Ncert Solutions Maths class 12th

LHS $\left|\begin{array}{ccc}\hfill x\hfill & \hfill x^2\hfill & \hfill yz\hfill \\ \hfill y\hfill & \hfill y^2\hfill & \hfill zx\hfill \\ \hfill z\hfill & \hfill z^2\hfill & \hfill xy\hfill \end{array}\right|$

$=\frac{1}{xyz}$ $\left|\begin{array}{ccc}\hfill x·x\hfill & \hfill x·x^2\hfill & \hfill x·yz\hfill \\ \hfill y·y\hfill & \hfill y·y^2\hfill & \hfill y·zx\hfill \\ \hfill z·z\hfill & \hfill z·z^2\hfill & \hfill z·xy\hfill \end{array}\right|$ Multiplying R1, R2& R3 by x, y&z.

$=\frac{1}{xyz}\left|\begin{array}{ccc}\hfill x^2\hfill & \hfill x^3\hfill & \hfill xyz\hfill \\ \hfill y^2\hfill & \hfill y^3\hfill & \hfill xyz\hfill \\ \hfill z^2\hfill & \hfill z^3\hfill & \hfill xyz\hfill \end{array}\right|.$

$=\frac{xyz}{xyz}\left|\begin{array}{ccc}\hfill x^2\hfill & \hfill x^3\hfill & \hfill 1\hfill \\ \hfill y^2\hfill & \hfill y^3\hfill & \hfill 1\hfill \\ \hfill z^2\hfill & \hfill z^3\hfill & \hfill 1\hfill \end{array}\right|$ Taking xyz common from c3.

$=\left|\begin{array}{ccc}\hfill x^2\hfill & \hfill x^3\hfill & \hfill 1\hfill \\ \hfill y^2-x^2\hfill & \hfill y^3-x^3\hfill & \hfill 1-1\hfill \\ \hfill z^2-x^2\hfill & \hfill 2^3-x^3\hfill & \hfill 1-1\hfill \end{array}\right|\begin{array}{cc}\hfill {}_2\to {}_2-{}_1\hfill & \hfill {}_3\to {}_3-{}_1.\hfill \end{array}$

= $\left|\begin{array}{ccc}\hfill x^2\hfill & \hfill x^3\hfill & \hfill 1\hfill \\ \hfill (y-x)(y+x)\hfill & \hfill (y-x)\left(y^2+x^2+xy\right)\hfill & \hfill 0\hfill \\ \hfill (z-x)(z+x)\hfill & \hfill (2-x)\left(z^2+x^2+2x\right)\hfill & \hfill 0\hfill \end{array}\right|$

Expanding along c3 we get,

= 1 $\left|\begin{array}{cc}\hfill (y-x)(y+x)\hfill & \hfill (y-x)\left(y^2+x^2+xy\right)\hfill \\ \hfill (z-x)(z+x)\hfill & \hfill (z-x)\left(z^2+x^2+zx\right)\hfill \end{array}\right|$

= (y-x)(z-x). $\left|\begin{array}{cc}\hfill y+x\hfill & \hfill y^2+x^2+xy\hfill \\ \hfill z+x\hfill & \hfill z^2+x^2+zx\hfill \end{array}\right|$ Taking (yx) & (zx) common from R₁ and R₂.

= (y-x)(z-x).  Taking (y-x) & (z-x) common from R₁ and R₂.

= (y-x)(z-x)[(y + x)(z² + x² + zx) - (z + x)(y²

=(b-a)(c-a)[(c+a)-(b+a)]

=(b-a)(c-a)[(c+a – b – a)]

=(b-a)(c-a)(c-b)

=(-1)(a-b) ×(-1)(b-c)(c-a)

=(a-b)(b-c)(c-a).

= R.H.S.

$=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1-1\hfill & \hfill 1-1\hfill \\ \hfill a\hfill & \hfill b-a\hfill & \hfill c-a\hfill \\ \hfill a^3\hfill & \hfill b^3-a^3\hfill & \hfill c^3-a^3\hfill \end{array}\right|\begin{array}{cc}\hfill c_2\to c_2-c_1\hfill & \hfill c_3\to c_3-c_1\hfill \end{array}$

= $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill a\hfill & \hfill b-a\hfill & \hfill c-a\hfill \\ \hfill a^3\hfill & \hfill (b-a)\left(b^2+a^2+ab\right)\hfill & \hfill (c-a)\left(c^2+a^2+ab\right)\hfill \end{array}\right|$ $\{\begin{array}{l}?x^3-y^3\\ =(x-y)(x^2+y^2+xy)\end{array}$

Expanding along R1

= $\left|\begin{array}{cc}\hfill b-a\hfill & \hfill c-a\hfill \\ \hfill \left(b-a(b^2+a^2+ab\right)\hfill & \hfill (c-a)\left(c^2+a^2+ab\right)\hfill \end{array}\right|$

= (ba)(ca) $\left|\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill b^2+a^2+ab\hfill & \hfill c^2+a^2+ab\hfill \end{array}\right|$ $\{\begin{array}{l}(b-a)\&(c-a)\\ c_1\&c_3\end{array}$

= (b-a)(c-a) {(c² + a² + ab)- (b² + a² + ab)}

= (b-a) (c -a) {c² + a² + ab-b²-a²-ab}

= (b-a) (c-a)(c²-b² + ac-ab)

Kindly go through the solution

=a²b²c²× (−2) $\left(1\times (-1)-1\times 1\right)$

= a²b²c²× (−2)× (−2)

= 4a²b²c²

= R.H.S.

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=0=RHS

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Kindly go through the solution

Kindly go through the solution