In Chapter 6 of RD Sharma's Class 12 Mathematics textbook determinants play a crucial role in the solving systems of linear equations and various matrix-related problems. Exercise 6.2 | Set 1 provides the practice problems designed to deepen students' understanding of the determinants and their properties. This set focuses on the fundamental determinant operations and their applications essential for mastering advanced topics in algebra.
Determinants
A determinant is a scalar value computed from the elements of a square matrix that provides useful information about the matrix's properties. It helps in solving linear equations finding the inverse of the matrices and understanding matrix transformations. The Determinants can be computed using various methods such as cofactor expansion or row reduction. The value of the determinant can indicate if a matrix is invertible or singular.
Question 1. Evaluate the following determinant:
(i) \begin{vmatrix} 1 & 3 &5 \\ 2 & 6 & 10 \\ 31 & 11 & 38 \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} 1 & 3 &5 \\ 2 & 6 & 10 \\ 31 & 11 & 38 \end{vmatrix} \\ \triangle = 2\begin{vmatrix} 1 & 3 &5 \\ 1 & 3 & 5 \\ 31 & 11 & 38 \end{vmatrix} As R1 and R2 are identical
Hence, △ = 0
(ii) \begin{vmatrix} 67 & 19 &21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} 67 & 19 &21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \end{vmatrix} C1⇢C1 - 3C3
\triangle = \begin{vmatrix} 4 & 19 &21 \\ -3 & 13 & 14 \\ 3 & 24 & 26 \end{vmatrix} R3⇢R3 + R2 and R1⇢R1 + R2
\triangle = \begin{vmatrix} 1 & 32 &35 \\ -3 & 13 & 14 \\ 0 & 37 & 40 \end{vmatrix} R2⇢R2 + 3R1
\triangle = \begin{vmatrix} 1 & 32 &35 \\ 0 & 109 & 119 \\ 0 & 37 & 40 \end{vmatrix} △ = 1(109 × 40 - 119 × 37)
Hence, △ = -43
(iii) \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}\\ \triangle = a\begin{vmatrix} b & f \\ f & c \end{vmatrix}-h\begin{vmatrix} h & f \\ g & c \end{vmatrix}+g\begin{vmatrix} h & b \\ g & f \end{vmatrix} △ = a(bc - f2) - h(hc - fg) + g(hf - gb)
△ = abc - af2 - h2c + fgh + fgh - g2b
Hence, △ = abc + 2fgh - af2 - ch2 - bg2
(iv) \begin{vmatrix} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \end{vmatrix}
\triangle = 1\begin{vmatrix} -1 & 2 \\ 5 & 2 \end{vmatrix}-(-3)\begin{vmatrix} 4 & 2 \\ 3 & 2 \end{vmatrix}+2\begin{vmatrix} 4 & -1 \\ 3 & 5 \end{vmatrix} △ = 1(-2 - 10) + 3(8 - 6) + 2(20 + 3)
△ = 1(-12) + 3(2) + 2(23)
△ = -12 + 6 + 46
Hence, △ = 40
(v) \begin{vmatrix} 1 & 4 & 9 \\ 4 & 9 &16 \\ 9 & 16 & 25 \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} 1 & 4 & 9 \\ 4 & 9 &16 \\ 9 & 16 & 25 \end{vmatrix}\\ \triangle = 1\begin{vmatrix} 9 & 16 \\ 16 & 25 \end{vmatrix}-4\begin{vmatrix} 4 & 16 \\ 9 & 25 \end{vmatrix}+9\begin{vmatrix} 4 & 9 \\ 9 & 16 \end{vmatrix} △ = 1(225-256) + 4(100-144) + 9(64-81)
△ = 1(-31) - 4(-44) + 9(-17)
△ = -31 + 176 - 153
Hence, △ = -8
(vi) \begin{vmatrix} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{vmatrix} Taking -2 common from C1, C2 and C3
\triangle = -2\begin{vmatrix} -3 & -3 & 2 \\ -1 & -1 & 2 \\ 5 & 5 & 2 \end{vmatrix} As C1 and C2 are identical
Hence, △ = 0
(vii) \begin{vmatrix} 1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9 \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} 1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9 \end{vmatrix}\\ \triangle = \begin{vmatrix} 1 & 3 & 3^2 & 3^3 \\ 3 & 3^2 & 3^3 & 1 \\ 3^2 & 3^3 & 1 & 3 \\ 3^3 & 1 & 3 & 3^2 \end{vmatrix} C1⇢C1 + C2 + C3
