System of Linear Equations MCQ Quiz in தமிழ் - Objective Question with Answer for System of Linear Equations - இலவச PDF ஐப் பதிவிறக்கவும்

CONCEPT:

Consider the system of m linear equations

a11 x1 + a12 x2 + … + a1n xn = 0

a21 x1 + a22 x2 + … + a2n xn = 0

am1 x1 + am2 x2 + … + amn xn = 0

• 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 matrix.

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

• A is the coefficient matrix of the given system of equations.

\(x = \frac{|A_x|}{|A|}\), \(y = \frac{|A_y|}{|A|}\), \(z = \frac{|A_z|}{|A|}\)

• Where, A_x, A_y, A_z is the coefficient matrix of the given system of equations after replacing the first, second, and third columns from the constant term column which will be having all the entries as 0.
• In the case of homogeneous equations, the determinants of, Ax, Ay, Az will be 0 definitely.
• So, for the system of homogeneous equations having the the-trivial solution, the determinant of A should be zero.
• The system of homogeneous equations has a unique solution (trivial solution) if and only if the determinant of A is non-zero.

CALCULATION:

For non - trivial solution, the |A| = 0

\(⇒ \;\left| {\begin{array}{*{20}{c}} 1&{ K}&3\\ 4&3&{ k}\\ 2&1&{ 2} \end{array}} \right|\; = \;0\)

⇒ 1(6 - K) - K(8 - 2K) + 3(4 - 6) = 0

⇒ 9K -  2K² = 0

⇒ k = 0 or \(\frac{9}{2}\)

So, the correct answer is option 1.

Explanation:

Since 1 and 3 are the eigen values of A so the characteristic equation of A is:

\({\lambda ^2} - 4\lambda + 3 = 0\)

Also by Caley-Hamilton theorem, every square matrix satisfies its own characteristic equation so

A² - 4A + 3I₂ = 0

A² = 4A - 3I₂

A³ = 4A² - 3A = 4(4A - 3I₂) - 3A

A³= 13A - 12I₂

Concept:

It is case of equal matrix

∴ L. H. S = R. H. S

Calculation:

\(\left[ {\begin{array}{*{20}{c}} 2&5\\ { - 4}&3 \end{array}} \right]\left\{ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right\} = \left\{ {\begin{array}{*{20}{c}} 2\\ { - 30} \end{array}} \right\}\)

2x + 5y = 2     ----(I)

-4x + 3y = -30   ----(II)

Multiply eq (1) by 4 and (II) by 2

∴8x + 20y = 8       ---(III)

-8x + 6y = -60       ---(IV)

Adding eq (I) and (II)

26y = -52 ⇒ y = -2

Putting the value of y is eq (I)

2x + 5 (-2) = 2 ⇒ x = 6

Gauss-Seidel Method:

In Gauss Seidel method, the value of x calculated is used in next calculation putting other variables as 0.

x1 + 2x2 + 3x3 = 5

Putting x2 = 0, x3 = 0 ⇒ x1 = 5

2x1 + 3x2 + x3 = 1

Putting x1 = 5, x3 = 0 ⇒ x2 = -3

3x1 + 2x2 + x3 = 3

Putting x1 = 5, x2 = -3 ⇒ x3 = 3 – 3(5) – 2 (-3)

x3 = 3 – 15 + 6

x3 = -6

Concept:

Gauss-Seidel iteration method: This is a modification of the Jacobi's iteration method. We start the initial approximation x₀, y₀, z₀, w₀ (each = 0).

The value of x calculated is used in next calculation putting other variable as 0.

Calculation:

Rewriting the given equation as 

x = 0.3 + 0.2y + 0.1z + 0.1w;

y = 1.5 + 0.2x + 0.1z + 0.1w;

z = 2.7 + 0.1x + 0.1y + 0.2w;

w = -0.9 + 0.1x + 0.1y + 0.2z;

Now the first iteration,

Putting y = 0, z = 0, w = 0 in 1st equation ⇒ x = 0.3;

Putting x = 0.3, z = 0, w = 0 in 2nd equation ⇒ y = 1.56;

Putting x = 0.3, y = 1.56, w = 0 in 3rd equation ⇒ z = 2.886;

