Systems of equations multiple choice test pdf

Latest Systems of Linear Equations MCQ Objective Questions

Systems of Linear Equations MCQ Question 1:

The solution of the following system of equations is

\(\frac{x}{a} = \frac{y}{b}\), ax + by = a2 + b2

1. -a, -b
2. a, b
3. ab, ab
4. \(\frac{b^2}{a}, \frac{a^2}{b}\)

Answer (Detailed Solution Below)

Option 2 : a, b

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Given:

The given system of equations:-

\(\frac{x}{a} = \frac{y}{b}\) and ax + by = a2 + b2

Calculation:

We have,

⇒ \(\frac{x}{a} = \frac{y}{b}\)     ---- (1)

⇒ ax + by = a2 + b2    ---- (2)

⇒ y = \(\frac{bx}{a}\)     [from (1)]

Substituting the value of y in equation (2), we get

⇒ \(ax + \frac{b \times bx}{a} = a^2 + b^2\)

⇒ a2x + b2x = (a2 + b2)a

⇒ x = \(\frac{(a^2\ + \ b^2) \times a }{a ^2 \ +\ b^2}\)

⇒ x = a

Now, put x = a in equation (1), we get

⇒ \(\frac{a}{a} = \frac{y}{b}\)

⇒ y = b

∴ The value of x is a and the value of y is b.

Systems of Linear Equations MCQ Question 2:

The system of linear equations in real (x, y) given by

\(\rm \begin{pmatrix} \rm x & \rm y \end{pmatrix} \begin{bmatrix} 2 & 5- 2 α \\\ α & 1 \end{bmatrix} = \rm \begin{pmatrix} \rm 0 & \rm 0 \end{pmatrix} \)

involves a real parameter α and has infinitely many non-trivial solutions for special value(s) of α. Which one or more among the following options is/are non-trivial solution(s) of (x, y) for such special value(s) of α ?

1. x = 2, y = −2
2. x = −1, y = 4
3. x = 1, y = 1
4. x = 4, y = −2

Answer (Detailed Solution Below)

Option :

Concept:

Solution of homogeneous equation AX = 0:

Consistent Infinite many solutions:  if the rank of a matrix is less than the number of variables then it has infinitely many solutions.

rank(r) < no. of variables(n) ⇒ |A| = 0

Calculation:

Given:

\(\rm \begin{pmatrix} \rm x & \rm y \end{pmatrix} \begin{bmatrix} 2 & 5- 2 α \\\ α & 1 \end{bmatrix} = \rm \begin{pmatrix} \rm 0 & \rm 0 \end{pmatrix} \)

for non-trivial infinite solution: |A| = 0

A = \(\begin{bmatrix} 2 & 5- 2 α \\\ α & 1 \end{bmatrix}\) ⇒ |A| = 2 - α(5 - 2α) = 0

2α2 - 5α + 2 = 0

(α - 2) (2α - 1) = 0

α = 2, 1/2

For α = 2 

A = \(\begin{bmatrix} 2 & 1 \\\ 2 & 1 \end{bmatrix}\)

\(\rm \begin{pmatrix} \rm x & \rm y \end{pmatrix} \begin{bmatrix} 2 & 1 \\\ 2 & 1 \end{bmatrix} = \rm \begin{pmatrix} \rm 0 & \rm 0 \end{pmatrix} \)

2x + 2 y = 0

x = -y

Substitute option in this equation.

For α = 1/2

A = \(\begin{bmatrix} 2 & 4 \\\ \frac{1}{2} & 1 \end{bmatrix} \)

\(\rm \begin{pmatrix} \rm x & \rm y \end{pmatrix} \begin{bmatrix} 2 & 4 \\\ \frac{1}{2} & 1 \end{bmatrix} = \rm \begin{pmatrix} \rm 0 & \rm 0 \end{pmatrix} \)

2x + 1/2 y = 0 

4x + y = 0 

from both equations we get,

y = - 4x

Substitute options in equation.

We get, 

option 1. x = 2, y = −2 and

option 2. x = −1, y = 4  satisfies equations.

Systems of Linear Equations MCQ Question 3:

The equation of the curve passing through (1, 0) and Satisfying the differential equation

(1 + y2) dx - xydy = 0 is -

1. x2 + y2 = 1
2. x2 - y2 = 1
3. y2 = 4x
4. 2x2 + y2 = 2

Answer (Detailed Solution Below)

Option 2 : x2 - y2 = 1

Given a differential equation,

(1 + y2) dx - xydy = 0

or, (1 + y2) dx = xydy

It can be written as,

\(\frac{(dx)}{x}=\frac{y\ dy}{(1+y^2)}\)

Integrating both sides,

ln (x) = \(\frac{1}{2}ln\ (1+y^2)+ln \ C\)

or, 2 ln (x) = ln (1 + y2) + 2 ln (C)

or, ln (x2) = ln [(1 + y2) (C2)]

