### Chapter 4 Determinants – Exercise 4.6 | Set 1

### Question 11. 2x + y + z = 1

### x – 2y – z = 3/2

### 3y – 5z = 9

**Solution:**

Matrix form of the given equation is AX = B

i.e.

∴ |A| =

∴ Solution is unique.

Now, X = A

^{-1}B = (adj.A)BTherefore, x=1, y=1/2, z=3/2

### Question 12. x – y + z = 4

### 2x + y – 3z = 0

### x + y + z = 2

**Solution:**

Matrix form of the given equation is AX = B

i.e

∴ |A| =

∴ Solution is unique.

Now, X = A

^{-1}B = (adj.A)BTherefore, x = 2, y = -1, z = 1

### Question 13. 2x + 3y +3 z = 5

### x – 2y + z = – 4

### 3x – y – 2z = 3

**Solution:**

Matrix form of given equation is AX = B

i.e.

∴ |A| =

∴ Solution is unique.

Now, X = A

^{-1}B = (adj.A)BTherefore, x = 1, y = 2, z = -1

### Question 14. x – y + 2z = 7

### 3x + 4y – 5z = – 5

### 2x – y + 3z = 12

**Solution:**

Matrix form of given equation is AX = B

i.e.

∴ |A| =

∴ Solution is unique.

Now, X = A

^{-1}B = (adj.A)BTherefore, x = 2, y = 1, z = 3

### Question 15. If A=, find A–1. Using A–1 solve the system of equations

### 2x – 3y + 5z = 11

### 3x + 2y – 4z = – 5

### x + y – 2z = – 3

**Solution:**

Given:A=Now, |A|=

∴ |A|=

Means, A

^{-1}exists.And A

^{-1}=(adj.A)……(1)Now,

∴ adj. A =

From eq. (1),

A

^{-1}=Now, Matrix form of given equation is AX = B

i.e.

∵ Solution is unique.

∴ X=A

^{-1}B⇒

Therefore, x = 1, y = 2, z = 3

### Question 16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is 60 rupees. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is 90 rupees. The cost of 6 kg onion 2 kg wheat and 3 kg rice is 70 rupees. Find cost of each item per kg by matrix method.

**Solution:**

Let Rs x, Rs y, Rs z per kg be the prices of onion, wheat and rice respectively.

A.T.Q.

4x+3y+2z=60

2x+4y+6z=90

6x+2y+3z=70

Matrix form of equation is AX = B

where, A=,B=and X=

=>

Now, |A|=

∴ Solution is unique

Now, X=A

^{-1}B=(adj. A)B……(1)Now,

∴ (adj.A)=

From eq

^{n}.(1)Therefore, x = 5, y = 8, z = 8

Hence, the cost of onion, wheat and rice are Rs. 5, Rs 8 and Rs 8 per kg.

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