Class 12 NCERT Solutions- Mathematics Part I – Chapter 4 Determinants – Exercise 4.6 | Set 1

Note: This exercise has been renumbered as Exercise 4.5 in the updated NCERT syllabus.

Examine the consistency of the system of equations in Exercises 1 to 6.

Question 1. x + 2y = 2

                            2x + 3y = 3

Solution:

Matrix form of the given equations is AX = B

where, A = , B =   and, X = 

∴  

Now, |A| = 

∵ Inverse of matrix exists, unique solution.

∴ System of equation is consistent.

Question 2. 2x – y = 5 

                            x + y = 4

Solution:

Matrix form of the given equations is AX = B

where, A =, B =  and, X =

∴

Now, |A| =

∵ Inverse of matrix exists, unique solution.

∴ System of equation is consistent.

Question 3. x + 3y = 5 

                            2x + 6y = 8

Solution:

Matrix form of the given equations is AX = B

where, A =, B =  and, X =

∴ 

Now, |A| =

And, adj. A =

∴ (adj. A) B = 

∵ Have no common solution.

∴ System of equation is inconsistent.

Question 4. x + y + z = 1 

                            2x + 3y + 2z = 2 

                            ax + ay + 2az = 4

Solution:

Matrix form of the given equations is AX = B

where, A =, B =and, X =

∴  

Now, |A| = 

∵ Inverse of matrix exists, unique solution.

∴ System of equation is consistent.

Question 5. 3x – y – 2z = 2 

                            2y – z = -1 

                            3x – 5y = 3 

Solution:

Matrix form of the given equations is AX = B

where, A =, B=and, X =

∴

Now, |A| =

And, adj. A =

∴ (adj. A) B =

∴ System of equation is inconsistent.

Question 6. 5x – y + 4z = 5

                            2x + 3y + 5z = 2 

                             5x – 2y + 6z = –1

Solution:

Matrix form of the given equations is AX = B

where, A =, B = and, X=

∴

Now, |A| =

∵ Inverse of matrix exists, unique solution.

∴ System of equation is consistent.

Solve system of linear equations, using matrix method, in Exercises 7 to 14.

Question 7. 5x + 2y = 4 

                            7x + 3y = 5

Solution:

Matrix form of the given equations is AX = B

where, A=, B=, X=

∴

Now, |A|=

∴Unique solution

Now, X = A^-1B =(adj.A)B

Therefore, x=2 and y=-3

Question 8. 2x – y = -2 

                            3x + 4y = 3

Solution:

Matrix form of the given equations is AX = B

where, A=, B=, X=

∴

Now, |A|=

∴Unique solution

Now, X = A^-1B (adj.A)B

Therefore, x=-5/11 and y=12/11

Question 9. 4x – 3y = 3

                            3x – 5y = 7

Solution:

Matrix form of the given equations is AX = B

where, A=, B=, X=

∴

Now, |A|=

∴Unique solutionn

Now, X =A^-1B A(adj.A)B

Therefore, x= -6/11 and y= -19/11

Question 10. 5x + 2y = 3 

                               3x + 2y = 5

Solution:

Matrix form of the given equations is AX = B

where, A=, B=, X=

∴

Now, |A|=

∴Unique solution

Now, X = A^-1BA(adj.A)B

Therefore, x= -1 and y= 4