Let, C2 = K

or, x2 = K (1 + y2) .... (1)

Since, the curve passing through (1, 0),

1 = K (1 + 0)

or, K = 1

From equation (1),

x2 = (1 + y2)

Hence,

x2 - y2 = 1

Systems of Linear Equations MCQ Question 4:

The pair of linear equations kx + 2y = 5 and 3x + y = 1 has a unique solution if:

1. k = 0
2. k = 6
3. k ≠ 6
4. k has any value

Answer (Detailed Solution Below)

Option 3 : k ≠ 6

Given:

The given equations are

kx + 2y = 5

3x + y = 1

Concept used:

If (a1/a2) ≠ (b1/b2), then the system of equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 has a unique solution.

Calculation:

kx + 2y - 5 = 0      ----(1)

a1 = k, b1 = 2, c1 = -5

3x + y - 1 = 0      ----(2)

a2 = 3, b2 = 1, c2 = -1

According to the question,

(a1/a2) ≠ (b1/b2)

⇒ (k/3) ≠ (2/1)

⇒ 6 ≠ k

⇒ k ≠ 6

∴ The system has a unique solution if k ≠ 6.

Systems of Linear Equations MCQ Question 5:

The value of k for which kx + 3y - k + 3 = 0 and 12x + ky = k have infinite solution is:

1. -6
2. 0
3. 6
4. 1

Answer (Detailed Solution Below)

Option 3 : 6

Concept:

Let the two equations be:

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Then,

• For unique solution, \(\frac{a_{1}}{a_{2}}≠ \frac{b_{1}}{b_{2}}\)
• For infinitely many solutions, \(\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)

• For no solution, \(\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq\frac{c_{1}}{c_{2}}\)

Formula used:

For any quadratic equation, ax2 + bx + c = 0, the quadratic formula is:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Calculation:

Given equations are:

kx + 3y - k + 3 = 0 and 12x + ky = k 

On comparing the equations with ax + by + c = 0, we get

a1 = k, b1 = 3, c1 = -k + 3

a2 = 12, b2 = k, c2 = -k

So, for infinitely many solutions,

\(\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)

\(⇒ \frac{k}{(12)}= \frac{3}{k}=\frac{-k+3}{-k}\)

Solving, \(\frac{k}{12}=\frac{3}{k}\)

k2 = 36

⇒ k = 6

Solving, \( \frac{3}{k}=\frac{-k+3}{-k}\)

⇒ -3k = -k2 + 3k

⇒ k2 - 6k = 0

⇒ k2 = 6k

k = 6

Hence, k = 6.

If the following system has non-trivial solution,

px + qy + rz = 0

qx + ry + pz = 0

rx + py + qz = 0,

then which one of the following options is TRUE?

1. p – q + r = 0 or p = q = -r
2. p + q – r = 0 or p = -q = r
3. p + q + r = 0 or p = q = r
4. p – q + r = 0 or p = -q = -r

Answer (Detailed Solution Below)

Option 3 : p + q + r = 0 or p = q = r

Concept:

If rank of 3 × 3 homogeneous matrix is less than 3 then the corresponding equations will have non-trivial solutions.

Explanation:

For non-trivial solution

\(\left| {\begin{array}{*{20}{c}} p&q&r\\ q&r&p\\ r&p&q \end{array}} \right| = 0\)

R1 = R1 + R2 + R3

\(\left| {\begin{array}{*{20}{c}} {p + q + r}&{\;q + r + p}&{r + p + q}\\ q&r&p\\ r&p&q \end{array}} \right| = 0\)

\(\left( {p + q + r} \right)\left| {\begin{array}{*{20}{c}} 1&{\;1}&1\\ q&r&p\\ r&p&q \end{array}} \right| = 0\)

∴ p + q + r = 0

(OR)

\(\left| {\begin{array}{*{20}{c}} 1&{\;1}&1\\ q&r&p\\ r&p&q \end{array}} \right| = 0\)

∴ p = q = r

The system of equations

x + 2y + z = 0

x – z = 0

x + y = 0

has

1. a trivial solution only
2. no solution
3. infinitely many solutions
4. non-trivial unique solution

Answer (Detailed Solution Below)

Option 3 : infinitely many solutions

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 contain 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.

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

(i) If the rank of matrix A is equal to the number of unknowns, then the system has only a trivial zero solution.

The rank of A = n

(ii) If the rank of matrix A is less than the number of unknowns, then the system has an infinite number of solutions.

The rank of A < n

Calculation:

Given:

The given system of equations can be represented in a matrix form as shown below.