\triangle = \begin{vmatrix} 1+3 + 3^2 + 3^3 & 3 & 3^2 & 3^3 \\ 1+3 + 3^2 + 3^3 & 3^2 & 3^3 & 1 \\ 1+3 + 3^2 + 3^3 & 3^3 & 1 & 3 \\ 1+3 + 3^2 + 3^3 & 1 & 3 & 3^2 \end{vmatrix}\\ \triangle = 1+3 + 3^2 + 3^3\begin{vmatrix} 1 & 3 & 3^2 & 3^3 \\ 1 & 3^2 & 3^3 & 1 \\ 1& 3^3 & 1 & 3 \\ 1 & 1 & 3 & 3^2 \end{vmatrix}\\ C2⇢C2 - C1
C3⇢C3 - C1
C4⇢C4 - C1
\triangle = 1+3 + 3^2 + 3^3\begin{vmatrix} 1 & 3 & 3^2 & 3^3 \\ 0 & 3^2-3 & 3^3-3^2 & 1-3^3 \\ 0 & 3^3-3 & 1-3^2 & 3-3^3 \\ 0 & 1-3 & 3-3^2 & 3^2-3^3 \end{vmatrix}\\ \triangle = (1+3 + 3^2 + 3^3)1\begin{vmatrix} 3^2-3 & 3^3-3^2 & 1-3^3 \\ 3^3-3 & 1-3^2 & 3-3^3 \\ 1-3 & 3-3^2 & 3^2-3^3 \end{vmatrix}\\ \triangle = 1+3 + 3^2 + 3^3\begin{vmatrix} 6 & 18 & -26 \\ 24 & -8 & -24 \\ -2 & -6 & -18 \end{vmatrix}\\ Taking 2, -2 and -2 common from C1, C2 and C3
\triangle = (1+3 + 3^2 + 3^3)(2^3)\begin{vmatrix} 3 & -9 & 13\\ 12 & 4 & 12\\ -1 & 3 & 9\end{vmatrix}\\ Taking 4 common from R2 and R1⇢R1+3R3
\triangle = (1+3 + 3^2 + 3^3)(2^2)(2^3)\begin{vmatrix} 0 & 0 & 40\\ 3 & 1 & 3\\ -1 & 3 & 9\end{vmatrix}\\ △ = (1 + 3 + 32 + 33)(4)(8)[40(9 - (-1))]
△ = (40)(4)(8)[40(9 + 1)]
△ = 40 × 4 × 8 × 40 × 10
Hence, △ = 512000
(viii) \begin{vmatrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix} Taking 6 common from R1, we get
\triangle = 6\begin{vmatrix} 17 & 3 & 6 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix} As R1 and R3 are identical
Hence, △ = 0
Question 2. Without expanding, show that the values of each of the following determinants are zero:
(i) \begin{vmatrix} 8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3 \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} 8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3 \end{vmatrix} Taking 4 common from C1, we get
\triangle = 4\begin{vmatrix} 2 & 2 & 7 \\ 3 & 3 & 5 \\ 4 & 4 & 3 \end{vmatrix} As C1 and C2 are identical
Hence, △ = 0
(ii) \begin{vmatrix} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{vmatrix} Taking -2 common from C1, we get
\triangle = (-2)\begin{vmatrix} -3 & -3 & 2 \\ -1 & -1 & 2 \\ 5 & 5 & 2 \end{vmatrix} As C1 and C2 are identical
Hence, △ = 0
(iii) \begin{vmatrix} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12 \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12 \end{vmatrix} R3⇢R3 - R2
\triangle = \begin{vmatrix} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 2 & 3 & 7 \end{vmatrix} As R1 and R3 are identical
Hence, △ = 0
(iv) \begin{vmatrix} \frac{1}{a} & a^2 & bc \\ \frac{1}{b} & b^2 & ac \\ \frac{1}{c} & c^2 & ab \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} \frac{1}{a} & a^2 & bc \\ \frac{1}{b} & b^2 & ac \\ \frac{1}{c} & c^2 & ab \end{vmatrix} Multiplying and dividing △ by abc, we get
\triangle = \frac{abc}{abc}\begin{vmatrix} \frac{1}{a} & a^2 & bc \\ \frac{1}{b} & b^2 & ac \\ \frac{1}{c} & c^2 & ab \end{vmatrix}\\ Multiplying R1, R2 and R3 by a, b and c respectively
\triangle = \frac{1}{abc}\begin{vmatrix} 1 & a^3 & abc \\ 1 & b^3 & abc \\ 1 & c^3 & abc \end{vmatrix} Taking abc common from C3, we get
\triangle = \frac{abc}{abc}\begin{vmatrix} 1 & a^3 & 1 \\ 1 & b^3 & 1 \\ 1 & c^3 & 1 \end{vmatrix} As C2 and C3 are identical
Hence, △ = 0
(v) \begin{vmatrix} a+b & 2a+b & 3a+b \\ 2a+b & 3a+b & 4a+b \\ 4a+b & 5a+b & 6a+b \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} a+b & 2a+b & 3a+b \\ 2a+b & 3a+b & 4a+b \\ 4a+b & 5a+b & 6a+b \end{vmatrix} C3⇢C3 - C2 and C2⇢C2 - C1
\triangle = \begin{vmatrix} a+b & a & a \\ 2a+b & a & a \\ 4a+b & a & a \end{vmatrix} As C2 and C3 are identical
Hence, △ = 0
