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

### 2x + 3y = 3

Solution:

Matrix form of the given equations is AX = B

where, A = , B =   and, X =

âˆ´

Now, |A| =

âˆµ Inverse of matrix exists, unique solution.

âˆ´ System of equation is consistent.

### x + y = 4

Solution:

Matrix form of the given equations is AX = B

where, A =, B =  and, X =

âˆ´

Now, |A| =

âˆµ Inverse of matrix exists, unique solution.

âˆ´ System of equation is consistent.

### 2x + 6y = 8

Solution:

Matrix form of the given equations is AX = B

where, A =, B =  and, X =

âˆ´

Now, |A| =

âˆµ Have no common solution.

âˆ´ System of equation is inconsistent.

### ax + ay + 2az = 4

Solution:

Matrix form of the given equations is AX = B

where, A =, B =and, X =

âˆ´

Now, |A| =

âˆµ Inverse of matrix exists, unique solution.

âˆ´ System of equation is consistent.

### 3x – 5y = 3

Solution:

Matrix form of the given equations is AX = B

where, A =, B=and, X =

âˆ´

Now, |A| =

âˆ´ System of equation is inconsistent.

### 5x – 2y + 6z = â€“1

Solution:

Matrix form of the given equations is AX = B

where, A =, B = and, X=

âˆ´

Now, |A| =

âˆµ Inverse of matrix exists, unique solution.

âˆ´ System of equation is consistent.

### 7x + 3y = 5

Solution:

Matrix form of the given equations is AX = B

where, A=, B=, X=

âˆ´

Now, |A|=

âˆ´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=

âˆ´

Now, |A|=

âˆ´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=

âˆ´

Now, |A|=

âˆ´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=

âˆ´

Now, |A|=

âˆ´Unique solution