\(A = \left[ {\begin{array}{*{20}{c}} 1&{ 2}&{ 1}\\ { 1}&0&{ -1}\\ { 1}&{ 1}&0 \end{array}} \right],\;X = \left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right],\;B = \left[ {\begin{array}{*{20}{c}} 0\\ 0\\ 0 \end{array}} \right] \)

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

C1 → C1 + C3,

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

C1 = C2, Therefore |A|3× 3 = 0, Rank ≠ 3

Determinant of the matrix order 2 ≠ 0, Therefore rank of matrix A = 2

Rank of matrix A = 2 < n = 3

As rank of matrix is less than the variables (n). Hence the given system of linear equations has infinitely many solutions.

The value of k for which kx + 3y - k + 3 = 0 and 12x + ky = k have infinite solution is:

1. -6
2. 0
3. 6
4. 1

Answer (Detailed Solution Below)

Option 3 : 6

Concept:

Let the two equations be:

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Then,

• For unique solution, \(\frac{a_{1}}{a_{2}}≠ \frac{b_{1}}{b_{2}}\)
• For infinitely many solutions, \(\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)

• For no solution, \(\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq\frac{c_{1}}{c_{2}}\)

Formula used:

For any quadratic equation, ax2 + bx + c = 0, the quadratic formula is:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Calculation:

Given equations are:

kx + 3y - k + 3 = 0 and 12x + ky = k 

On comparing the equations with ax + by + c = 0, we get

a1 = k, b1 = 3, c1 = -k + 3

a2 = 12, b2 = k, c2 = -k

So, for infinitely many solutions,

\(\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)

\(⇒ \frac{k}{(12)}= \frac{3}{k}=\frac{-k+3}{-k}\)

Solving, \(\frac{k}{12}=\frac{3}{k}\)

k2 = 36

⇒ k = 6

Solving, \( \frac{3}{k}=\frac{-k+3}{-k}\)

⇒ -3k = -k2 + 3k

⇒ k2 - 6k = 0

⇒ k2 = 6k

k = 6

Hence, k = 6.

The following simultaneous equations:

x + y + z = 3

x + 2y + 3z = 4

x + 4y + kz = 6

Will not have a unique solution for k equal to

1. 7
2. 6
3. 0
4. 5

Answer (Detailed Solution Below)

Option 1 : 7

Concept:

For unique solution:

Determinant of A, |A|≠ 0

where A is any matrix

Calculation:

\(D = \begin{vmatrix} 1 &1 &1 \\ 1&2&3 \\ 1 &4 &k \end{vmatrix}\)

|A| ≠ 0

⇒ 1 (2k - 12) - (k - 3) + 1 (4 - 2) = 0

⇒ k - 7 = 0

⇒ k = 7

So, The following simultaneous equations will not have a unique solution for k equal to 7

Consider the following system of equations:

3x + 2y = 1

4x + 7z = 1

x + y + z = 3

x – 2y + 7z = 0

The number of solutions for this system is _________

Answer (Detailed Solution Below) 1

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.

Explanation:

Adding first two equations, we get the set equations as,

7x + 2y + 7z = 2

x + y + z = 3

x – 2y + 7z = 0

Augmented matrix will look like,

\(\begin{array}{*{20}{c}} 7&2&7\\ 1&1&1\\ 1&{ - 2}&7 \end{array}\;\left| {\;\;\begin{array}{*{20}{c}} 2\\ 3\\ 0 \end{array}\;} \right|\)

R1 → R1 – 7R2

\(\begin{array}{*{20}{c}} 0&{ - 5}&0\\ 1&1&1\\ 1&{ - 2}&7 \end{array}\;\left| {\;\;\begin{array}{*{20}{c}} { - 19}\\ 3\\ 0 \end{array}\;} \right|\)

R3 → R3 – R2

\(\begin{array}{*{20}{c}} 0&{ - 5}&0\\ 1&1&1\\ 0&{ - 3}&6 \end{array}\;\left| {\;\;\begin{array}{*{20}{c}} { - 19}\\ 3\\ { - 3} \end{array}\;} \right|\)

R3 → 5R3 – 3R1

\(\begin{array}{*{20}{c}} 0&{ - 5}&0\\ 1&1&1\\ 0&0&{30} \end{array}\;\left| {\;\;\begin{array}{*{20}{c}} { - 19}\\ 3\\ {42} \end{array}\;} \right|\)

No further reductions are going to be useful.

Hence, Rank of Augmented matrix = Rank of Coefficient matrix = Number of variables = 3.

No of solutions = 1.

Alternate Method:

\(\left| {\begin{array}{*{20}{c}} 7&2&7\\ 1&1&1\\ 1&{ - 2}&7 \end{array}} \right| = 30 \ne 0\)

There Rank of Augmented matrix = Rank of Coefficient matrix = 3 and hence it has unique solution

Consider the following system of equations

2x1 + x2 + x3 = 0,

x2 – x3 = 0,

x1 + x2 = 0,

This system has

1. A unique solution
2. No solution
3. Infinite number of solutions
4. Five solutions

Answer (Detailed Solution Below)

Option 3 : Infinite number of solutions

Concept:

For a Homogenous system, AX = O

[A] is the Coefficient matrix

[A/O] be Augmented matrix

[O] is a null matrix and

n = total number of variables

Case 1: ρ(A) = ρ(A/O) = n

In this case, the system possesses only a zero solution (or Trivial solution) i.e unique solution.

