Linear Independent MCQ Quiz in मल्याळम - Objective Question with Answer for Linear Independent - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

\(A = \left[ {\begin{array}{*{20}{c}} 2&2&3&3\\ 0&1&1&1\\ {\begin{array}{*{20}{c}} 0\\ 0 \end{array}}&{\begin{array}{*{20}{c}} 0\\ 0 \end{array}}&{\begin{array}{*{20}{c}} 3\\ 0 \end{array}}&{\begin{array}{*{20}{c}} 3\\ 2 \end{array}} \end{array}} \right]\)

The given matrix is a upper triangular matrix

For upper triangular matrix, eigen values are the diagonal elements

Hence,

λ₁ = 2 = λ₄

λ₂ = 1

λ₃ = 3

Hence, there are three distinct eigen values.

\(A\left[ {\begin{array}{*{20}{c}} { - 1}\\ 1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 1}\\ 1 \end{array}} \right]\) 

It is in the form of AX = λX

Where, λ is eigen value and X is eigen vector.

\(\begin{array}{l} {\lambda _1} = 1,{X_1}\left[ {\begin{array}{*{20}{c}} { - 1}\\ 1 \end{array}} \right]\\ A\left[ {\begin{array}{*{20}{c}} { - 1}\\ 2 \end{array}} \right] = 2\left[ {\begin{array}{*{20}{c}} { - 1}\\ 1 \end{array}} \right]\\ A\left[ {\begin{array}{*{20}{c}} { - 1}\\ 2 \end{array}} \right] = - 2\left[ {\begin{array}{*{20}{c}} 1\\ { - 2} \end{array}} \right]\\ \Rightarrow A\left[ {\begin{array}{*{20}{c}} { - 1}\\ 2 \end{array}} \right] = - 2\left[ {\begin{array}{*{20}{c}} { - 1}\\ 2 \end{array}} \right] \end{array}\) 

Now, it is in the form of AX = λX

\({\lambda _2} = - 2,{X_2} = \left[ {\begin{array}{*{20}{c}} { - 1}\\ 2 \end{array}} \right]\) 

We can represent any matrix A;

A = PDP^-1

Where, \(P = \left[ {{X_1}{X_2}} \right]D = \left[ {\begin{array}{*{20}{c}} {{\lambda _1}}&0\\ 0&{{\lambda _2}} \end{array}} \right]\)

Now, \(P = \left[ {\begin{array}{*{20}{c}} { - 1}&{ - 1}\\ 1&2 \end{array}} \right],D = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&{ - 2} \end{array}} \right]\)

\(\begin{array}{l} {P^{ - 1}} = \frac{1}{{\left| P \right|}}\left[ {\begin{array}{*{20}{c}} 2&1\\ { - 1}&{ - 1} \end{array}} \right]\\ = \frac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}} 2&1\\ { - 1}&{ - 1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 2}&{ - 1}\\ 1&1 \end{array}} \right]\\ = \frac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}} 2&1\\ { - 1}&{ - 1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 2}&{ - 1}\\ 1&1 \end{array}} \right] \end{array}\)

Concept:

Determinant of a matrix is the product of its eigenvalues.

Determinant of a matrix is same as its transpose.

Determinant of a matrix is reciprocal to its inverse.

Calculation:

Eigenvalues of matrix A are 10 and y.

Determinant of matrix A = 10 × y

50 = 10 × y

∴ y = 5

 the other Eigen value is 5

Concept:

Consider the system of m linear equations

a11 x1 + a12 x2 + … + a1n xn = b1

a21 x1 + a22 x2 + … + a2n xn = b2

…

am1 x1 + am2 x2 + … + amn xn = bm

The above equations containing the n unknowns x1, x2, …, xn. To determine whether the above system of equations is consistent or not, we need to find the rank of the following matrices.

\(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}& \ldots &{{a_{1n}}}\\ {{a_{21}}}&{{a_{22}}}& \ldots &{{a_{2n}}}\\ \ldots & \ldots & \ldots & \ldots \\ {{a_{m1}}}&{{a_{m2}}}& \ldots &{{a_{mn}}} \end{array}} \right]\) and \(\left[ {A{\rm{|}}B} \right] = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}& \ldots &{{a_{1n}}}&{{b_1}}\\ {{a_{21}}}&{{a_{22}}}& \ldots &{{a_{2n}}}&{{b_2}}\\ \ldots & \ldots & \ldots & \ldots & \ldots \\ {{a_{m1}}}&{{a_{m2}}}& \ldots &{{a_{mn}}}&{{b_m}} \end{array}} \right]\)

A is the coefficient matrix and [A|B] is called an augmented matrix of the given system of equations.

We can find the consistency of the given system of equations as follows:

(i) If the rank of matrix A is equal to rank of an augmented matrix and it is equal to the number of unknowns, then the system is consistent and there is a unique solution.

The rank of A = Rank of augmented matrix = n

(ii) If the rank of matrix A is equal to rank of an augmented matrix and it is less than the number of unknowns, then the system is consistent and there are an infinite number of solutions.

The rank of A = Rank of augmented matrix < n

(iii) If the rank of matrix A is not equal to rank of the augmented matrix, then the system is inconsistent, and it has no solution.

