Chapter 6 Determinants Ex 6.1 Solutions

Balbharati Maharashtra State Board11th Commerce Maths Solution Book PdfChapter 6 Determinants Ex 6.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.1

= 3(0 – 63) – 5(0 – 27) + 2(7 – 24)
= 3(-63) + 5 (-27) + 2(-17)
= – 189 – 135 – 34
= -358

= 1(2 – 10) – i(-i – 15) + 3(-2i – 6)
= -8 + i²+ 15i – 6i – 18
= i²– 26 + 9i
= -1 – 26 + 9i …[∵ i²= -1]
= -27 + 9i

= 5(32 – 16) – 5(40 – 20) + 5(20 – 20)
= 5(16) – 5(20) + 5(0)
= 80 – 100
= -20

(v) ≤ft|{array}{cc}
2 i & 3 \\
4 & -i
{array}
= 2i(-i) – 3(4)
= -2i2– 12
= -2(-1) – 12 …[∵ i2= -1]
= 2 – 12
= -10

= 3(1 + 6) + 4(1 + 4) + 5(3 – 2)
= 3(7)+ 4(5) + 5(1)
= 21 + 20 + 5
= 46

= a(bc – f²) – h(hc – gf) + g(hf – gb)
= abc – af²– h²c + fgh + fgh – g²b
= abc + 2fgh – af²– bg²– ch²

= 0 – a(0 + bc) – b(-ac – 0)
= -a(bc) – b(-ac)
= -abc + abc
= 0

Check:
We can check if our answer is right or wrong.
In order to do so, substitute x = 2 in the given determinant.
For x = 2,
L.H.S. = ≤ft|{array}{ll}
2 & 3 \\
4 & 5
{array}
= 10 – 12
= -2
R.H.S. =≤ft|{array}{cc}
x & 3 \\
2 x & 5
{array}
= ≤ft|{array}{ll}
2 & 3 \\
4 & 5
{array}
= 10 – 12
= -2
Thus, our answer is correct.

∴ 2(9 – 20) – 1 (-3 – 0) + (x + 1) (5 – 0) = 0
∴ 2(-11) – 1(-3) + (x + 1)(5) = 0
∴ -22 + 3 + 5x + 5 = 0
∴ 5x = 14
∴ x =

∴ (x – 1)[(x – 2)(x – 3) – 0] – x(0 – 0) + (x – 2)(0 – 0) = 0
∴ (x – 1)(x – 2)(x – 3) = 0
∴ x – 1 = 0 or x – 2 = 0 or x – 3 = 0
∴ x = 1 or x = 2 or x = 3

∴ 1(-10x²– 10x) – 4(5x²– 5) + 20(2x + 2) = 0
∴ -10x²– 10x – 20x²+ 20 + 40x + 40 = 0
∴ -30x²+ 30x + 60 = 0
∴ x²– x – 2 = 0 …..[Dividing throughout by (-30)]
∴ x²– 2x + x – 2 = 0
∴ (x – 2)(x + 1) = 0
∴ x – 2 = 0 or x + 1 = 0 x = 2 or x = -1