Case 2: ρ(A) = ρ(A/O) < n

In this case, the system has an infinite number of non-zero solutions (or Trivial solutions).

Case 3: ρ(A) = ρ(A/O)

Hence, inconsistency does not arise, moreover, zero solution is always a solution to it.

Calculation:

Given:

2x1 + x2 + x3 = 0

x2 – x3 = 0

x1 + x2 = 0

Here n = 3

Now, we know that

For a Homogenous system, AX = O

Augmented matrix is:

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

R3 → R3 - (R1/2)

\(\left[ {\left. {\begin{array}{*{20}{c}} 2&1&1\\ 0&1&{ - 1}\\ 0&{1/2}&{-1/2} \end{array}} \right|\begin{array}{*{20}{c}} 0\\ 0\\ 0 \end{array}} \right]\)

R3 → R3 - (R2/2)

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

As, ρ(A) = ρ(A/O) = 2 < 3

∴ The system is consistent and will have infinite number of solutions.

The value of λ for which the system of equations 2x - y - z = 2, x - 2y + z = -4, x + y + λz = 4 has no solution, is:

1. 0
2. -2
3. 3
4. -3

Answer (Detailed Solution Below)

Option 2 : -2

Given:

2x - y - z = 2,

x - 2y + z = -4,

x + y + λz = 4

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.

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

Rank of A ≠ Rank of augmented matrix. Therefore if det (A) = 0 system is inconsistent (no solution)

Calculation:

\( A = \left[ {\begin{array}{*{20}{c}} 2&{ - 1}&{ - 1}\\ 1&{ - 2}&1\\ 1&1&λ \end{array}} \right]\;\;B = \left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right]\;C = \left[ {\begin{array}{*{20}{c}} 2\\ { - 4}\\ 4 \end{array}} \right] \)

Hence, for no solution, |A| = 0

\( ⇒ \left| {\begin{array}{*{20}{c}} 2&{ - 1}&{ - 1}\\ 1&{ - 2}&1\\ 1&1&λ \end{array}} \right|\; = 0 \)

⇒ 2(-2λ - 1) + 1(λ - 1) - 1(1 + 2) = 0

⇒ -4λ - 2 + λ - 1 - 3 = 0

⇒ -3λ = 6

⇒ λ  = - 2

• If the rank of matrix A is equal to the rank of an augmented matrix and it is equal to a number of unknowns, then system is consistent and there is a unique solution.
• Rank of A = Rank of augmented matrix = n
• If the rank of matrix A is equal to rank of augmented matrix and it is less than the number of unknowns, then system is consistent and there are infinite number of solutions.
• Rank of A = Rank of augmented matrix < n

If the following equations gives non-trivial, then value of k is
4x – 3y + 1 = 0
kx – 8y + 10 = 0
x + y – 5 = 0

1. -6
2. 7
3. 6
4. -7

Answer (Detailed Solution Below)

Option 2 : 7

Concept: 

If the system of equation a1x + b1y + c1 = 0,  a2x + b2y + c2 = 0, a3x + b3y + c3 = 0 be non-trivial, then

\(\begin{vmatrix} a_1 &b_1 & c_1\\ a_2 &b_2 & c_2 \\ a_3 &b_3 & c_3 \end{vmatrix} = 0\)

Calculation:

Given:

\(\begin{vmatrix} 4 &-3 & 1\\ k&-8 & 10 \\ 1 &1 & -5 \end{vmatrix} = 0\)

⇒ 4(40 - 10) - (-3)(-5k - 10) + 1(k + 8) = 0

⇒ 120 - 15k - 30 + k + 8 = 0

⇒ 98 - 14k = 0

∴ k = 7

a11 x1 + a12 x2 + … + a1n xn = b1

a21 x1 + a22 x2 + … + a2n xn = b2

…

am1 x1 + am2 x2 + … + amn xn = bm

The above equations contain 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

|A| = 0

(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

The pair of linear equations in two variables x and y, \({29x+37y+7 = 0}\atop{87x + 74y + 5 = 0}\) will have:

1. infinite solution
2. trivial solution
3. unique solution
4. no solution

Answer (Detailed Solution Below)

Option 3 : unique solution

Concept:

Let us consider the system of linear equations:

a11 × x + a12 × y = b1

a21 × x + a22 × y = b2

We can write these equations in matrix form as: A X = B,

where \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right],\;X = \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right]\;and\;B = \left[ {\begin{array}{*{20}{c}} {{b_1}}\\ {{b_2}} \end{array}} \right]\)

In order to say that the given system of linear equations is consistent and has a unique solution, |A| ≠ 0.

