Prove That A B C 2a 2a 2b B C A 2b 2c 2c C A Babc3

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. 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:.

Q If A2 B C B2 C A C2 A B X A B B C C A Then \(\begin{vmatrix} a b c & 2a & 2a \\ 2b & b c a & 2b \\ 2c & 2c & c a b \end{vmatrix} \). 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. 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. 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.

Prove Using Properties Of Determinants A B C 2a 2a 2b B C A 2a 2c 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. 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. 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. Define $f(a,b,c)=a^2b b^2c c^2a ab bc ca$. since the triangle $a b c=3$ with $a,b,c \ge 0$ is compact, f has a global minimum on this domain. if this minimum lies on the boundary, wlog at $a=0$, it's easy to see that $f \ge 0$ on this line. As the equation is somehow symmetrical, you can use $$a^3 b^3 c^3=a^2b b^2c c^2a (c^2 b^2)(c a) (a^2 b^2)(a b)$$ in this case. $\endgroup$ –. 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.