System of Linear Equations MCQ Quiz - Objective Question with Answer for System of Linear Equations - Download Free PDF

Explanation:

Augmented matrix will be

\(\left[ {\begin{array}{*{20}{c}} 2&{ - 3}&7&5\\ 3&1&{ - 3}&{13}\\ 2&{19}&{ - 47}&{32} \end{array}} \right] \)

R₃ → R₃ – R₁

R₂ → 2R₂ – 3R₁

\(\left[ {\begin{array}{*{20}{c}} 2&{ - 3}&7&5\\ 0&{11}&{ - 27}&{11}\\ 0&{22}&{ - 54}&{27} \end{array}} \right]\)

R₃ → R₃ – 2R₂

\(\left[ {\begin{array}{*{20}{c}} 2&{ - 3}&7&5\\ 0&{11}&{ - 27}&{11}\\ 0&0&0&5 \end{array}} \right]\)

Rank of A ≠ Rank of Augmented matrix.

Hence given system of equations has no solution.

Hence Option(5) is the correct answer.

Explanation:

Consider the augmented matrix  

\([A|b] = \begin{bmatrix} 1 & 1 & 5 & 3 \\ 1 & 2 & m & 5 \\ 1 & 4 & k & 5 \end{bmatrix} \)  

Applying  row operations: 
 

\(R_3 \to R_3 - R_1, \quad R_2 \to R_2 - R_1 \)

\(\begin{bmatrix} 1 & 1 & 5 & 3 \\ 0 & 1 & m-5 & 2 \\ 0 & 3 & k-5 & 2 \end{bmatrix} \)  

\( R_3 \to R_3 - R_2 : \)  

\(\begin{bmatrix} 1 & 1 & 5 & 3 \\ 0 & 1 & m-5 & 2 \\ 0 & 0 & 4 - m(k - 5) & -5 \end{bmatrix}\)  

The system is consistent if and only if  

\(\rho([A|b]) = \rho(A) \)

If  \(m \neq 4\)  , then  \(\rho([A|b]) = \rho(A) = 3 \)  for any value of k 

Hence, option (1) is true

If k = 5 , then  \(\rho([A|b]) = \rho(A) \)  for any value of m

Hence, option (4) is true.  

If  \(k \neq 5 \)  and m = 4 , then \(\rho([A|b]) = 3 > \rho(A) = 2 \) ,   

and the system will be inconsistent.  

Hence, options (1) and (4) are  correct.

Concept:

Consistency means, the system of equations should have either infinite solutions or unique solution

Explanation:

For Consistent solution, the system of equations should have either infinite solutions or unique solution

\(\left[ {\begin{array}{*{20}{c}} 1&2&{ - 3}&:&p\\ 2&6&{ - 11}&:&q\\ 1&{ - 2}&7&:&r \end{array}} \right]\) 

R₂ → R₂ - 2R₁

R₃ → R₃ – R₁

\(\left[ {\begin{array}{*{20}{c}} 1&2&{ - 3}&:&p\\ 0&2&{ - 5}&:&{q - 2p}\\ 0&{ - 4}&{10}&:&{r - p} \end{array}} \right]\) 

R₃ → R₃ + 2R₂

\(\left[ {\begin{array}{*{20}{c}} 1&2&{ - 3}&:&p\\ 0&2&{ - 5}&:&{q - 2p}\\ 0&0&0&:&{r - p + 2q - 4p} \end{array}} \right]\) 

\(\left[ {\begin{array}{*{20}{c}} 1&2&{ - 3}&:&p\\ 0&2&{ - 5}&:&{q - 2p}\\ 0&0&0&:&{ - 5p + 2q + r} \end{array}} \right]\) 

For consistency ρ [A/b] = ρ [A] < n

⇒ -5p + 2q + r = 0 

Hence Option(3) is the correct answer.