Analysis:

89x + 37y = -7

87x + 74y = -5

\(A = \left[ {\begin{array}{*{20}{c}} {29}&{37} \\ {87}&{74} \end{array}} \right]\;\;\;B = \left[ {\begin{array}{*{20}{c}} { - 7} \\ { - 5} \end{array}} \right]\)

|A| = 29 × 74 – 87 × 37

= -1073 ≠ 0

∴ Solution is unique

The system of algebraic equations given below has

x  + 2y + z = 4

2x + y + 2z = 5

x – y + z = 1

1. A unique solution of x = 1, y = 1 and z = 1
2. Only the two solutions of (x = 1, y = 1 and z = 1) and (x = 2, y = 1 and z = 0)
3. Infinite number of solutions
4. No feasible solution

Answer (Detailed Solution Below)

Option 3 : Infinite number of solutions

Concept:

Let

[A] is the Coefficient matrix

[A/B] be Augmented matrix

n = total number of variables

Case 1: ρ(A) = ρ(A/B) = n

In this case, the system will be consistent and will have a unique solution.

Case 2: ρ(A) = ρ(A/B) < n

In this case, the system will be consistent and will have infinite solutions.

Case 3: ρ(A) < ρ(A/B)

In this case, the system will be inconsistent and will have no solution.

Calculation:

Given:

x  + 2y + z = 4

2x + y + 2z = 5

x – y + z = 1

Here n = 3

Augmented matrix is:

\(\left[ {A/B} \right] = \left[ {\left. {\begin{array}{*{20}{c}} 1&2&1\\ 2&1&2\\ 1&{ - 1}&1 \end{array}} \right|\begin{array}{*{20}{c}} 4\\ 5\\ 1 \end{array}} \right]\)

R2 → R2 – 2R1 and R3 → R3 – R1

\(\left[ {A/B} \right] = \left[ {\left. {\begin{array}{*{20}{c}} 1&2&1\\ 0&{ - 3}&0\\ 0&{ - 3}&0 \end{array}} \right|\begin{array}{*{20}{c}} {\;\;\;4}\\ { - 3}\\ { - 3} \end{array}} \right]\)

R3 → R3 – R2

\($\left[ {A/B} \right] = \left[ {\left. {\begin{array}{*{20}{c}} 1&2&1\\ 0&{ - 3}&0\\ 0&0&0 \end{array}} \right|\begin{array}{*{20}{c}} {\;\;4}\\ { - 3}\\ {\;\;0} \end{array}} \right]\)

ρ(A) = ρ(A/B)  = 2 < 3

∴ system will be consistent and will have infinite solutions.

While testing by Cramer’s Rule, we find that the system 3x + 6y = 9, 6x + 12y = 18 has

1. No solution
2. A unique solution
3. Infinite number of solutions
4. Zero

Answer (Detailed Solution Below)

Option 3 : Infinite number of solutions

We know that,

a1x + b1y = c1

a2x + b2y = c2 

\(Let\;\Delta = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}\\ {{a_2}}&{{b_2}} \end{array}} \right|and\;\;{\Delta _1} = \left| {\begin{array}{*{20}{c}} {{c_1}}&{{b_1}}\\ {{c_2}}&{{b_2}} \end{array}} \right|\)

So,

\(\begin{array}{l} \Delta = \left| {\begin{array}{*{20}{c}} 3&6\\ 6&{12} \end{array}} \right| = 0\\ \Delta = 36 - 36 = 0\\ {\Delta _1}\; = \left| {\begin{array}{*{20}{c}} 9&6\\ {18}&{12} \end{array}} \right|\\ {\Delta _1}\; = 108 - 108 = 0 \end{array}\)  

We know that the result will be infinite number of solutions, if

\(\Delta = 0,\;\;{\Delta _1} = 0\)

Thus, the given system has infinite number of solutions.​​​

Consider the systems, each consisting of m linear equations in n variables.

I. If m < n, then all such systems have a solution

II. If m > n, then none of these systems has a solution

III. If m = n, then there exists a system which has a solution

Which one of the following is CORRECT? 

1. I, II and III are true
2. Only II and III are true
3. Only III is true
4. None of them is true

Answer (Detailed Solution Below)

Option 3 : Only III is true

Statement I:

If m < n, then all such systems have a solution [incorrect]

Let us suppose m = 2, n = 3

x + 2y + z = 3

x + y + z = 3

here, this will not give any solution. Because, when we have 2 equations with 3 variables, we can’t find the solution for this.

Statement II:

If m > n, then none of these systems has a solution [incorrect]

Consider, m = 3, n = 2

System of equation will be like:

x + 2y = 2

x + y = 1

2x + 5y = 3

But here we can easily find the value of x and y.

Statement III:

If m = n, then there exists a system which has a solution [Correct]

Consider m = 2, n = 2

System of equation will be like:

x + 2y = 3

2x + 4y = 4

Here, x and y can be calculated. These systems of equations have a solution.