(vi) \begin{vmatrix} 1 & a & a^2-bc \\ 1 & b & b^2-ac \\ 1 & c & c^2-ab \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} 1 & a & a^2-bc \\ 1 & b & b^2-ac \\ 1 & c & c^2-ab \end{vmatrix} Splitting the determinant, we have
\triangle = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}-\begin{vmatrix} 1 & a & bc \\ 1 & b & ac \\ 1 & c & ab \end{vmatrix} R2⇢R2-R1 and R3⇢R3-R1
\triangle = \begin{vmatrix} 1 & a & a^2 \\ 0 & b-a & b^2-a^2 \\ 0 & c-a & c^2-a^2 \end{vmatrix}-\begin{vmatrix} 1 & a & bc \\ 0 & b-a & ac-bc \\ 0 & c-a & ab-bc \end{vmatrix} \\ \triangle = \begin{vmatrix} 1 & a & a^2 \\ 0 & b-a & (b-a)(b+a) \\ 0 & c-a & (c-a)(c+a) \end{vmatrix}-\begin{vmatrix} 1 & a & bc \\ 0 & b-a & -(b-a)c \\ 0 & c-a & -(c-a)b \end{vmatrix} Taking (b-a) and (c-a) common from R2 and R3, we have
\triangle = (b-a)(c-a)\begin{vmatrix} 1 & a & a^2 \\ 0 & 1 & b+a \\ 0 & 1 & c+a \end{vmatrix}-(b-a)(c-a)\begin{vmatrix} 1 & a & bc \\ 0 & 1 & -c \\ 0 & 1 & -b \end{vmatrix} △ = (b - a)(c - a)(c + a - (b + a)) - (b - a)(c - a)(-b - (-c))
△ = (b - a)(c - a)(c + a - b - a) - (b - a)(c - a)(-b + c)
△ = (b - a)(c - a)(c - b) - (b - a)(c - a)(c - b)
Hence, △ = 0
(vii) \begin{vmatrix} 49 & 1 & 6 \\ 39 & 7 & 4 \\ 26 & 2 & 3 \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} 49 & 1 & 6 \\ 39 & 7 & 4 \\ 26 & 2 & 3 \end{vmatrix} C1⇢C1 - 8C3
\triangle = \begin{vmatrix} 1 & 1 & 6 \\ 7 & 7 & 4 \\ 2 & 2 & 3 \end{vmatrix} As C1 and C2 are identical
Hence, △ = 0
(viii) \begin{vmatrix} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{vmatrix} Multiplying and dividing by xyz, we have
\triangle = \frac{xyz}{xyz}\begin{vmatrix} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{vmatrix} Multiplying C1, C2 and C3 by z, y and x respectively
\triangle = \frac{1}{xyz}\begin{vmatrix} 0 & xy & yx \\ -xz & 0 & zx \\ -yz & -zy & 0 \end{vmatrix} Taking y, x and z common in R1, R2 and R3 respectively
\triangle = \frac{xyz}{xyz}\begin{vmatrix} 0 & x & x \\ -z & 0 & z \\ -y & -y & 0 \end{vmatrix} C2⇢C2 - C3
\triangle = \frac{xyz}{xyz}\begin{vmatrix} 0 & 0 & x \\ -z & -z & z \\ -y & -y & 0 \end{vmatrix} As C1 and C2 are identical
Hence, △ = 0
(ix) \begin{vmatrix} 1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2 \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} 1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2 \end{vmatrix} C2⇢C2 - 7C3
\triangle = \begin{vmatrix} 1 & 1 & 6 \\ 7 & 7 & 4 \\ 3 & 3 & 2 \end{vmatrix} As C1 and C2 are identical
Hence, △ = 0
(x) \begin{vmatrix} 1^2 & 2^2 & 3^2 & 4^2 \\ 2^2 & 3^2 & 4^2 & 5^5 \\ 3^3 & 4^4 & 5^2 & 6^2 \\ 4^2 & 5^2 & 6^2 &7^2 \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} 1^2 & 2^2 & 3^2 & 4^2 \\ 2^2 & 3^2 & 4^2 & 5^5 \\ 3^3 & 4^4 & 5^2 & 6^2 \\ 4^2 & 5^2 & 6^2 &7^2 \end{vmatrix} C3⇢C3 - C2 and C4⇢C4 - C1
\triangle = \begin{vmatrix} 1^2 & 2^2 & 3^2-2^2 & 4^2-1^2 \\ 2^2 & 3^2 & 4^2-3^2 & 5^2-2^2 \\ 3^2 & 4^2 & 5^2-4^2 & 6^2-3^2 \\ 4^2 & 5^2 & 6^2-5^2 &7^2-4^2 \end{vmatrix} \\ \triangle = \begin{vmatrix} 1^2 & 2^2 & 5 & 15 \\ 2^2 & 3^2 & 7 & 21 \\ 3^2 & 4^2 & 9 & 27 \\ 4^2 & 5^2 & 11 &33 \end{vmatrix} Taking 3 common from C3, we get
\triangle = 3\begin{vmatrix} 1^2 & 2^2 & 5 & 5 \\ 2^2 & 3^2 & 7 & 7 \\ 3^2 & 4^2 & 9 & 9 \\ 4^2 & 5^2 & 11 &11 \end{vmatrix} As C3 and C4 are identical
Hence, △ = 0
(xi) \begin{vmatrix} a & b & c \\ a+2x & b+2y & c+2z \\ x & y & z \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} a & b & c \\ a+2x & b+2y & c+2z \\ x & y & z \end{vmatrix} R3⇢R3 + R1 and R2⇢R2 + R1
