties of Determinant:

Given two square matrices A and B of order n × n and their determinants Det(A) and Det(B) respectively, determinant of their product i.e Det( AB) = Det(A) * Det(B). However, the same does not hold for the addition of the given matrices.

Example:

Consider two square matrices A and B each of order 2×2.

$M=\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]Rank\left(M\right)=2,Det\left(M\right)=5$

$N=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]Rank\left(N\right)=2,Det\left(N\right)=1$

$MN=\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]\times \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]Rank\left(MN\right)=2,Det\left(MN\right)=5$

$M+N=\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}3& 1\\ 3& 5\end{array}\right]Rank\left(M+N\right)=2,Det\left(M+N\right)=12$

Option 1 is TRUE.  

Rank(M + N) = 2. Sum of rank of M and N is: 2 + 2 = 4. Therefore, the relation: Rank( M + N) ≤ Rank M) + Rank (N) holds true

Therefore, the rank of the addition matrix is less than or equal to the sum of the rank of the individual matrices.

Option 2: FALSE.

The rank of product matrix MN is 2. Product of Rank(M) and Rank(N) is: 2*2 = 4. Therefore, the rank of the product matrix is not equal to the product of the rank of individual matrices.

Option 3 is TRUE. 

Det(MN)= 5 = Det(M) * Det(N)

Option 4 is FALSE.  

Det(M+N)= 12, which is greater than the sum of the determinants of individual matrices.

$f\left(x\right)=\{\begin{array}{c}{e}^x,x<1\\ \log x+a{x}^2+bx,x\ge 1\end{array}$

$f^{\prime }\left(x\right)=\{\begin{array}{c}{e}^x,x<1\\ \frac{1}{x}+2ax+b,x\ge 1\end{array}$

f’(1) = e, x < 1

f’ (1) = 1 + 2a + b, x ≥ 1

since f(x) is differentiable at x = 1,

e = 1 + 2a + b → (1)

At x = 1,

f(1) = e, x < 1

f(1) = a + b, x ≥ 1

since f(x) is continuous at x = 1,

e = a + b → (2)

From (1) and (2)

⇒ 1 + 2a + b = a + b

⇒ a = -1

⇒ b = e + 1

$f(x)=x^3-18x^2+105x-10$

Diff w.r.t to x

$f^{\prime }\left(x\right)=3x^2-36x+105$

$f^{″}\left(x\right)=6x-36$

To find maxima or minima

f'(x) = 0

$3x^2-36x+105=0$

$x^2-12x+35=0$

(x - 5)(x - 7) = 0

∴ x = 5 or x = 7

If x = 5

f'’(5) = -6

f'’(5) < 0

∴ maxima may fall at 5 in the given domain

If x = 7

f'’(7) = 6

f'’(7) > 0

∴ minima may fall at 7 in the given domain

Check the boundary condition along with x = 1 and x = 2

The fundamental property of Definite Integral.

 $\underset{0}{\overset{\mathbf{a}}{\int }}f\left(x\right)dx$  = $\underset{0}{\overset{\mathbf{a}}{\int }}f\left(a-x\right)dx$     

Calculation:    

Given:

$I=\underset{0}{\overset{\pi /2}{\int }}sin2x\mathrm{l}\mathrm{o}\mathrm{g}(\tan x)dx$

Then, by using the above property, we get –

 $I=\underset{0}{\overset{\pi /2}{\int }}sin\left(2\left(\frac{\pi }{2}-x\right)\right)\mathrm{l}\mathrm{o}\mathrm{g}\left(tan\left(\frac{\pi }{2}-x\right)\right)dx$ 

 $\Rightarrow I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\mathrm{s}\mathrm{i}\mathrm{n}\left(\pi -2x\right)\mathrm{l}\mathrm{o}\mathrm{g}(\cot x)dx$ 

 $\Rightarrow I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sin \left(2x\right)\mathrm{l}\mathrm{o}\mathrm{g}\left({\left(\tan x\right)}^{-1}\right)dx$ 