Let c1 … cn be scalars, not all zero, such that \(\mathop \sum \limits_{i = 1}^n {c_i}{a_i} = 0\) where ai are column vectors in Rn. Consider the set of linear equations

Ax = b

where A = \(\left[ {{a_1} \ldots {a_n}} \right]\;and\;b = \mathop \sum \limits_{i = 1}^n {a_i}\) The set of equations has

1. a unique solution at x = Jn where Jn denotes a n-dimensional vector of all 1
2. no solution
3. infinitely many solutions 
4. finitely many solutions

Answer (Detailed Solution Below)

Option 3 : infinitely many solutions 

Explanation:

c1, c2, … cn are scalars i.e. not all zero. \(\mathop \sum \limits_{i = 1}^n {c_i}{a_i} = 0\). Here ci is scalar and ai vector.

And when a scalar multiplied with vector, result will be zero, scalar value will be zero.

c1a1 + c2 a2 + … + cn an = 0 So, here ci value will be 0.

Here we can say that A is linearly dependent, and determinant of A is 0. (Linearly Dependent: Vectors being linearly independent means they represent independent directions in your vector spaces, while linearly dependent vectors mean they don’t. So, for example if you have a set of vectors {v1, v2, v3, v4, v5} and you can walk some distance in the v1 direction, then a difference distance in v2 and then in direction v3. If in the end, you are back at the point where you started then the vectors are linearly dependent otherwise not). A= [a1, a2, a3, … an]

b = \(\mathop \sum \limits_{i = 1}^n {a_i}\)

Ax = b

a1x1 + a2 x2 + … + an xn = a1 + a2 + … + an

x1 = x2 = x3 = … = xn = 1 (It is one of the infinite solutions) and all other solution are on that same line.

So, the answer will be infinitely many solutions.

The number of values of k for which the system of equations x + y = 2, kx + y = 4, x + ky = 5 has atleast one solution is:

1. 1
2. 0
3. 3
4. 2

Answer (Detailed Solution Below)

Option 1 : 1

Concept:

For at least one solution solve the given equation for the unique value.

Given:

x + y = 2   ------ (1) 

kx + y = 4   ------ (2) 

x + ky = 5   ------ (3)

Calculation:

After adding equation (2) & (3),

kx + y + x + ky = 4 + 5

k(x + y) + (x + y) = 9

(k + 1)(x + y) = 9   ------ (4)

And, We have 

x + y = 2 

From equation (4)

(k + 1) × 2 = 9

k = 3.5

Then,

For the unique solution K have only one value.

Let us consider a system of equations in three variables:

a1 × x + b1 × y + c1 × z = d1

a2 × x + b2 × y + c2 × z = d2

a3 × x + b3 × y + c3 × z = d3

Then, \({\rm{\Delta }} = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right|,\;\;{{\rm{\Delta }}_1} = \left| {\begin{array}{*{20}{c}} {{d_1}}&{{b_1}}&{{c_1}}\\ {{d_2}}&{{b_2}}&{{c_2}}\\ {{d_3}}&{{b_3}}&{{c_3}} \end{array}} \right|,\;{{\rm{\Delta }}_2} = \;\left| {\begin{array}{*{20}{c}} {{a_1}}&{{d_1}}&{{c_1}}\\ {{a_2}}&{{d_2}}&{{c_2}}\\ {{a_3}}&{{d_3}}&{{c_3}} \end{array}} \right|\;and\;{{\rm{\Delta }}_3} = \;\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{d_1}}\\ {{a_2}}&{{b_2}}&{{d_2}}\\ {{a_3}}&{{b_3}}&{{d_3}} \end{array}} \right|\)

By cramer’s rule:

• If Δ ≠ 0, then the system of the equation has a unique solution and it is given by: \(x = \frac{{{{\rm{\Delta }}_1}}}{{\rm{\Delta }}},\;y = \frac{{{{\rm{\Delta }}_2}}}{{\rm{\Delta }}}\;and\;z = \frac{{{{\rm{\Delta }}_3}}}{{\rm{\Delta }}}\)
• If Δ = 0 and at least one of the determinants Δ, Δ1, Δ2, and Δ3 is non-zero, then the given system is inconsistent.
• If Δ = 0 and Δ1 =  Δ2 =  Δ3 = 0, then the system is consistent and has infinitely many solutions.

 If 𝑎 and 𝑏 are constants, the most general solution of the differential equation\(\frac{{{d^2}x}}{{d{t^2}}} + 2\frac{{dx}}{{dt}} + x = 0\)

1. ae-t
2. ae-t + bte-t
3. aet + bte-t
4. ae-2t

Answer (Detailed Solution Below)

Option 2 : ae-t + bte-t

Concept:

Differential equation: It is an equation which involves differential coefficients or differentials.

