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

• Difficulty Level : Expert
• Last Updated : 12 Oct, 2021

### 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.

### 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.

### 2x + 6y = 8

Solution:

Matrix form of the given equations is AX = B

where, A =, B =  and, X =

∴

Now, |A| =

∵ Have no common solution.

∴ System of equation is inconsistent.

### 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.

### 3x – 5y = 3

Solution:

Matrix form of the given equations is AX = B

where, A =, B=and, X =

Now, |A| =

∴ System of equation is inconsistent.

### 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.

### 7x + 3y = 5

Solution:

Matrix form of the given equations is AX = B

where, A=, B=, X=

Now, |A|=

∴Unique solution

Therefore, x=2 and y=-3

### 3x + 4y = 3

Solution:

Matrix form of the given equations is AX = B

where, A=, B=, X=

Now, |A|=

∴Unique solution

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

### 3x – 5y = 7

Solution:

Matrix form of the given equations is AX = B

where, A=, B=, X=

Now, |A|=

∴Unique solutionn

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

### 3x + 2y = 5

Solution:

Matrix form of the given equations is AX = B

where, A=, B=, X=

Now, |A|=

∴Unique solution