 $\Rightarrow I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\mathrm{s}\mathrm{i}\mathrm{n}\left(2x\right)\left(-1\right)\mathrm{l}\mathrm{o}\mathrm{g}(\tan x)dx$                    

 $I=-\underset{0}{\overset{\pi /2}{\int }}sin2x\mathrm{l}\mathrm{o}\mathrm{g}(\tan x)dx$ 

I = -I , 2I = 0

$A=\left[\begin{array}{ccc}1& 0& -1\\ -1& 2& 0\\ 0& 0& -2\end{array}\right]$

B = A³ – A² - 4A + 5I

|A - λI| = 0

$\Rightarrow \left|\begin{array}{ccc}1-\lambda & 0& -1\\ -1& 2-\lambda & 0\\ 0& 0& -2-\lambda \end{array}\right|=0$

⇒ (1- λ) (2 - λ) (-2 - λ) – 1 (0) = 0

⇒ λ = 1, λ = 2, λ = -2

If λ is an Eigen value pf A, then λⁿ will be an Eigen value pf Aⁿ.

If λ is an Eigen value of A, then A, then k λ will be an Eigen value of kA where k is a scalar.

Eigen values of A are 1, 2, -2

Eigen values of A² are 1, 4, 4

Eigen values of A³ are 1, 8, -8

Eigen values of 4A are 4, 8, -8

Eigen values of 5I are 5, 5, 5

B = A3 – A2 - 4A + 5I

λ_1B = λ1³– λ12– 4λ1 + 5 = 1 - 1 - 4 + 5 = 1

λ2B = λ23– λ22– 4λ2 + 5 = 8 - 4 - 8 + 5 = 1

λ3B = λ33– λ32– 4λ3 + 5 = -8 - 4 + 8 + 5 = 1

Eigen values of B are 1, 1, 1

Let matrix $M=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

$v_1=\left[\begin{array}{c}a\\ c\end{array}\right],v_2=\left[\begin{array}{c}b\\ d\end{array}\right]$

$M^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$

$u_1^T=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\end{array}\right]$

$u_2^T=\frac{1}{ad-bc}\left[\begin{array}{cc}-c& a\end{array}\right]$

$u_1^T{v}_1=\frac{1}{ab-bc}\left[\begin{array}{cc}d& -b\end{array}\right]\left[\begin{array}{c}a\\ c\end{array}\right]=1$

$u_2^T{v}_2=\frac{1}{ab-bc}\left[\begin{array}{cc}-c& a\end{array}\right]\left[\begin{array}{c}b\\ d\end{array}\right]=1$

$u_1^T{v}_2=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\end{array}\right]\left[\begin{array}{c}b\\ d\end{array}\right]=0$

$u_2^T{v}_1=\frac{1}{ad-bc}\left[\begin{array}{cc}-c& a\end{array}\right]\left[\begin{array}{c}a\\ c\end{array}\right]=0$

The coefficient matrix of the given linear equation is

$\left[\begin{array}{ccc}1& 3& -2\\ 0& 5& 3\\ 1& -2& -5\end{array}\right]\left[\begin{array}{c}\begin{array}{c}a\\ b\end{array}\\ c\end{array}\right]=\left[\begin{array}{c}\begin{array}{c}-1\\ -8\end{array}\\ 7\end{array}\right]$

$letA=\left[\begin{array}{ccc}1& 3& -2\\ 0& 5& 3\\ 1& -2& -5\end{array}\right]$

order = 3

Let the augmented matrix be

$X=\left[\begin{array}{ccc}1& 3& -2\\ 0& 5& 3\\ 1& -2& -5\end{array}|\begin{array}{c}-1\\ -8\\ 7\end{array}\right]$

R₃ → R₃ – R₁

$X\sim \left[\begin{array}{ccc}1& 3& -2\\ 0& 5& 3\\ 0& -5& -3\end{array}|\begin{array}{c}-1\\ -8\\ 8\end{array}\right]$