The solution of differential equations consists of two parts Complementary function (C.F) and Particular Integral (P.I)

To solve the linear second-order differential equation which is in a general form as shown:

  \(\frac{{{d^n}y}}{{d{x^n}}} + {k_1}\frac{{{d^{n - 1}}y}}{{d{x^{n - 1}}}} + {k_2}\frac{{{d^{n - 2}}y}}{{d{x^{n - 2}}}} + \cdots + {k_n}y = 0\)

 this equation in symbolic form is

\(\left( {{D^n} + {k_1}{D^{n - 1}} + \cdots + {k_n}} \right)y = 0\)

We assume operator ‘D’ as d/dx and form auxiliary Equation.

To find the roots of the differential equation, make symbolic coefficient to zero.

  \({D^n} + {k_1}{D^{n - 1}} + \cdots + {k_n} = 0\) is called auxiliary equation and (A.E.)

Let m1, m2, ⋯ , mn be its roots.

There are four different types of roots.

1. If roots are different m1 ≠ m2 then solution is of the form

\({c_1}{e^{{m_1}x}} + {c_2}{e^{{m_2}x}} \)

2. If the roots are equal, m1 = m2 = m

\(\left( {{c_1} + {c_2}} \right){e^{mx}} \)

3. If the roots are complex and not equal m1 = α + ι β and m2 = α - ι β

\({e^{\alpha x}}\left( {{C_1}cos\beta x + {C_2}sin\beta x} \right) \)

4. If roots are complex and equal, i.e, m1 = m2 = α + ι β and m3 = m4 = α - ι β

\({e^{\alpha x}}\left( {\left( {{c_1}x + {c_2}} \right)cos\beta x + \left( {{c_3}x + {c_4}} \right){\bf{cos\beta x}}} \right) \)

P.I part is also found in different methods based on the X function which is R.H.S of differential function.

Calculation:

Given DE is \(\frac{{{d^2}x}}{{d{t^2}}} + 2\frac{{dx}}{{dt}} + x = 0\) 

A.E is D2 + 2D + 1 = 0 and roots of AE are (D+1)(D+1) = 0

so roots are equal and m = -1

∴ from the above table we can write the solution for given equation as

 \(\left( {a + bt} \right){e^{mt}} = \left( {a + bt} \right){e^{ - t}}\)

Consider the following system of equations in three real variables x1, x2 and x3

2x1 – x2 + 3x3 = 1

3x2 – 2x2 + 5x3 = 2

-x1 – 4x2 + x3 = 3

The system of equations has:

1. No solutions
2. A unique solution
3. More than one but a finite number of solutions
4. An infinite number of solutions

Answer (Detailed Solution Below)

Option 2 : A unique solution

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 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 as augmented matrix of the given system of equations.

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

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

Rank of A = Rank of augmented matrix = n

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

Rank of A = Rank of augmented matrix < n

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

Rank of A ≠ Rank of augmented matrix

Calculation:

The given system of equations can be represented in a matrix form as shown below.

\(A = \left[ {\begin{array}{*{20}{c}} 2&{ - 1}&3\\ 3&{ - 2}&5\\ { - 1}&{ - 4}&1 \end{array}} \right],\;X = \left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right],\;B = \left[ {\begin{array}{*{20}{c}} 1\\ 2\\ 3 \end{array}} \right]\)

The Augmented matrix can be written by

\(\left[ {A|B} \right] = \left[ {\left. {\begin{array}{*{20}{c}} 2&{ - 1}&3\\ 3&{ - 2}&5\\ { - 1}&{ - 4}&1 \end{array}} \right|\begin{array}{*{20}{c}} 1\\ 2\\ 3 \end{array}} \right]\)

R3 → 2R3 + R1

\(= \left[ {\left. {\begin{array}{*{20}{c}} 2&{ - 1}&3\\ 3&{ - 2}&5\\ 0&{ - 9}&5 \end{array}} \right|\begin{array}{*{20}{c}} 1\\ 2\\ 7 \end{array}} \right]\)

R2 → 2R2 - 3R1

\(= \left[ {\left. {\begin{array}{*{20}{c}} 2&{ - 1}&3\\ 0&{ - 1}&1\\ 0&{ - 9}&5 \end{array}} \right|\begin{array}{*{20}{c}} 1\\ 1\\ 7 \end{array}} \right]\)

R3 → R3 - 9R2

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

Rank of matrix A = Rank of Augmented matrix = 3

Hence the given system has unique solution.