\triangle = \begin{vmatrix} a & b & c \\ a+a+2x & b+b+2y & c+c+2z \\ a+x & b+y & c+z \end{vmatrix} \\ \triangle = \begin{vmatrix} a & b & c \\ 2a+2x & 2b+2y & 2c+2z \\ a+x & b+y & c+z \end{vmatrix} Taking 2 common from R2, we get
\triangle = 2\begin{vmatrix} a & b & c \\ a+x & b+y & c+z \\ a+x & b+y & c+z \end{vmatrix} As R2 and R3 are identical
Hence, △ = 0
Question 3. \begin{vmatrix} a & b+c & a^2 \\ b & c+a & b^2 \\ c & a+b & c^2 \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} a & b+c & a^2 \\ b & c+a & b^2 \\ c & a+b & c^2 \end{vmatrix} C2⇢C2+C1
\triangle = \begin{vmatrix} a & b+c+a & a^2 \\ b & c+a+b & b^2 \\ c & a+b+c & c^2 \end{vmatrix} Taking (a+b+c) common from C2, we get
\triangle = (a+b+c)\begin{vmatrix} a & 1 & a^2 \\ b & 1 & b^2 \\ c & 1 & c^2 \end{vmatrix} R3⇢R3-R1 and R2⇢R2-R1
\triangle = (a+b+c)\begin{vmatrix} a & 1 & a^2 \\ b-a & 0 & b^2-a^2 \\ c-a & 0 & c^2-a^2 \end{vmatrix} \\ \triangle = (a+b+c)\begin{vmatrix} a & 1 & a^2 \\ b-a & 0 & (b-a)(b+a) \\ c-a & 0 & (c-a)(c+a) \end{vmatrix} Taking (b - a) and (c - a) from R2 and R3, we have
\triangle = (a+b+c)\begin{vmatrix} a & 1 & a^2 \\ b-a & 0 & b^2-a^2 \\ c-a & 0 & c^2-a^2 \end{vmatrix} \\ \triangle = (a+b+c)(b-a)(c-a)\begin{vmatrix} a & 1 & a^2 \\ 1 & 0 & b+a \\ 1 & 0 & c+a \end{vmatrix} △ = (a + b + c)(b - a)(c - a)[1(b + a - (c + a))]
△ = (a + b + c)(b - a)(c - a)(b + a - c - a)
Hence, △ = (a + b + c)(b - a)(c - a)(b - c)
Question 4. \begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix} R3⇢R3-R1 and R2⇢R2-R1
\triangle = \begin{vmatrix} 1 & a & bc \\ 0 & b-a & ca-bc \\ 0 & c-a & ab-bc \end{vmatrix}\\ \triangle = \begin{vmatrix} 1 & a & bc \\ 0 & b-a & -c(b-a) \\ 0 & c-a & -b(c-a) \end{vmatrix}\\ Taking (b-a) and (c-a) from R2 and R3, we have
\triangle = (b-a)(c-a)\begin{vmatrix} 1 & a & bc \\ 0 & 1 & -c \\ 0 & 1 & -b \end{vmatrix}\\ △ = (b - a)(c - a)[1((1)(-b) - (1)(-c))]
△ = (b - a)(c - a)[-b - (-c)]
△ = (b - a)(c - a)[-b + c]
Hence, △ = (a - b)(b - c)(c - a)
Question 5. \begin{vmatrix} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{vmatrix} C1⇢C1+C2+C3
\triangle = \begin{vmatrix} 3x+\lambda & x & x \\ 3x+\lambda & x+\lambda & x \\ 3x+\lambda & x & x+\lambda \end{vmatrix} Taking (3x+λ) common from C1, we get
\triangle = (3x+\lambda)\begin{vmatrix} 1 & x & x \\ 1 & x+\lambda & x \\ 1 & x & x+\lambda \end{vmatrix} R3⇢R3-R1 and R2⇢R2-R1
\triangle = (3x+\lambda)\begin{vmatrix} 1 & x & x \\ 0 & x+\lambda-x & x-x \\ 0 & x-x & x+\lambda-x \end{vmatrix} \\ \triangle = (3x+\lambda)\begin{vmatrix} 1 & x & x \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{vmatrix} △ = (3x + λ)[λ(λ(1) - 0)]
△ = (3x + λ)[λ(λ)]
Hence, △ = λ2(3x + λ)
Question 6. \begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix} C1⇢C1 + C2 + C3
\triangle = \begin{vmatrix} a+b+c & b & c \\ c+a+b & a & b \\ b+c+a & c & a \end{vmatrix} Taking (a + b + c) common from C1, we get
\triangle = (a+b+c)\begin{vmatrix} 1 & b & c \\ 1 & a & b \\ 1 & c & a \end{vmatrix} R3⇢R3 - R1 and R2⇢R2 - R1
\triangle = (a+b+c)\begin{vmatrix} 1 & b & c \\ 0 & a-b & b-c \\ 0 & c-b & a-c \end{vmatrix} △ = (a + b + c)[1((a - b)(a - c) - (c - b)(b - c))]
△ = (a + b + c)[(a2 - ac - ab + bc) - (cb - c2 - b2 + bc)]
△ = (a + b + c)[a2 - ac - ab + bc + c2 + b2 - 2bc]
Henfce, △ = (a + b + c)[a2 + b2 + c2 - ac - ab - bc]
Question 7. \begin{vmatrix} sin\hspace{0.1cm}\alpha & cos\hspace{0.1cm}\alpha & cos(\alpha+\delta) \\ sin\hspace{0.1cm}\beta & cos\hspace{0.1cm}\beta & cos(\beta+\delta) \\ sin\hspace{0.1cm}\gamma & cos\hspace{0.1cm}\gamma & cos(\gamma+\delta) \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} sin\hspace{0.1cm}\alpha & cos\hspace{0.1cm}\alpha & cos(\alpha+\delta) \\ sin\hspace{0.1cm}\beta & cos\hspace{0.1cm}\beta & cos(\beta+\delta) \\ sin\hspace{0.1cm}\gamma & cos\hspace{0.1cm}\gamma & cos(\gamma+\delta) \end{vmatrix}