The rank of A ≠ Rank of an augmented matrix

Explanation:

If determinant of coefficient matrix is non-zero then there is a unique solution. If determinant is non – zero. So all the Eigen values are non-zero. If Eigen values are non – zero then question are linearly independent.

we have a set of three linear equations with three variables.

i.e., [A]_3×3 [X]_3×1 = [B]_3×1

Statement P:

In this case,

P(A) = P(A : B) = Number of variables

⇒ P(A) = P(A : B) = 3

∴ three independent rows exist in matrix A irrespective of matrix B.

Also, |A| ≠ 0     ---(i)

Hence, P = Q = S

Statement R:

Let λ_1, λ₂ & λ₃ are eigen values of matrix A

using the properties,

λ1 λ2 λ₃ = |A|   ----(ii)

from equation (i) & (ii),

λ1 λ2 λ3 ≠ 0

⇒ λ1 ≠ 0, λ2 ≠ 0 & λ3 ≠ 0

Therefore, from above discussion, it is inferred that,

P = Q = R = S

Concept:

From the property of eigenvalues

Product of eigenvalues = |P| = Determinant of matrix

Sum of Eigenvalues = Sum of Principal elements of Matrix

Calculation:

\(P = \left[ {\begin{array}{*{20}{c}} 2&0&1\\ 4&{ - 3}&3\\ 0&2&{ - 1} \end{array}} \right]\)

Product of Eigenvalues = |P| = 2(3 - 6) + 1(8 - 0) = 2

\(\begin{array}{l} {\rm{AX\;}} = {\rm{\;\lambda X}}\\ \left[ {\begin{array}{*{20}{c}} 4&1&2\\ {\rm{p}}&2&1\\ {14}&{ - 4}&{10} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ 2\\ 3 \end{array}} \right] = {\rm{\lambda }}\left[ {\begin{array}{*{20}{c}} 1\\ 2\\ 3 \end{array}} \right]\\ \therefore {\rm{\;}}4{\rm{\;}} + {\rm{\;}}2{\rm{\;}} + {\rm{\;}}6{\rm{\;}} = {\rm{\;\lambda \;\;\;}} \Rightarrow {\rm{\;\lambda \;}} = {\rm{\;}}12\\ {\rm{p\;}} + {\rm{\;}}4{\rm{\;}} + {\rm{\;}}3{\rm{\;}} = {\rm{\;}}2{\rm{\lambda \;\;\;\;}} \Rightarrow {\rm{\;P\;}} = {\rm{\;}}24{\rm{\;}}-{\rm{\;}}7{\rm{\;}} = {\rm{\;}}17 \end{array}\)

Given engine values = -3, -5

For λ = -3

Eigen value of the matrix (X + I)^-1 (X + 7I) is

= (-3 + 1)^-1 (-3 + 7)

\(= \frac{{ - 1}}{2}\left( 4 \right) = - 2\)

For λ = -5

Eigen value of the matrix (X + I)^-1 (X + 7I) is

= (-5 + 1)^-1 (-5 + 7)

\(= \frac{{ - 1}}{4}\left( 2 \right) = \frac{{ - 1}}{2}\)

The eigen values are\(= - \frac{1}{2}, - 2\)

Concept:

The eigenvalues of a matrix are obtained when we find the values λ which satisfies the characteristic equation of the matrix A, namely those values of λ for which;

det (A – λI) = 0

Calculation:

For the Given matrix:-

\(A=\left[ \begin{matrix} 4 & 1 \\ 3 & 2 \\ \end{matrix} \right]\) 

⇒ Solving for det (A – λ I) = 0, we find the value of λ which satisfies it.

det [A - λ I] = 0; where I is a 2 × 2 Identify Matrix;

\(\Rightarrow \left[ \begin{matrix} 4 & 1 \\ 3 & 2 \\ \end{matrix} \right]-\left[ \begin{matrix} \lambda & 0 \\ 0 & \lambda \\ \end{matrix} \right]=0\) 

\(\Rightarrow det~\left[ \begin{matrix} 4-\lambda & 1 \\ 3 & 2-\lambda \\ \end{matrix} \right]=0\) 

⇒ (4 - λ) (2 - λ) – 3 = 0

⇒ 8 – 4 λ – 2 λ + λ ² – 3 = 0

⇒ λ ² – 6 λ + 5 = 0

λ ² – 5 λ – λ + 5 = 0

⇒ λ (λ - 5) – 1 (λ - 5) = 0

⇒ (λ - 1) (λ - 5) = 0

Concept:

The given matrix is a upper Triangular matrix.

For upper/lower/Diagonal matrix the Eigen values are the diagonal elements.

Calculation:

The diagonal elements are 3, 0, 2

hence eigen values are

λ₁ = 2

λ₂ = 0

λ₃ = 3

\(A = \left[ {\begin{array}{*{20}{c}} a&{ - 1}&4\\ 0&b&7\\ 0&0&3 \end{array}} \right]\)

Sum of Eigen values = 10

Product of Eigen values = 30

From the properties of Eigen values,

Sum of Eigen values = Trace = 10

⇒ a + b + 3 = 10

⇒ a + b = 7

Product of Eigen values = Determinant = 30

\(\Rightarrow \left[ {\begin{array}{*{20}{c}} a&{ - 1}&4\\ 0&b&7\\ 0&0&3 \end{array}} \right] = 30\)

⇒ 3ab = 30 ⇒ ab = 10

Now, we know that

(a + b)² = a² + b² + 2ab

⇒ a² + b² = (a + b)² – 2ab = (7)² – 2(10) = 29.