Solved A C 2a 2a Prove That 2b B A 2b A B Cp 2c 2c C A B

Prove That A B C 2a 2a 2b B C A 2b 2c 2c C A B A (a b c)³. step by step explanation: given matrix is . now, we solve the matrix from the row and column method = (a b c)[(b c a)(c a b) 4bc] 2a[2b(c a b) 4bc] 2a[4bc (b c a)2c] = this is the formula of . hence, it is proved. It presents comprehensive coverage of questions along with solutions that are helpful in preparation of jee main and jee advanced . the students can also evaluate their problem solving skills by.

Solved A C 2a 2a Prove That 2b B A 2b A B Cp 2c 2c C A B To ask unlimited maths doubts download doubtnut from goo.gl 9wzjcw prove that `[ [a b c , 2a , 2a ] , [2b , b c a , 2b ] ,[2c ,2c,c a b]]`= `(a b c. Prove that: \begin {vmatrix} a b c & 2a & 2a \ 2b & b c a & 2b \ 2c & 2c & c a b \end {vmatrix} = (a b c)^3. 🤔 not the exact question you're looking for? the determinant of a matrix can be calculated using various properties, including cofactor expansion and row column operations. Using properties of determinants, prove that |(a, a b, a b c), (2a, 3a 2b, 4a 3b 2c), (3a, 6a 3b, 10a 6b 3c)| = a^3. Using properties of determinants, show that Δabc is isosceles if:`|[1,1,1],[1 cosa,1 cosb,1 cosc],[cos^2a cosa,cos^b cosb,cos^2c cosc]|=0 ` using the property of determinants and without expanding, prove that: `|(a b,b c,c a),(b c,c a,a b),(a a,a b,b c)| = 0` by using properties of determinants, show that:.

Prove That Determinant B C 2a2 A2 B2 C A 2 B2 2abc A B C 3 C2 C2 A Using properties of determinants, prove that |(a, a b, a b c), (2a, 3a 2b, 4a 3b 2c), (3a, 6a 3b, 10a 6b 3c)| = a^3. Using properties of determinants, show that Δabc is isosceles if:`|[1,1,1],[1 cosa,1 cosb,1 cosc],[cos^2a cosa,cos^b cosb,cos^2c cosc]|=0 ` using the property of determinants and without expanding, prove that: `|(a b,b c,c a),(b c,c a,a b),(a a,a b,b c)| = 0` by using properties of determinants, show that:. If | (a b c,2a,2a), (2b, b c a,2b), (2c,2c, c a b)| = (a b c) (x a b c)2, x ≠ 0 and a b c ≠ 0, then x is equal to : (1) 2 (a b c) (2) –2 (a b c) (3) abc. (4) – (a b c) if | (a b c,2a,2a), (2b, b c a,2b), (2c,2c, c a b)| = (a b c) (x a b c)2, x ≠ 0 and a b 2) 2 (a b c) (3) abc (4) (a b c). If [(2a,x1,y1),(2b,x2,y2),(2c,x3,y3)] = abc 2 ≠ 0, then the area of the triangle whose vertices are ((x1 a,y1 a)), ((x2 b,y2 b)), ((x3 c,y3 c)) is. Question 6 prove that | 8(𝑎&𝑎 𝑏&𝑎 𝑏 𝑐@2𝑎&3𝑎 2𝑏&4𝑎 3𝑏 2𝑐@3𝑎&6𝑎 3𝑏&10𝑎 6𝑏 3𝑐)| = a3 let Δ = | 8(𝑎&𝑎 𝑏&𝑎 𝑏 𝑐@2𝑎&3𝑎 2𝑏&4𝑎 3𝑏 2𝑐@3𝑎&6𝑎 3𝑏&10𝑎 6𝑏 3𝑐)| applying r2 → r2 – 2r1 = | 8(𝑎&𝑎 𝑏&𝑎 𝑏 𝑐@𝟐𝒂−𝟐(𝒂)&3𝑎. Expanding this determinant along column 1, we get: hence proved. was this answer helpful? click a picture with our app and get instant verified solutions.

Prove The Following Identities A B C 2a 2a 2b B C A 2b 2c 2c C A If | (a b c,2a,2a), (2b, b c a,2b), (2c,2c, c a b)| = (a b c) (x a b c)2, x ≠ 0 and a b c ≠ 0, then x is equal to : (1) 2 (a b c) (2) –2 (a b c) (3) abc. (4) – (a b c) if | (a b c,2a,2a), (2b, b c a,2b), (2c,2c, c a b)| = (a b c) (x a b c)2, x ≠ 0 and a b 2) 2 (a b c) (3) abc (4) (a b c). If [(2a,x1,y1),(2b,x2,y2),(2c,x3,y3)] = abc 2 ≠ 0, then the area of the triangle whose vertices are ((x1 a,y1 a)), ((x2 b,y2 b)), ((x3 c,y3 c)) is. Question 6 prove that | 8(𝑎&𝑎 𝑏&𝑎 𝑏 𝑐@2𝑎&3𝑎 2𝑏&4𝑎 3𝑏 2𝑐@3𝑎&6𝑎 3𝑏&10𝑎 6𝑏 3𝑐)| = a3 let Δ = | 8(𝑎&𝑎 𝑏&𝑎 𝑏 𝑐@2𝑎&3𝑎 2𝑏&4𝑎 3𝑏 2𝑐@3𝑎&6𝑎 3𝑏&10𝑎 6𝑏 3𝑐)| applying r2 → r2 – 2r1 = | 8(𝑎&𝑎 𝑏&𝑎 𝑏 𝑐@𝟐𝒂−𝟐(𝒂)&3𝑎. Expanding this determinant along column 1, we get: hence proved. was this answer helpful? click a picture with our app and get instant verified solutions.