NCERT Solutions Exercise 4.3: Determinants

Q1: Write Minors and Cofactors of the elements of following determinants:

Ans: (i) The given determinant is
Minor of element a_ij is M_ij.


Explanation:
For any element a_ij of a determinant, its minor M_ij is the determinant obtained by deleting the i-th row and j-th column. The corresponding cofactor A_ij is given by A_ij = (-1)^i+jM_ij.
The results of these deletions and the computed minors and cofactors are shown in the images above ( and ). Use the sign factor (-1)^i+j to obtain each cofactor from the corresponding minor.
Q2: 

Ans:






Explanation:
The minors and cofactors for the given determinant are obtained by deleting the appropriate row and column for each element and then applying the sign (-1)^i+j to the computed minor to get the cofactor. The intermediate minors and the final cofactors are displayed in the images through . Follow the same rule for every element: compute the 2×2 determinant after deletion, then multiply by (-1)^i+j.
Q3: Using Cofactors of elements of second row, evaluate Δ =.
Ans: The given determinant is
 .
We have:



We know that ∆ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.
Solution:
Expanding along the second row,
∆ = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃.
From the determinant (see images), the second row elements and their cofactors are:
a₂₁ = 2,   A₂₁ = 7
a₂₂ = 0,   A₂₂ = 7
a₂₃ = 1,   A₂₃ = -7
Therefore,
∆ = 2 × 7 + 0 × 7 + 1 × (-7) = 14 - 7 = 7.
Q4: Using Cofactors of elements of third column, evaluate Δ=  

Ans: The given determinant is .

We know that Δ is equal to the sum of the product of the elements of the second row
with their corresponding cofactors.

Hence,
Clarification and Correction:
To evaluate Δ by expanding along the third column, use the formula
Δ = a₁₃A₁₃ + a₂₃A₂₃ + a₃₃A₃₃.
The necessary minors and cofactors for the third column and the numerical substitution are provided in the images -. The final computed value of Δ is shown in .
Q5: If ∆ =

and A_ij is Cofactors of a_ij , then value of ∆ is given by
(A) a₁₁ A₃₁+ a₁₂ A₃₂ + a₁₃ A₃₃ 
(B) a₁₁ A₁₁ + a₁₂ A₂₁ + a₁₃ A₃₁ 
(C) a₂₁ A₁₁ + a₂₂ A₁₂ + a₂₃ A₁₃ 
(D) a₁₁ A₁₁+ a₂₁ A₂₁ + a₃₁ A₃₁
Ans: D

Given : Δ =

Δ = Sum of products of elements of row (or column) with their corresponding cofactors.
Δ=a₁₁ A₁₁+ a₂₁ A₂₁ + a₃₁ A₃₁
So, option D is correct.

MCQ (Mathematical) - Finalised solution:
Ans: (D)
Sol:
The determinant ∆ can be expanded along any column. Expanding along the first column gives
∆ = a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁,
which matches option (D). Hence (D) is correct.

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Old NCERT Questions

Q1: For the matrices A and B, verify that (AB)′ = B'A' where

Ans:

Hence,
Explanation:
The transpose of a product satisfies the rule (AB)' = B'A'. A brief index-wise justification is:
((AB)')_ij = (AB)_ji = Σ_k A_jkB_ki = (B'A')_ij.
The explicit multiplication and transposition carried out for the given matrices are displayed in the images and , which verify the equality.
Q2: Examine the consistency of the system of equations.
 x + 2y = 2
 2x + 3y = 3
Ans:
The given system of equations is:
x + 2y = 2
2x + 3y = 3
The given system of equations can be written in the form of AX = B, where

 A is non-singular.
Therefore, A^-1 exists.
Hence, the given system of equations is consistent.
Explanation:
Form A and compute det(A) (see ). Since det(A) ≠ 0, matrix A is invertible and a unique solution X = A^-1B exists. Thus the system is consistent with a unique solution.
Q3: Examine the consistency of the system of equations.
 2x - y = 5 x
 + y = 4
Ans: The given system of equations is:
2x - y = 5 x
+ y = 4
The given system of equations can be written in the form of AX = B, where

 A is non-singular.
Therefore, A^-1 exists.
Hence, the given system of equations is consistent.
Explanation:
Write A and compute its determinant (refer ). As det(A) ≠ 0, A is invertible and the system has a unique solution X = A^-1B. Hence the system is consistent.
Q4: Examine the consistency of the system of equations.
 x + 3y = 5
 2x + 6y = 8
Ans: The given system of equations is:
x + 3y = 5
2x + 6y = 8
The given system of equations can be written in the form of AX = B, where

Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.
Explanation:
The second equation is 2 times the left-hand side of the first equation, but the right-hand side is 8 rather than 10. That is, the coefficient rows are proportional while the constants are not proportional. Therefore the equations are inconsistent - there is no common solution.
Q5: Examine the consistency of the system of equations.
 x + y + z = 1 
2x + 3y + 2z = 2 
ax + ay + 2az = 4
Ans: The given system of equations is:
x + y + z = 1 2x
+ 3y + 2z = 2 ax
+ ay + 2az = 4
This system of equations can be written in the form AX = B, where

 A is non-singular.
Therefore, A^-1 exists.
Hence, the given system of equations is consistent.
Explanation:
Form the coefficient matrix A and compute det(A) (shown in ). Since det(A) ≠ 0 for the given value(s) of parameter a (as computed), A is invertible and the system has a unique solution. Thus it is consistent.
Q6: Examine the consistency of the system of equations.
 3x - y - 2z = 2
 2y - z = -1
 3x - 5y = 3
Ans: The given system of equations is:
3x - y - 2z = 2
2y - z = -1
3x - 5y = 3
This system of equations can be written in the form of AX = B, where

A is a singular matrix.



Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.
Explanation:
The determinant of the coefficient matrix A is zero (see ), so A is singular. Comparing the ranks of the coefficient matrix and the augmented matrix (shown in and ) shows that the augmented matrix has a larger rank. Hence the system is inconsistent - there is no solution.
Q7: Examine the consistency of the system of equations.
 5x - y + 4z = 5
 2x + 3y + 5z = 2
 5x - 2y + 6z = -1
Ans: The given system of equations is:
5x - y + 4z = 5
2x + 3y + 5z = 2
5x - 2y + 6z = -1
This system of equations can be written in the form of AX = B, where

Therefore, A^-1 exists.
Hence, the given system of equations is consistent.
Explanation:
Determinant det(A) ≠ 0 (see ), so A is invertible and the unique solution X = A^-1B exists. Thus the system is consistent.
Q8: Solve system of linear equations, using matrix method.

Ans: The given system of equations can be written in the form of AX = B, where

Thus, A is non-singular. Therefore, its inverse exists.

Solution Outline:
Write the system as AX = B. Compute A^-1 (shown in ) and then multiply to get X = A^-1B. The detailed arithmetic steps and final values are displayed in the images.
Q9: Solve system of linear equations, using matrix method.

Ans:The given system of equations can be written in the form of AX = B, where

Thus, A is non-singular. Therefore, its inverse exists.

Solution Outline:
As before, compute X = A^-1B. The inverse and multiplication steps are carried out in to produce the solution vector.
Q10: Solve system of linear equations, using matrix method.

Ans: The given system of equations can be written in the form of AX = B, where

Thus, A is non-singular. Therefore, its inverse exists.




Solution Outline:
Compute A^-1 and then X = A^-1B. The intermediate calculations and the final solution are presented in the shown images -.
Q11: Solve system of linear equations, using matrix method.
 5x + 2y = 3
 3x + 2y = 5
Ans: The given system of equations can be written in the form of AX = B, where

Thus, A is non-singular. Therefore, its inverse exists.
Solution Outline:
For a 2×2 system, compute det(A) and A^-1 = (1/det(A)) adj(A). Then X = A^-1B. The computations and the final numeric values are shown in .
Q12: Solve system of linear equations, using matrix method.

Ans: The given system of equations can be written in the form of AX = B, where

Thus, A is non-singular. Therefore, its inverse exists.

Solution Outline:
Compute A^-1 and then X = A^-1B. The detailed steps and result are provided in the images.
Q13: Solve system of linear equations, using matrix method.
 x - y + z = 4 
2x + y - 3z = 0 
x + y + z = 2
Ans: The given system of equations can be written in the form of AX = B, where

Thus, A is non-singular. Therefore, its inverse exists.

Solution Outline:
Compute A^-1 (shown in ) and multiply by B to obtain the solution vector X.
Q14: Solve system of linear equations, using matrix method.
 2x + 3y + 3z = 5
 x - 2y + z = -4
 3x - y - 2z = 3
Ans: The given system of equations can be written in the form of AX = B, where

Thus, A is non-singular. Therefore, its inverse exists.

Solution Outline:
Find A^-1 and compute X = A^-1B. The computational details are presented in .
Q15: Solve system of linear equations, using matrix method.
x - y + 2z = 7
3x + 4y - 5z = -5
2x - y + 3z = 12
Ans: The given system of equations can be written in the form of AX = B, where

Thus, A is non-singular. Therefore, its inverse exists.

Solution Outline:
Compute A^-1 and multiply by B. The inverse and multiplication steps are shown in yielding the solution.
Q16: If   find A^-1. Using A^-1 solve the system of equations
Ans:

Now, the given system of equations can be written in the form of AX = B, where


Solution Outline:
Compute A^-1 as shown in . Then use X = A^-1B (with A^-1 and B given in and ) to find the unknowns. The numeric steps are provided in the images.
Q17: The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method.
Ans: Let the cost of onions, wheat, and rice per kg be Rs x, Rs y,and Rs z respectively.
Then, the given situation can be represented by a system of equations as:

This system of equations can be written in the form of AX = B, where

Now,
X = A^-1 B

Hence, the cost of onions is Rs 5 per kg, the cost of wheat is Rs 8 per kg, and the cost of rice is Rs 8 per kg.

Explanation (final):
Set up the coefficient matrix A and constant vector B from the given information. Compute A^-1 and then X = A^-1B. The inverse and multiplication steps are shown in , yielding x = 5, y = 8, z = 8 (Rs per kg).