Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Determinants Ex 6.2 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.2

Question 1.
Without expanding, evaluate the following determinants.
(i) $$\left|\begin{array}{lll} 1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b \end{array}\right|$$
Solution:

(ii) $$\left|\begin{array}{ccc} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 6 x & 9 x & 12 x \end{array}\right|$$
Solution:

(iii) $$\left|\begin{array}{lll} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{array}\right|$$
Solution:

Question 2.
Using properties of determinants, show that $$\left|\begin{array}{ccc} a+b & a & b \\ a & a+c & c \\ b & c & b+c \end{array}\right|$$ = 4abc
Solution:

Question 3.
Solve the following equation.
$$\left|\begin{array}{ccc} x+2 & x+6 & x-1 \\ x+6 & x-1 & x+2 \\ x-1 & x+2 & x+6 \end{array}\right|=0$$
Solution:
$$\left|\begin{array}{ccc} x+2 & x+6 & x-1 \\ x+6 & x-1 & x+2 \\ x-1 & x+2 & x+6 \end{array}\right|=0$$
Applying R2 → R2 – R1 and R3 → R3 – R1, we get
$$\left|\begin{array}{ccc} x+2 & x+6 & x-1 \\ 4 & -7 & 3 \\ -3 & -4 & 7 \end{array}\right|=0$$
∴ (x + 2)(-49 + 12) – (x + 6)(28 + 9) + (x – 1)(-16 – 21) = 0
∴ (x + 2) (-37) – (x + 6) (37) + (x – 1) (-37) = 0
∴ -37(x + 2 + x + 6 + x – 1) = 0
∴ 3x + 7 = 0
∴ x = $$\frac{-7}{3}$$

Question 4.
If $$\left|\begin{array}{lll} 4+x & 4-x & 4-x \\ 4-x & 4+x & 4-x \\ 4-x & 4-x & 4+x \end{array}\right|=0$$, then find the values of x.
Solution:

∴ (12 – x)[1(4x2 – 0) – (4 – x)(0 – 0) + (4 – x)(0 – 0)] = 0
∴ (12 – x)(4x2) = 0
∴ x2(12 – x) = 0
∴ x = 0 or 12 – x = 0
∴ x = 0 or x = 12

Question 5.
Without expanding determinants, show that
$$\left|\begin{array}{ccc} 1 & 3 & 6 \\ 6 & 1 & 4 \\ 3 & 7 & 12 \end{array}\right|+4\left|\begin{array}{lll} 2 & 3 & 3 \\ 2 & 1 & 2 \\ 1 & 7 & 6 \end{array}\right|=10\left|\begin{array}{lll} 1 & 2 & 1 \\ 3 & 1 & 7 \\ 3 & 2 & 6 \end{array}\right|$$
Solution:

Question 6.
Without expanding determinants, find the value of
(i) $$\left|\begin{array}{lll} 10 & 57 & 107 \\ 12 & 64 & 124 \\ 15 & 78 & 153 \end{array}\right|$$
Solution:

(ii) $$\left|\begin{array}{lll} 2014 & 2017 & 1 \\ 2020 & 2023 & 1 \\ 2023 & 2026 & 1 \end{array}\right|$$
Solution:

Question 7.
Without expanding determinants, prove that
(i) $$\left|\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right|=\left|\begin{array}{lll} b_{1} & c_{1} & a_{1} \\ b_{2} & c_{2} & a_{2} \\ b_{3} & c_{3} & a_{3} \end{array}\right|=\left|\begin{array}{lll} c_{1} & a_{1} & b_{1} \\ c_{2} & a_{2} & b_{2} \\ c_{3} & a_{3} & b_{3} \end{array}\right|$$
Solution:

(ii) $$\left|\begin{array}{lll} 1 & y z & y+z \\ 1 & z x & z+x \\ 1 & x y & x+y \end{array}\right|=\left|\begin{array}{lll} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}\right|$$
Solution:

In 1st determinant, taking (x + y + z) common from C3 and in 2nd determinant, taking $$\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$$ common from R1, R2, R3 respectively, we get