Putting x = 0.3, y = 1.56, z = 2.886 in 4th equation ⇒ w = - 0.1368

Concept Used:-

Let consider a standard pair of linear equation such that,

\(\begin{aligned} & a_1 x+b_1 y+c_1=0 \\ & a_2 x+b_2 y+c_2=0 \end{aligned}\)

For the pair of linear equations, the condition for the unique solution is,

\(\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\)

Explanation:-

Given a pair of linear equation is,

kx - y = 2

6x - 2y = 3

These equations can be written as,

⇒ kx - y - 2 = 0

⇒ 6x - 2y - 3 = 0

On comparing it with standard pair of linear equation, we get, 

⇒ a₁ = k, b₁ = -1, c₁ = -2

⇒ a2 = 6, b2 = -2, c2 = -3

Using the condition for unique solution,

\(\Rightarrow \frac{a_1}{a_2}\neq\frac{b_1}{b_2}\\ \Rightarrow \frac{k}{6}\neq\frac{-1}{-2}\\ \Rightarrow \frac{k}{6}\neq\frac{1}{2}\\ \Rightarrow k\neq\frac{6}{2}\\ \Rightarrow k\neq3\)

So, the value of k for which the system of equations has unique solution,

k ≠ 3

Hence, the correct option is 2.

Concept:

The number of solutions can be determined by finding out the rank of the Augmented matrix and the rank of the Coefficient matrix.

• If rank(Augmented matrix) = rank(Coefficient matrix) = no. of variables then no of solutions = 1.
• If rank(Augmented matrix)  ≠ rank(Coefficient matrix) then no of solutions = 0.
• If rank(Augmented matrix) = rank(Coefficient matrix) < no. of variables, no of solutions = infinite.

Calculation:

The Augmented matrix

\({\rm{[}}A\;{\rm{|}}\;B] = \left[ {\left. {\begin{array}{*{20}{c}} 2&1&1\\ 0&1&{ - 1}\\ 1&1&0 \end{array}} \right|\begin{array}{*{20}{c}} 0\\ 0\\ 0 \end{array}} \right]\)

Performing gauss elimination on [A | B] we get

\(\left[ {\left. {\begin{array}{*{20}{c}} 2&1&1\\ 0&1&{ - 1}\\ 1&1&0 \end{array}} \right|\begin{array}{*{20}{c}} 0\\ 0\\ 0 \end{array}} \right]\;\mathop \to \limits^{{R_3} - \frac{1}{2}{R_1}} \;\left[ {\left. {\begin{array}{*{20}{c}} 2&1&1\\ 0&1&{ - 1}\\ 0&{\frac{1}{2}}&{-\frac{1}{2}} \end{array}} \right|\begin{array}{*{20}{c}} 0\\ 0\\ 0 \end{array}} \right]\;\)

\(\left[ {\left. {\begin{array}{*{20}{c}} 2&1&1\\ 0&1&{ - 1}\\ 0&{\frac{1}{2}}&{-\frac{1}{2}} \end{array}} \right|\begin{array}{*{20}{c}} 0\\ 0\\ 0 \end{array}} \right]\;\mathop \to \limits^{{R_3} - \frac{1}{2}{R_2}} \left[ {\left. {\begin{array}{*{20}{c}} 2&1&1\\ 0&1&{ - 1}\\ 0&0&0 \end{array}} \right|\begin{array}{*{20}{c}} 0\\ 0\\ 0 \end{array}} \right]\)

Rank (A) = Rank (A | B) = 2 < 3

So an infinite number of solutions are obtained.

Concept:

If |A| ≠ 0 ⇒ It has a unique solution.

If |A| =  0 ⇒ It has either no solution or infinite solution.