\triangle = \begin{vmatrix} sin\hspace{0.1cm}\alpha sin\hspace{0.1cm}\delta & cos\hspace{0.1cm}\alpha cos\hspace{0.1cm}\delta & cos(\alpha+\delta) \\ sin\hspace{0.1cm}\beta sin\hspace{0.1cm}\delta & cos\hspace{0.1cm}\beta cos\hspace{0.1cm}\delta & cos(\beta+\delta) \\ sin\hspace{0.1cm}\gamma sin\hspace{0.1cm}\delta & cos\hspace{0.1cm}\gamma cos\hspace{0.1cm}\delta & cos(\gamma+\delta) \end{vmatrix} C2⇢C2 - C1
\triangle = \begin{vmatrix} sin\hspace{0.1cm}\alpha sin\hspace{0.1cm}\delta & cos\hspace{0.1cm}\alpha cos\hspace{0.1cm}\delta - sin\hspace{0.1cm}\alpha sin\hspace{0.1cm}\delta & cos(\alpha+\delta) \\ sin\hspace{0.1cm}\beta sin\hspace{0.1cm}\delta & cos\hspace{0.1cm}\beta cos\hspace{0.1cm}\delta - sin\hspace{0.1cm}\beta sin\hspace{0.1cm}\delta & cos(\beta+\delta) \\ sin\hspace{0.1cm}\gamma sin\hspace{0.1cm}\delta & cos\hspace{0.1cm}\gamma cos\hspace{0.1cm}\delta - sin\hspace{0.1cm}\gamma sin\hspace{0.1cm}\delta & cos(\gamma+\delta) \end{vmatrix} Using the trigonometric identity,
cos a cos b - sin a sin b = cos (a + b)
\triangle = \begin{vmatrix} sin\hspace{0.1cm}\alpha sin\hspace{0.1cm}\delta & cos(\alpha+\delta) & cos(\alpha+\delta) \\ sin\hspace{0.1cm}\beta sin\hspace{0.1cm}\delta & cos(\beta+\delta) & cos(\beta+\delta) \\ sin\hspace{0.1cm}\gamma sin\hspace{0.1cm}\delta & cos(\gamma+\delta) & cos(\gamma+\delta) \end{vmatrix} As C2 and C3 are identical
Hence, △ = 0
Prove the following identities:
Question 8. \begin{vmatrix} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{vmatrix} = a3 + b2 + c3 - 3abc
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{vmatrix} R3⇢R3 + R1 and R2⇢R2 + R1
\triangle = \begin{vmatrix} a & b & c \\ a-b & b-c & c-a \\ b+c+a & c+a+b & a+b+c \end{vmatrix} Taking (a + b + c) common from R3, we get
\triangle = (a+b+c)\begin{vmatrix} a & b & c \\ a-b & b-c & c-a \\ 1 & 1 & 1 \end{vmatrix} R2⇢R2 - R1
\triangle = (a+b+c)\begin{vmatrix} a & b & c \\ a-b-a & b-c-b & c-a-c \\ 1 & 1 & 1 \end{vmatrix}\\ \triangle = (a+b+c)\begin{vmatrix} a & b & c \\ -b & -c & -a \\ 1 & 1 & 1 \end{vmatrix} Taking (-1) common from R2, we get
\triangle = (a+b+c)(-1)\begin{vmatrix} a & b & c \\ b & c & a \\ 1 & 1 & 1 \end{vmatrix} C1⇢C1 - C2 and C2⇢C2 - C3
\triangle = (a+b+c)(-1)\begin{vmatrix} a-b & b-c & c \\ b-c & c-a & a \\ 0 & 0 & 1 \end{vmatrix} △ = (-1)(a + b + c)[1((a - b)(c - a) - (b - c)(b - c))]
△ = (-1)(a + b + c)[(a - b)(c - a) - (b - c)2]
△ = (-1)(a + b + c)[(ac - a2 - bc + ab) - (b2 - 2cb + c2)]
△ = (-1)(a + b + c)(ac - a2 - bc + ab - b2 + 2cb - c2)
△ = (a + b + c)(-ac + a2 - bc - ab + b2 + c2)
△ = (a + b + c)(a2 + b2 + c2 - ac - ab - cb)
△ = a3 + b3 + c3 - 3abc
Hence proved
Question 9. \begin{vmatrix}b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{vmatrix} = 3abc - a3 - b2 - c3
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix}b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{vmatrix} C1⇢C1 + C3
\triangle = \begin{vmatrix}b+c+a & a-b & a \\ c+a+b & b-c & b \\ a+b+c & c-a & c \end{vmatrix} Taking (a + b + c) common from C1, we get
\triangle = (a+b+c)\begin{vmatrix}1 & a-b & a \\ 1 & b-c & b \\ 1 & c-a & c \end{vmatrix} △ = (a + b + c)[1((b - c)c - b(c - a)) - 1((a - b)c - a(c - a)) + 1(b(a - b) - a(b - c))]
