Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Determinants Ex 6.1 Questions and Answers.

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

Question 1.

Evaluate the following determinants:

Solution:

(i) \(\left|\begin{array}{cc}

4 & 7 \\

-7 & 0

\end{array}\right|\)

= 4(0) – (-7)(7)

= 0 + 49

= 49

= 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 + i^{2} + 15i – 6i – 18

= i^{2} – 26 + 9i

= -1 – 26 + 9i …[∵ i^{2} = -1]

= -27 + 9i

= 5(32 – 16) – 5(40 – 20) + 5(20 – 20)

= 5(16) – 5(20) + 5(0)

= 80 – 100

= -20

(v) \(\left|\begin{array}{cc}

2 \mathrm{i} & 3 \\

4 & -\mathrm{i}

\end{array}\right|\)

= 2i(-i) – 3(4)

= -2i^{2} – 12

= -2(-1) – 12 …[∵ i^{2} = -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 – f^{2}) – h(hc – gf) + g(hf – gb)

= abc – af^{2} – h^{2}c + fgh + fgh – g^{2}b

= abc + 2fgh – af^{2} – bg^{2} – ch^{2}

= 0 – a(0 + bc) – b(-ac – 0)

= -a(bc) – b(-ac)

= -abc + abc

= 0

Question 2.

Find the value(s) of x, if

Solution:

(i) \(\left|\begin{array}{ll}

2 & 3 \\

4 & 5

\end{array}\right|=\left|\begin{array}{cc}

x & 3 \\

2 x & 5

\end{array}\right|\)

∴ 10 – 12 = 5x – 6x

∴ -2 = -x

∴ x = 2

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. = \(\left|\begin{array}{ll}

2 & 3 \\

4 & 5

\end{array}\right|\)

= 10 – 12

= -2

R.H.S. =\(\left|\begin{array}{cc}

x & 3 \\

2 x & 5

\end{array}\right|\)

= \(\left|\begin{array}{ll}

2 & 3 \\

4 & 5

\end{array}\right|\)

= 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 = \(\frac{14}{5}\)

∴ (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

Question 3.

Solve the following equations.

Solution:

∴ x(x^{2} – 4) – 2(2x – 4) + 2(4 – 2x) = 0

∴ x(x^{2} – 4) – 2(2x – 4) – 2(2x – 4) = 0

∴ x(x + 2)(x – 2) – 4(2x – 4) = 0

∴ x(x + 2)(x – 2) – 8(x – 2) = 0

∴ (x – 2)[x(x + 2) – 8] = 0

∴ (x – 2)(x^{2} + 2x – 8) = 0

∴ (x – 2)(x^{2} + 4x – 2x – 8) = 0

∴ (x – 2)(x + 4)(x – 2) = 0

∴ (x – 2)^{2} (x + 4) = 0

∴ (x – 2)^{2} = 0 or x + 4 = 0

∴ x – 2 = 0 or x = -4

∴ x = 2 or x = -4

∴ 1(-10x^{2} – 10x) – 4(5x^{2} – 5) + 20(2x + 2) = 0

∴ -10x^{2} – 10x – 20x^{2} + 20 + 40x + 40 = 0

∴ -30x^{2} + 30x + 60 = 0

∴ x^{2} – x – 2 = 0 …..[Dividing throughout by (-30)]

∴ x^{2} – 2x + x – 2 = 0

∴ (x – 2)(x + 1) = 0

∴ x – 2 = 0 or x + 1 = 0 x = 2 or x = -1

Question 4.

Find the value of x, if

\(\left|\begin{array}{ccc}

x & -1 & 2 \\

2 x & 1 & -3 \\

3 & -4 & 5

\end{array}\right|\) = 29

Solution:

∴ x(5 – 12) + 1(10x + 9) + 2(-8x – 3) = 29

∴ -7x + 10x + 9 – 16x – 6 = 29

∴ -13x + 3 = 29

∴ -13x = 26

∴ x = -2

Question 5.

Find x and y if \(\left|\begin{array}{ccc}

4 i & i^{3} & 2 i \\

1 & 3 i^{2} & 4 \\

5 & -3 & i

\end{array}\right|\) = x + iy, where i = √-1.

Solution:

= 4i(-3i + 12) + i(i – 20) + 2i(-3 + 15)

= -12i^{2} + 48i + i^{2} – 20i + 24i

= -11i^{2} + 52i

= -11(-1) + 52i …..[∵ i^{2} = -1]

= 11 + 52i

Comparing with x + iy, we get

x = 11, y = 52