Explanation:

Augmented matrix will be

\(\left[ {\begin{array}{*{20}{c}} 2&{ - 3}&7&5\\ 3&1&{ - 3}&{13}\\ 2&{19}&{ - 47}&{32} \end{array}} \right] \)

R₃ → R₃ – R₁

R₂ → 2R₂ – 3R₁

\(\left[ {\begin{array}{*{20}{c}} 2&{ - 3}&7&5\\ 0&{11}&{ - 27}&{11}\\ 0&{22}&{ - 54}&{27} \end{array}} \right]\)

R₃ → R₃ – 2R₂

\(\left[ {\begin{array}{*{20}{c}} 2&{ - 3}&7&5\\ 0&{11}&{ - 27}&{11}\\ 0&0&0&5 \end{array}} \right]\)

Rank of A ≠ Rank of Augmented matrix.

Hence given system of equations has no solution.

Hence Option(1) is the correct answer.

Given:

x + 4y + z = a,

2x - 3y – z = – 3, and

5x – 13y + k z = 9

Concept Used:

The given system of equation will not have unique solution if determinant of the coefficient of x, y, and z is equal to zero.

Solution:

D = \(\begin{vmatrix}1 & 4 & 1\\ 2& -3 & -1\\5 & -13 & k\end{vmatrix} \)  = 0

⇒ (-3k - 13) - (4) (2k + 5) +  (- 26 + 15) = 0

⇒ -3k - 13 - 8k - 20  - 11 = 0

⇒ -11k - 44 = 0

⇒ k = -4

\(\therefore\) Option 1 is correct.

Given:

x + 4y + z = a,

2x - 3y – z = – 3, and

5x – 13y + k z = 9

Concept Used:

The given system of equation will not have unique solution if determinant of the coefficient of x, y, and z is equal to zero.

Solution:

D = \(\begin{vmatrix}1 & 4 & 1\\ 2& -3 & -1\\5 & -13 & k\end{vmatrix} \)  = 0

⇒ (-3k - 13) - (4) (2k + 5) +  (- 26 + 15) = 0

⇒ -3k - 13 - 8k - 20  - 11 = 0

⇒ -11k - 44 = 0

⇒ k = -4

\(\therefore\) Option 1 is correct.

Concept:

Consistency means, the system of equations should have either infinite solutions or unique solution

Explanation:

For Consistent solution, the system of equations should have either infinite solutions or unique solution

\(\left[ {\begin{array}{*{20}{c}} 1&2&{ - 3}&:&p\\ 2&6&{ - 11}&:&q\\ 1&{ - 2}&7&:&r \end{array}} \right]\) 

R₂ → R₂ - 2R₁

R₃ → R₃ – R₁

\(\left[ {\begin{array}{*{20}{c}} 1&2&{ - 3}&:&p\\ 0&2&{ - 5}&:&{q - 2p}\\ 0&{ - 4}&{10}&:&{r - p} \end{array}} \right]\) 

R₃ → R₃ + 2R₂

\(\left[ {\begin{array}{*{20}{c}} 1&2&{ - 3}&:&p\\ 0&2&{ - 5}&:&{q - 2p}\\ 0&0&0&:&{r - p + 2q - 4p} \end{array}} \right]\) 

\(\left[ {\begin{array}{*{20}{c}} 1&2&{ - 3}&:&p\\ 0&2&{ - 5}&:&{q - 2p}\\ 0&0&0&:&{ - 5p + 2q + r} \end{array}} \right]\) 

For consistency ρ [A/b] = ρ [A] < n

⇒ -5p + 2q + r = 0 

Hence Option(3) is the correct answer.

Explanation:

Augmented matrix will be

\(\left[ {\begin{array}{*{20}{c}} 2&{ - 3}&7&5\\ 3&1&{ - 3}&{13}\\ 2&{19}&{ - 47}&{32} \end{array}} \right] \)

R₃ → R₃ – R₁

R₂ → 2R₂ – 3R₁

\(\left[ {\begin{array}{*{20}{c}} 2&{ - 3}&7&5\\ 0&{11}&{ - 27}&{11}\\ 0&{22}&{ - 54}&{27} \end{array}} \right]\)

R₃ → R₃ – 2R₂

\(\left[ {\begin{array}{*{20}{c}} 2&{ - 3}&7&5\\ 0&{11}&{ - 27}&{11}\\ 0&0&0&5 \end{array}} \right]\)

Rank of A ≠ Rank of Augmented matrix.