△ = (a + b + c)[(b - c)c - b(c - a) - (a - b)c + a(c - a) + b(a - b) - a(b - c)]
△ = (a + b + c)[(bc - c2-bc + ab) - (ac - bc) + ac - a2 + ab - b2 - (ab - ac)]
△ = (a + b + c)[bc - c2 - bc + ab - ac + bc + ac - a2 + ab - b2 - ab + ac]
△ = (a + b + c)[bc - c2 + ab + ac - a2 - b2]
△ = (a + b + c)[bc + ab + ac - a2 - b2 - c2]
△ = 3abc - a3 - b3 - c3
Hence proved
Question 10. \begin{vmatrix}a+b & b+c & c+a \\ b+c & c+a & a+b \\ c+a & a+b& b+c \end{vmatrix} = 2\begin{vmatrix}a & b & c \\ b & c & a \\ c & a& b \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix}a+b & b+c & c+a \\ b+c & c+a & a+b \\ c+a & a+b& b+c \end{vmatrix} C1⇢C1 + C2 + C3
\triangle = \begin{vmatrix}a+b+b+c+c+a & b+c & c+a \\ b+c+c+a+a+b & c+a & a+b \\ c+a+a+b+b+c & a+b& b+c \end{vmatrix}\\ \triangle = \begin{vmatrix}2(a+b+c) & b+c & c+a \\ 2(a+b+c) & c+a & a+b \\ 2(a+b+c) & a+b& b+c \end{vmatrix} Taking 2 common from C1, we get
\triangle = 2\begin{vmatrix}a+b+c & b+c & c+a \\ a+b+c & c+a & a+b \\ a+b+c & a+b& b+c \end{vmatrix} C2⇢C2 - C1 and C3⇢C3 - C1
\triangle = 2\begin{vmatrix}a+b+c & b+c-(a+b+c) & c+a-(a+b+c) \\ a+b+c & c+a-(a+b+c) & a+b-(a+b+c) \\ a+b+c & a+b-(a+b+c) & b+c-(a+b+c) \end{vmatrix}\\ \triangle = 2\begin{vmatrix}a+b+c & -a & -b \\ a+b+c & -b & -c \\ a+b+c & -c & -a \end{vmatrix}\\ Taking (-1) and (-1) common from C2 and C3,
\triangle = 2(-1)(-1)\begin{vmatrix}a+b+c & a & b \\ a+b+c & b & c \\ a+b+c & c & a \end{vmatrix}\\ \triangle = 2\begin{vmatrix}a+b+c & a & b \\ a+b+c & b & c \\ a+b+c & c & a \end{vmatrix}\\ By splitting the determinant, we get
\triangle = 2\begin{vmatrix}c & a & b \\ a & b & c \\ b & c & a \end{vmatrix}+2\begin{vmatrix}a & a & b \\ b & b & c \\ c & c & a \end{vmatrix}+2\begin{vmatrix}b & a & b \\ c & b & c \\ a & c & a \end{vmatrix}\\ \triangle = 2\begin{vmatrix}c & a & b \\ a & b & c \\ b & c & a \end{vmatrix}+2(0)+2(0)\\ \triangle = 2\begin{vmatrix}c & a & b \\ a & b & c \\ b & c & a \end{vmatrix}+0+0\\ \triangle = 2\begin{vmatrix}c & a & b \\ a & b & c \\ b & c & a \end{vmatrix} Hence proved
Question 11. \begin{vmatrix}a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix} = 2(a + b + c)3
Considering the determinant, we have
\triangle = \begin{vmatrix}a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix} C1⇢C1 + C2 + C3
\triangle = \begin{vmatrix}2a+2b+2c & a & b \\ 2b+2c+2a & b+c+2a & b \\ 2a+2b+2c & a & c+a+2b \end{vmatrix} Taking (2a + 2b + 2c) common from C1, we get
\triangle = (2a+2b+2c)\begin{vmatrix}1 & a & b \\ 1 & b+c+2a & b \\ 1 & a & c+a+2b \end{vmatrix}\\ \triangle = 2(a+b+c)\begin{vmatrix}1 & a & b \\ 1 & b+c+2a & b \\ 1 & a & c+a+2b \end{vmatrix} R2⇢R2 - R1 and R3⇢R3 - R1
\triangle = 2(a+b+c)\begin{vmatrix}1 & a & b \\ 0 & b+c+2a-a & b-b \\ 0 & a-a & c+a+2b-b \end{vmatrix}\\ \triangle = 2(a+b+c)\begin{vmatrix}1 & a & b \\ 0 & b+c+a & 0 \\ 0 & 0 & c+a+b \end{vmatrix} △ = 2(a + b + c)[1((a + b + b)(a + b + c) - 0)]
△ = 2(a + b + c)[(a + b + b)2]
△ = 2(a + b + c)3
Hence proved
Question 12. \begin{vmatrix}a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} = (a + b + c)3
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix}a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} R1⇢R1 + R2 + R3
\triangle = \begin{vmatrix}a-b-c+2b+2c & 2a+b-c-a+2c & 2a+2b+c-a-b \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}\\ \triangle = \begin{vmatrix}a+b+c & a+b+c & a+b+c \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} Taking (a + b + c) common from R1, we get
\triangle = (a+b+c)\begin{vmatrix}1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} C2⇢C2 - C1 and C3⇢C3 - C1
