# Class 12 RD Sharma Solutions – Chapter 6 Determinants – Exercise 6.5

### 5x + 4y – 9z = 0

Solution:

Given:

x + y – 2z = 0

2x + y – 3z =0

5x + 4y – 9z = 0

This system of equations can be expressed in the form of a matrix AX = B

Now find the determinant,

= 1(1 Ã— (-9) – 4 Ã— (-3)) – 1(2 Ã— (-9) – 5 Ã— (-3)) – 2(4 Ã— 2 – 5 Ã— 1)

= 1(-9 + 12) – 1(-18 + 15) – 2(8 – 5)

= 1 Ã— 3 – 1 Ã— (-3) – 2 Ã— 3

= 3 + 3 – 6

= 0

So, D = 0, that means this system of equations has infinite solution.

Now,

Let z = k

â‡’ x + y = 2k

And 2x + y = 3k

Now using the Cramerâ€™s rule

x =

x =

x =

x = k

Similarly,

y =

y =

y =

y = k

Therefore,

x = y = z = k.

### 2x + 5y – 2z = 0

Solution:

2x + 3y + 4z = 0

x + y + z = 0

2x + 5y – 2z = 0

This system of equations can be expressed in the form of a matrix AX = B

Find the determinant

= 2(1 Ã— (-2) – 1 Ã— 5) – 3(1 Ã— (-2) – 2 Ã— 1) + 4(1 Ã— 5 – 2 Ã— 1)

= 2(-2 – 5) – 3(-2 – 2) + 4(5 – 2)

= 2 Ã— (-7) – 3 Ã— (-4) + 4 Ã— 3

= -14 + 12 + 12

= -10

Hence, D â‰  0, so the system of equation has trivial solution.

Therefore, the system of equation has only solution as x = y = z = 0.

### 2x +5y – 2z = 0

Solution:

Given:

3x + y + z = 0

x – 4y + 3z = 0

2x +5y – 2z = 0

This system of equations can be expressed in the form of a matrix AX = B

Find the determinant

= 3(8 – 15) – 1(-2 – 6) + 1(13)

= -21 + 8 + 13

= 0

So, the system has infinite solutions:

Let z = k,

So,

3x + y = -k

x – 4y = -3k

Now,

x =

y =

x =

y =

z = k

and there values satisfy equation 3

Hence, x = -7k, y = 8k, z = 13k

### 2x + Î»z = 0

Solution:

Finding the determinant

= 3Î»3 + 2Î» – 8 – 6Î»

= 2Î»3 – 4Î» – 8

Which is satisfied by Î» = 2 {for non-trivial solutions Î» =2}

Now let z = k

4x – 2y = -3k

x + 2y = -3k

x =

y =

Hence, the solution is x = -k, y = , z = k

### has a non-trivial solution, then prove that ab + bc + ca = abc

Solution:

Finding the determinant

Now for non-trivial solution, D = 0

0 = (a – 1)[(b – 1)(c – 1) – 1]+1[-c + 1 – 1] + [-c + 1 – 1] – [ 1 + b – 1]

0 = (a – 1)[bc – b – c + 1 – 1] – c – b

0 = abc – ab -ac + b + c – c – b

ab + bc + ac = abc

Hence proved

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