Hence given system of equations has no solution.

Hence Option(5) is the correct answer.

Explanation:

Consider the augmented matrix  

\([A|b] = \begin{bmatrix} 1 & 1 & 5 & 3 \\ 1 & 2 & m & 5 \\ 1 & 4 & k & 5 \end{bmatrix} \)  

Applying  row operations: 
 

\(R_3 \to R_3 - R_1, \quad R_2 \to R_2 - R_1 \)

\(\begin{bmatrix} 1 & 1 & 5 & 3 \\ 0 & 1 & m-5 & 2 \\ 0 & 3 & k-5 & 2 \end{bmatrix} \)  

\( R_3 \to R_3 - R_2 : \)  

\(\begin{bmatrix} 1 & 1 & 5 & 3 \\ 0 & 1 & m-5 & 2 \\ 0 & 0 & 4 - m(k - 5) & -5 \end{bmatrix}\)  

The system is consistent if and only if  

\(\rho([A|b]) = \rho(A) \)

If  \(m \neq 4\)  , then  \(\rho([A|b]) = \rho(A) = 3 \)  for any value of k 

Hence, option (1) is true

If k = 5 , then  \(\rho([A|b]) = \rho(A) \)  for any value of m

Hence, option (4) is true.  

If  \(k \neq 5 \)  and m = 4 , then \(\rho([A|b]) = 3 > \rho(A) = 2 \) ,   

and the system will be inconsistent.  

Hence, options (1) and (4) are  correct.

Explanation:

Augmented matrix will be

\(\left[ {\begin{array}{*{20}{c}} 2&{ - 3}&7&5\\ 3&1&{ - 3}&{13}\\ 2&{19}&{ - 47}&{32} \end{array}} \right] \)

R₃ → R₃ – R₁

R₂ → 2R₂ – 3R₁

\(\left[ {\begin{array}{*{20}{c}} 2&{ - 3}&7&5\\ 0&{11}&{ - 27}&{11}\\ 0&{22}&{ - 54}&{27} \end{array}} \right]\)

R₃ → R₃ – 2R₂

\(\left[ {\begin{array}{*{20}{c}} 2&{ - 3}&7&5\\ 0&{11}&{ - 27}&{11}\\ 0&0&0&5 \end{array}} \right]\)

Rank of A ≠ Rank of Augmented matrix.

Hence given system of equations has no solution.

Hence Option(1) is the correct answer.

Explanation:

A system of linear equations is considered consistent if it has at least one solution.

Augmented Matrix:

\( \begin{pmatrix} 1 & 1 & 5 & 3 \\ 1 & 2 & m & 5 \\ 1 & 2 & 4 & k \\ \end{pmatrix} \)   

Row Operation:

\(R_2 → R_2 - R_1 \)  

\(R_3 → R_3 - R_1 \)   

This gives us:

\(\begin{pmatrix} 1 & 1 & 5 & 3 \\ 0 & 1 & m-5 & 2 \\ 0 & 1 & -1 & k-3 \\ \end{pmatrix} \)   

\(R_3 = R_3 - R_2 \)  

\(\begin{pmatrix} 1 & 1 & 5 & 3 \\ 0 & 1 & m-5 & 2 \\ 0 & 0 & 4-m & k-5 \\ \end{pmatrix} \)  

For the system to be consistent, the last row must not represent an inconsistent equation 

If -m + 4 ≠ 0 (i.e., m ≠ 4), then we can solve for z in the last row and back-substitute to find x and y

If -m + 4 = 0 and k - 5 = 0 (i.e., m = 4 and k = 5), then the last row becomes 0 = 0, which is always true

The system has infinitely many solutions

If -m + 4 = 0 and k - 5 ≠ 0 (i.e., m = 4 and k ≠ 5),

then the last row represents an inconsistent equation (0 = a non-zero value), and the system has no solutions

Therefore, the system is consistent if:

1. m ≠ 4 

2. m = 4 and k = 5

⇒ the correct options are (A) and (D)

Hence Option(2) is the correct answer.