\triangle = (a+b+c)\begin{vmatrix}1 & 0 & 0 \\ 2b & b-c-a-2b & 2b-2b \\ 2c & 2c-2c & c-a-b-2c \end{vmatrix}\\ \triangle = (a+b+c)\begin{vmatrix}1 & 0 & 0 \\ 2b & -c-a-b & 0 \\ 2c & 0 & -a-b-c \end{vmatrix} △ = (a + b + c)[1((-b - c - a)(-b - c - a) - 0)]
△ = (a + b + c)[(b + c + a)(b + c + a)]
△ = (a + b + c)[(b + c + a)2]
△ = (a + b + c)3
Hence proved
Question 13. \begin{vmatrix}1 & b+c & b^2+c^2 \\ 1 & c+a & c^2+a^2 \\ 1 & a+b & a^2+b^2 \end{vmatrix} = (a - b)(b - c)(c - a)
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix}1 & b+c & b^2+c^2 \\ 1 & c+a & c^2+a^2 \\ 1 & a+b & a^2+b^2 \end{vmatrix} R2⇢R2 - R1 and R3⇢R3 - R1
\triangle = \begin{vmatrix}1 & b+c & b^2+c^2 \\ 0 & a-b & a^2-b^2 \\ 0 & a-c & a^2-c^2 \end{vmatrix}\\ \triangle = \begin{vmatrix}1 & b+c & b^2+c^2 \\ 0 & a-b & (a-b)(a+b) \\ 0 & a-c & (a-c)(a+c) \end{vmatrix} Taking (a - b) and (a - c) common from R2 and R3 respectively, we get
\triangle =(a-b)(a-c) \begin{vmatrix}1 & b+c & b^2+c^2 \\ 0 & 1 & a+b \\ 0 & 1 & a+c \end{vmatrix} △ = (a - b)(a - c)[1(1(a + c) - 1(a + b))]
△ = (a - b)(a - c)[(a + c) - (a + b)]
△ = (a - b)(a - c)[a + c - a - b]
△ = (a - b)(a - c)[c - b]
△ = (a - b)(a - c)(c - b)
△ = (a - b)(b - c)(c - a)
Hence proved
Question 14. \begin{vmatrix}a & a+b & a+2b \\ a+2b & a & a+b \\ a+b & a+2b & a \end{vmatrix} = 9(a + b)b2
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix}a & a+b & a+2b \\ a+2b & a & a+b \\ a+b & a+2b & a \end{vmatrix} C1⇢C1 + C2 + C3
\triangle = \begin{vmatrix}3a+3b & 3a+3b & 3a+3b \\ a+2b & a & a+b \\ a+b & a+2b & a \end{vmatrix} Taking (3a + 3b) common from R1, we get
\triangle = (3a+3b)\begin{vmatrix}1 & 1 & 1 \\ a+2b & a & a+b \\ a+b & a+2b & a \end{vmatrix} C2⇢C2 - C1 and C3⇢C3 - C2
\triangle = (3a+3b)\begin{vmatrix}1 & 0 & 0 \\ a+2b & a-(a+2b) & a+b-a \\ a+b & a+2b-(a+b) & a-(a+2b) \end{vmatrix}\\ \triangle = (3a+3b)\begin{vmatrix}1 & 0 & 0 \\ a+2b & -2b & b \\ a+b & b & -2b \end{vmatrix} △ = 3(a + b)[1((-2b)(-2b) - b(b))]
△ = 3(a + b)[4b2 - b2]
△ = 3(a + b)[3b2]
△ = 9(a + b)b2
Hence proved
Question 15. \begin{vmatrix}1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix} = \begin{vmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix}1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix} R1⇢aR1, R2⇢bR2 and R3⇢cR3
\triangle = \frac{1}{abc}\begin{vmatrix}a & a^2 & abc \\ b & b^2 & bca \\ c & c^2 & cab \end{vmatrix} Taking (abc) common from C3, we get
\triangle = \frac{abc}{abc}\begin{vmatrix}a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix}\\ \triangle = -\begin{vmatrix}a & 1 & a^2 \\ b & 1 & b^2 \\ c & 1 & c^2 \end{vmatrix}\\ \triangle = \begin{vmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} Hence proved
Question 16. \begin{vmatrix}z & x & y \\ z^2 & x^2 & y^2 \\ z^4 & x^4 & y^4 \end{vmatrix} = \begin{vmatrix}x & y & z \\ x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4 \end{vmatrix}=\begin{vmatrix} z^2 & x^2 & y^2 \\ x^4 & y^4 & z^4 \\x & y & z \end{vmatrix} = xyz(x - y)(y - z)(z - x)(x + y + z)
Solution:
Considering the determinant, we have
C1↔C2 and then
\triangle = \begin{vmatrix}z & x & y \\ z^2 & x^2 & y^2 \\ z^4 & x^4 & y^4 \end{vmatrix}\\ \triangle = - \begin{vmatrix}x & z & y \\ x^2 & z^2 & y^2 \\ x^4 & z^4 & y^4 \end{vmatrix} C2↔C3
\triangle = \begin{vmatrix}x & y & z \\ x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4 \end{vmatrix} R1↔R2
\triangle = -\begin{vmatrix}x^2 & y^2 & z^2 \\ x & y & z \\ x^4 & y^4 & z^4 \end{vmatrix} R2↔R3
