Question

[tex] If D₁=\left|\begin{array}{ccc}1&yz&x\\1&zx&y\\1&xy&z\end{array}\right| and D₂=\left|\begin{array}{ccc}1&1&1\\x&y&z\\x²&y²&z²\end{array}\right| then......,Select Proper option from the given options.(a) D₁+2D₂=0(b) 2D₁+D₂=0(c) D₁+D₂=0(d) D₁=+D₂ [\tex]

Answer 1

Answer:

Option C is correct.

Step-by-step explanation:

$D_1=\left|\begin{array}{ccc}1&yz&x\\1&zx&y\\1&xy&z\end{array}\right|$

R1 -> R1 - R2

R2 -> R2 - R3

$D_1=\left|\begin{array}{ccc}1-1&yz-zx&x-y\\1-1&zx-xy&y-z\\1&xy&z\end{array}\right|$

Take common (x-y) from R1 and (y-z) from R2

$D_1=(x-y)(y-z)\left|\begin{array}{ccc}0&-z&1\\0&-x&1\\1&xy&z\end{array}\right|$

Expand the determinant along C1

$D_1 = (x - y)(y - z)(x - z)...eq1$

$D_2=\left|\begin{array}{ccc}1&1&1\\x&y&z\\x^2&y^2&z^2\end{array}\right|$

C1 -> C1 - C2

C2 -> C2 - C3

$D_2=\left|\begin{array}{ccc}0&0&1\\x-y&y-z&z\\(x+y)(x-y)&(y+z)(y-z)&z^2\end{array}\right|$

Take common (x-y) from C1 and (y-z) from C2

$D_2=(x-y)(y-z)\left|\begin{array}{ccc}0&0&1\\1&1&z\\(x+y)&(y+z)&z^2\end{array}\right|$

C1 -> C1-C2

$D_2=(x-y)(y-z)\left|\begin{array}{ccc}0&0&1\\0&1&z\\x-z&y+z&z^2\end{array}\right|$

Expand the determinant along R1

$D_2 = - (x - y)(y - z)(x - z)...eq2$

So, it is clear from eq1 and eq2,that

$D_1 + D_2 = 0 \\ \\(x - y)(y - z)(x - z)+[-(x - y)(y - z)(x - z)]=0$

Option C is correct.

Hope it helps you.