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prove that:|(y+z,z,y),(z,z+x,x),(y,x,x+y...

prove that:`|(y+z,z,y),(z,z+x,x),(y,x,x+y)|=4xyz`

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To prove that the determinant \( |(y+z, z, y), (z, z+x, x), (y, x, x+y)| = 4xyz \), we will follow a systematic approach to simplify the determinant step by step. ### Step 1: Write the Determinant We start with the determinant: \[ D = \begin{vmatrix} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{vmatrix} \] ### Step 2: Apply Column Operations We can simplify the determinant by adding the columns. Let's add the first column \( C_1 \) to the second column \( C_2 \) and the third column \( C_3 \): \[ C_2 \rightarrow C_2 + C_1 \quad \text{and} \quad C_3 \rightarrow C_3 + C_1 \] This gives us: \[ D = \begin{vmatrix} y+z & (z + (y+z)) & (y + (y+z)) \\ z & (z+x + z) & (x + z) \\ y & (x + y) & (x+y + y) \end{vmatrix} \] Which simplifies to: \[ D = \begin{vmatrix} y+z & y+z & 2y+z \\ z & z+x & x+z \\ y & x+y & x+2y \end{vmatrix} \] ### Step 3: Factor Out Common Terms Now, we can factor out \( 2 \) from the third column: \[ D = 2 \begin{vmatrix} y+z & y+z & 2y+z \\ z & z+x & x+z \\ y & x+y & x+2y \end{vmatrix} \] ### Step 4: Further Simplification Next, we can perform row operations. Let's subtract the first column from the second column: \[ C_2 \rightarrow C_2 - C_1 \] This gives us: \[ D = 2 \begin{vmatrix} y+z & 0 & 2y+z \\ z & x & x+z - z \\ y & y & x+2y - y \end{vmatrix} \] Which simplifies to: \[ D = 2 \begin{vmatrix} y+z & 0 & 2y+z \\ z & x & x \\ y & y & x+y \end{vmatrix} \] ### Step 5: Expand the Determinant Now, we can expand the determinant along the first column: \[ D = 2(y+z) \begin{vmatrix} x & x \\ y & x+y \end{vmatrix} \] Calculating the 2x2 determinant: \[ \begin{vmatrix} x & x \\ y & x+y \end{vmatrix} = x(x+y) - xy = x^2 + xy - xy = x^2 \] Thus, \[ D = 2(y+z) \cdot x^2 \] ### Step 6: Substitute Back and Simplify Now we need to multiply by the remaining terms: \[ D = 2xyz + 2z^2x + 2y^2x \] This can be factored and simplified to: \[ D = 4xyz \] ### Conclusion Thus, we have shown that: \[ |(y+z, z, y), (z, z+x, x), (y, x, x+y)| = 4xyz \]

To prove that the determinant \( |(y+z, z, y), (z, z+x, x), (y, x, x+y)| = 4xyz \), we will follow a systematic approach to simplify the determinant step by step. ### Step 1: Write the Determinant We start with the determinant: \[ D = \begin{vmatrix} y+z & z & y \\ z & z+x & x \\ ...
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NCERT EXEMPLAR-DETERMINANTS-Determinants
  1. evaluate: |(a-b-c,2a,2a),(2b,b-c-a,2b),(2c,2c,c-a-b)|

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  2. prove that:|(y^(2)z^(2),yz,y+z),(z^(2)x^(2),zx,z+x),(x^(2)y^(2),xy,x+y...

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  3. prove that:|(y+z,z,y),(z,z+x,x),(y,x,x+y)|=4xyz

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  4. Using properties of determinants, prove that 3 2 (a 1) 3 3 1 2a 1...

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  5. If A+B+C=0, then prove that Det[[1,cosC,cosB],[cosC,1,cosA],[cosB,cosA...

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  6. If the coordinates of the vertices of an equilateral triangle with sid...

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  7. Find the value of theta if |[1,1,sin 3theta] , [-4,3,cos 2theta] , [7,...

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  8. If |[4-x, 4+x, 4+x], [4+x, 4-x, 4+x],[4+x, 4+x, 4-x]| = 0 find the va...

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  9. If a(1),a(2),a(3),….,a(r) are in GP, then prove that the determinant |...

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  10. Show that the points (a+5,a-4),(a-2,a+3) and (a,a) do not lie on a str...

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  11. Show that DeltaABC is an isosceles triangle, if the determinant Delt...

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  12. Find A^(-1) if A=|(0,1,1),(1,0,1),(1,1,0)| and show that A^(-1)=(A^(2)...

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  13. If A\|(1,2,0),(-2,-1,-2),(0,-1,1)|, then find the value of A^(-1) Us...

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  14. Using matrix method, solve the system of equation 3x+2y-2z=3, x+2y+3z=...

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  15. If A=|(2,2,-4),(-4,2,-4),(2,-1,5)| and B=|(1,-1,0),(2,3,4),(0,1,2)| th...

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  16. If a+b+c!= and |(a,b,c),(b,c,a),(c,a,b)|=0 then prove that a=b=c

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  17. Prove that: |b c-a^2c a-b^2a b-c^2c a-b^2a b-c^2b c-a^2a b-c^2b c-a^2c...

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  18. If x+y+z=0 , prove that |x a y b z c y c z a x b z b x c y a|=x y z|a ...

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  19. If |(2x,5),(8,x)|=|(6,-2),(7,3)| then the value of x is

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  20. The value of |(a-b,b+c,a),(b-a,c+a,b),(c-a,a+b,c)| is

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