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

### 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

^{-1}B =(adj.A)BTherefore, 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

^{-1}B (adj.A)BTherefore, 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 solution

Now, X =A

^{-1}B A(adj.A)BTherefore, 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

^{-1}BA(adj.A)BTherefore, x= -1 and y= 4

### Chapter 4 Determinants – Exercise 4.6 | Set 2

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