The equation of the curve passing through (1, 0) and Satisfying the differential equation

(1 + y2) dx - xydy = 0 is -

1. x2 + y2 = 1
2. x2 - y2 = 1
3. y2 = 4x
4. 2x2 + y2 = 2

Answer (Detailed Solution Below)

Option 2 : x2 - y2 = 1

Given a differential equation,

(1 + y2) dx - xydy = 0

or, (1 + y2) dx = xydy

It can be written as,

\(\frac{(dx)}{x}=\frac{y\ dy}{(1+y^2)}\)

Integrating both sides,

ln (x) = \(\frac{1}{2}ln\ (1+y^2)+ln \ C\)

or, 2 ln (x) = ln (1 + y2) + 2 ln (C)

or, ln (x2) = ln [(1 + y2) (C2)]

Let, C2 = K

or, x2 = K (1 + y2) .... (1)

Since, the curve passing through (1, 0),

1 = K (1 + 0)

or, K = 1

From equation (1),

x2 = (1 + y2)

Hence,

x2 - y2 = 1

If a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 have infinite number of solutions, then which one of the following is correct:

1. \(\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}\)
2. \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}\)
3. \(\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\)
4. \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\)

Answer (Detailed Solution Below)

Option 4 : \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\)

Explanation:

If both equation a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 have the infinite solution, then the graph will be the pair of coincidents line, and the relation between the coefficient of both the equation is 

\(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\)

A = [aij]m × n = coefficient of matrix, X = Column matrix of the variables

B = column matrix of constants

The system AX = B has

1) A unique solution, If Rank of A = Rank [A|B] and is equal to the number of variables.

2) Infinitely many solutions, If Rank of A = Rank of [A|B] < number of variables

3) No solution, If Rank of A ≠ Rank of [A|B], i.e. Rank of A < Rank of [A|B].

The system of linear equations in real (x, y) given by

\(\rm \begin{pmatrix} \rm x & \rm y \end{pmatrix} \begin{bmatrix} 2 & 5- 2 α \\\ α & 1 \end{bmatrix} = \rm \begin{pmatrix} \rm 0 & \rm 0 \end{pmatrix} \)

involves a real parameter α and has infinitely many non-trivial solutions for special value(s) of α. Which one or more among the following options is/are non-trivial solution(s) of (x, y) for such special value(s) of α ?

1. x = 2, y = −2
2. x = −1, y = 4
3. x = 1, y = 1
4. x = 4, y = −2

Answer (Detailed Solution Below)

Option :

Concept:

Solution of homogeneous equation AX = 0:

Consistent Infinite many solutions:  if the rank of a matrix is less than the number of variables then it has infinitely many solutions.

rank(r) < no. of variables(n) ⇒ |A| = 0

Calculation:

Given:

\(\rm \begin{pmatrix} \rm x & \rm y \end{pmatrix} \begin{bmatrix} 2 & 5- 2 α \\\ α & 1 \end{bmatrix} = \rm \begin{pmatrix} \rm 0 & \rm 0 \end{pmatrix} \)

for non-trivial infinite solution: |A| = 0

A = \(\begin{bmatrix} 2 & 5- 2 α \\\ α & 1 \end{bmatrix}\) ⇒ |A| = 2 - α(5 - 2α) = 0

2α2 - 5α + 2 = 0

(α - 2) (2α - 1) = 0

α = 2, 1/2

For α = 2 

A = \(\begin{bmatrix} 2 & 1 \\\ 2 & 1 \end{bmatrix}\)

\(\rm \begin{pmatrix} \rm x & \rm y \end{pmatrix} \begin{bmatrix} 2 & 1 \\\ 2 & 1 \end{bmatrix} = \rm \begin{pmatrix} \rm 0 & \rm 0 \end{pmatrix} \)

2x + 2 y = 0

x = -y

Substitute option in this equation.

For α = 1/2

A = \(\begin{bmatrix} 2 & 4 \\\ \frac{1}{2} & 1 \end{bmatrix} \)

\(\rm \begin{pmatrix} \rm x & \rm y \end{pmatrix} \begin{bmatrix} 2 & 4 \\\ \frac{1}{2} & 1 \end{bmatrix} = \rm \begin{pmatrix} \rm 0 & \rm 0 \end{pmatrix} \)

2x + 1/2 y = 0 

4x + y = 0 

from both equations we get,

y = - 4x

Substitute options in equation.

We get, 

option 1. x = 2, y = −2 and

option 2. x = −1, y = 4  satisfies equations.

The pair of linear equations kx + 2y = 5 and 3x + y = 1 has a unique solution if:

1. k = 0
2. k = 6
3. k ≠ 6
4. k has any value

Answer (Detailed Solution Below)

Option 3 : k ≠ 6

Given:

The given equations are

kx + 2y = 5

3x + y = 1

Concept used:

If (a1/a2) ≠ (b1/b2), then the system of equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 has a unique solution.

Calculation:

kx + 2y - 5 = 0      ----(1)

a1 = k, b1 = 2, c1 = -5

3x + y - 1 = 0      ----(2)

a2 = 3, b2 = 1, c2 = -1

According to the question,

(a1/a2) ≠ (b1/b2)

⇒ (k/3) ≠ (2/1)

⇒ 6 ≠ k

⇒ k ≠ 6

∴ The system has a unique solution if k ≠ 6.