\triangle = \begin{vmatrix}x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4 \\ x & y & z \end{vmatrix} Taking,
\triangle = \begin{vmatrix}x & y & z \\ x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4 \end{vmatrix} Taking x, y and z common from C1, C2 and C3 respectively
\triangle = (xyz)\begin{vmatrix}1 & 1 & 1 \\ x & y & z \\ x^3 & y^3 & z^3 \end{vmatrix} C1⇢C1 - C2 and C3⇢C3 - C2
\triangle = (xyz)\begin{vmatrix}0 & 1 & 0 \\ x-y & y & z-y \\ x^3-y^3 & y^3 & z^3-y^3 \end{vmatrix}\\ \triangle = (xyz)\begin{vmatrix}0 & 1 & 0 \\ x-y & y & z-y \\ (x-y)(x^2+xy+y^2) & y^3 & (z-y)(z^2+zy+y^2) \end{vmatrix} Taking (x - y) and (z - y) common from C1 and C3 respectively, we get
\triangle = (xyz)(x-y)(z-y)\begin{vmatrix}0 & 1 & 0 \\ 1 & y & 1 \\ x^2+xy+y^2 & y^3 & z^2+zy+y^2 \end{vmatrix} △ = (xyz)(x - y)(z - y)[1(1(z2 + zy + y2) - 1(x2 + xy + y2))]
△ = (xyz)(x - y)(z - y)[z2 + zy + y2 - (x2 + xy + y2)]
△ = (xyz)(x - y)(z - y)[z2 + zy + y2 - x2 - xy - y2]
△ = (xyz)(x - y)(z - y)[z2 + zy - x2 - xy]
△ = (xyz)(x - y)(z - y)[z2 - x2 + zy - xy]
△ = (xyz)(x - y)(z - y)[(z - x)(z + x) + y(z - x)]
△ = (xyz)(x - y)(z - y)(z - x)[z + x + y]
△ = (xyz)(x - y)(z - y)(z - x)(x + y + z)
\triangle=\begin{vmatrix}z & x & y \\ z^2 & x^2 & y^2 \\ z^4 & x^4 & y^4 \end{vmatrix} = \begin{vmatrix}x & y & z \\ x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4 \end{vmatrix}=\begin{vmatrix} z^2 & x^2 & y^2 \\ x^4 & y^4 & z^4 \\x & y & z \end{vmatrix}=xyz(x-y)(y-z)(z-x)(x+y+z) Hence proved
Question 17. \begin{vmatrix}(b+c)^2 & a^2 & bc \\ (c+a)^2 & b^2 & ca \\ (a+b)^2 & c^2 & ab \end{vmatrix} = (a - b)(b - c)(c - a)(a + b + c)(a2 + b2 + c2)
Solution:
Considering the determinant, we have
\triangle = \begin{vmatrix}(b+c)^2 & a^2 & bc \\ (c+a)^2 & b^2 & ca \\ (a+b)^2 & c^2 & ab \end{vmatrix} C1⇢C1 + C2 - 2C3
\triangle = \begin{vmatrix}(b+c)^2+a^2-2bc & a^2 & bc \\ (c+a)^2+b^2-2ca & b^2 & ca \\ (a+b)^2+c^2-2ab & c^2 & ab \end{vmatrix}\\ \triangle = \begin{vmatrix}a^2+b^2+c^2 & a^2 & bc \\ a^2+b^2+c^2 & b^2 & ca \\ a^2+b^2+c^2 & c^2 & ab \end{vmatrix} Taking (a2 + b2 + c2) common from C1, we get
\triangle = (a^2+b^2+c^2)\begin{vmatrix}1 & a^2 & bc \\ 1 & b^2 & ca \\ 1 & c^2 & ab \end{vmatrix} C2⇢C2-C1 and C3⇢C3-C1
\triangle = (a^2+b^2+c^2)\begin{vmatrix}1 & a^2 & bc \\ 0 & b^2-a^2 & ca-bc \\ 0 & c^2-a^2 & ab-bc \end{vmatrix}\\ \triangle = (a^2+b^2+c^2)\begin{vmatrix}1 & a^2 & bc \\ 0 & (b-a)(b+a) & -c(b-a) \\ 0 & (c-a)(c+a) & -b(c-a) \end{vmatrix} Taking (b - a) and (c - a) common from R2 and R3, we get
\triangle = (a^2+b^2+c^2)(b-a)(c-a)\begin{vmatrix}1 & a^2 & bc \\ 0 & b+a & -c \\ 0 & c+a & -b \end{vmatrix} △ = (a2 + b2 + c2)(b - a)(c - a)[1((b + a)(-b) - (c + a)(-c))]
△ = (a2 + b2 + c2)(b - a)(c - a)[(b + a)(-b) + (c + a)c]
△ = (a2 + b2 + c2)(b - a)(c - a)[(-b2 - ab) + (c2 + ac)]
△ = (a2 + b2 + c2)(b - a)(c - a)[c2 - b2 - ab + ac]
△ = (a2 + b2 + c2)(b - a)(c - a)[(c - b)(c + b) + a(c - b)]
△ = (a2 + b2 + c2)(b - a)(c - a)(c - b)[c + b + a]
△ = (a2 + b2 + c2)(a + b + c)(a - b)(b - c)(c - a)
Hence proved
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Summary
Exercise 6.2 in RD Sharma's Class 12 Mathematics textbook focuses on properties and applications of determinants. This exercise covers:
- Evaluation of 3x3 determinants
- Properties of determinants (e.g., effect of row/column interchange)
- Expansion of determinants
- Solving equations using determinants
- Proving identities involving determinants
- Application of determinant properties to solve problems