ρ(A) ≠ ρ(A|B) ⇒ It has either no solution

ρ(A) = ρ(A|B) r < n⇒ It has infinite solution

Calculation:

For given system of equations to have no solution, the determinant of the coefficient matrix should be equal to 0

\(\left| {\begin{array}{*{20}{c}} 1&1&1\\ 2&4&6\\ 2&4&x \end{array}} \right| = 0\)

\(1 \times \left( {4x - 24} \right) - 1 \times \left( {2x - 12} \right) + 1\left( {8 - 8} \right) = 0\)

⇒ 2x - 12 = 0;

⇒ x = 6;

Augmented matrix is given by

\(\left[ {\begin{array}{*{20}{c}} 1&1&1&{18}\\ 2&4&6&{12}\\ 2&4&6&y \end{array}} \right]\)

\(R3 \to R3 - R2\)

\(\left[ {\begin{array}{*{20}{c}} 1&1&1&{18}\\ 2&4&6&{12}\\ 0&0&0&{y - 12} \end{array}} \right]\)

For given system of equations to have no solution, the rank of the augmented matrix should not be equal to the coefficient matrix

ρ(A) = 2

∴ y - 12 ≠ 0;

⇒ y ≠ 12;

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

Trace (A) = 1 + 1 + 1 = 3

\(\begin{array}{l} \left| {\rm{A}} \right|{\rm{}} = {\rm{}}1{\rm{}}\left( {1{\rm{}}-{\rm{}}0} \right){\rm{}}-{\rm{}}1{\rm{}}\left( 0 \right){\rm{}} = {\rm{}}1\\ {{\rm{M}}_{11}}{\rm{}} + {{\rm{M}}_{22}}{\rm{}} + {\rm{}}{{\rm{M}}_{33}}{\rm{}} = {\rm{}}1{\rm{}} + {\rm{}}1{\rm{}} + {\rm{}}1{\rm{}} = {\rm{}}3 \end{array}\)

∴ Characteristic of given matrix is,

\(\begin{array}{l} - {{\rm{\lambda }}^3} + {\rm{trace}}\left( {\rm{A}} \right){{\rm{\lambda }}^2} - \left( {{{\rm{M}}_{11}} + {{\rm{M}}_{22}} + {{\rm{M}}_{33}}} \right){\rm{\lambda }} + \left| {\rm{A}} \right| = 0\\ - {{\rm{\lambda }}^3} + 3{{\rm{\lambda }}^2} - 3{\rm{\lambda }} + 1 = 0 \Rightarrow {{\rm{\lambda }}^3} - 3{{\rm{\lambda }}^2} + 3{\rm{\lambda }} - 1 = 0 \end{array}\)

From the Cauley Hamilton theorem, we have

\(\begin{array}{l} {{\rm{A}}^3} - 3{{\rm{A}}^2} + 3{\rm{A}} - {\rm{I}} = 0\\ {{\rm{A}}^3} = 3{{\rm{A}}^2} - 3{\rm{A}} + {\rm{I}}\\ {{\rm{A}}^4} = 3{{\rm{A}}^3} - 3{{\rm{A}}^2} + {\rm{A}}\\ \therefore {{\rm{A}}^4} - 3{{\rm{A}}^3} = - 3{{\rm{A}}^2} + {\rm{A}} \end{array}\)

Explanation:

By using Gaussian elimination method:

\(\begin{array}{*{20}{c}} {{R_1}}\\ {{R_2}}\\ {{R_3}} \end{array}\left[ {\begin{array}{*{20}{c}} 1&1&1\\ 1&2&3\\ 2&3&4 \end{array}\left| {\begin{array}{*{20}{c}} {150}\\ {100}\\ {200} \end{array}} \right.} \right]\)

\(\Rightarrow {R_2} - {R_1}\ {\rm{and}}\ {R_3} - 2{R_1} \Rightarrow\)

\(\begin{array}{l} = \left[ {\begin{array}{*{20}{c}} 1&1&1\\ 0&1&2\\ 0&1&2 \end{array}\left| {\begin{array}{*{20}{c}} {150}\\ { - 50}\\ { - 100} \end{array}} \right.} \right]\\ \Rightarrow {R_3} - {R_2}\\ \left[ {\begin{array}{*{20}{c}} 1&1&1\\ 0&1&2\\ 0&0&0 \end{array}\left| {\begin{array}{*{20}{c}} {150}\\ { - 50}\\ { - 50} \end{array}} \right.} \right] \end{array}\)

1x + 1y + 1z = 150

1y + 2z = -50

0 = -50

Here, two equations and 3 unknowns as third equation is invalid

and also rank of [A] < rank of [A : B]

So inconsistent and having no solution