मराठी

By Using Properties of Determinants, Show That: |(X,Xsqrt2,Yz),(Y,Ysqrt2,Zx),(Z,Zsqrt2,Xy)| = (X-y)(Y-z)(Z-x)(Xy+Yz+Zx) - Mathematics

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प्रश्न

By using properties of determinants, show that:

`|(x,x^2,yz),(y,y^2,zx),(z,z^2,xy)| = (x-y)(y-z)(z-x)(xy+yz+zx)`

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उत्तर

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पाठ 4: Determinants - Exercise 4.2 [पृष्ठ १२०]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
पाठ 4 Determinants
Exercise 4.2 | Q 9 | पृष्ठ १२०

संबंधित प्रश्‍न

 

If ` f(x)=|[a,-1,0],[ax,a,-1],[ax^2,ax,a]| ` , using properties of determinants find the value of f(2x) − f(x).

 

Using the property of determinants and without expanding, prove that:

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Using the property of determinants and without expanding, prove that:

`|(1, bc, a(b+c)),(1, ca, b(c+a)),(1, ab, c(a+b))| = 0`


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`|(y+k,y, y),(y, y+k, y),(y, y, y+k)| = k^2(3y + k)`


By using properties of determinants, show that:

`|(a-b-c, 2a,2a),(2b, b-c-a,2b),(2c,2c, c-a-b)| = (a + b + c)^2`


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Prove the following using properties of determinants :

\[\begin{vmatrix}a + b + 2c & a & b \\ c & b + c + 2a & b \\ c & a & c + a + 2b\end{vmatrix} = 2\left( a + b + c \right) {}^3\]


Using properties of determinants, prove the following:

\[\begin{vmatrix}x^2 + 1 & xy & xz \\ xy & y^2 + 1 & yz \\ xz & yz & z^2 + 1\end{vmatrix} = 1 + x^2 + y^2 + z^2\] .

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Without expanding evaluate the following determinant:

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Without expanding determinants find the value of `|(10,57,107),(12,64,124),(15,78,153)|`


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