Show that \(\Delta=\left|\begin{array}{ccc} {(y+z)^{2}} & {x y} & {z x} \\ {x y} & {(x+z)^{2}} & {y z} \\ {x z} & {y z} & {(x+y)^{2}} \end{array}\right|=2 x y z(x+y+z)^{3}\)
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Show that \(\Delta=\left|\begin{array}{ccc} {(y+z)^{2}} & {x y} & {z x} \\ {x y} & {(x+z)^{2}} & {y z} \\ {x z} & {y z} & {(x+y)^{2}} \end{array}\right|=2 x y z(x+y+z)^{3}\)
Answer
Applying R1 \(\rightarrow\) xR1, R2 \(\rightarrow\) yR2, R3 \(\rightarrow\) zR3 to \(\Delta\) and dividing by xyz, we get
\(\Delta=\frac{1}{x y z}\left|\begin{array}{ccc} {x(y+z)^{2}} & {x^{2} y} & {x^{2} z} \\ {x y^{2}} & {y(x+z)^{2}} & {y^{2} z} \\ {x z^{2}} & {y z^{2}} & {z(x+y)^{2}} \end{array}\right|\)
Taking common factors x, y, z from C1 C2 and C3, respectively, we get
\(\Delta=\frac{x y z}{x y z}\left|\begin{array}{ccc} {(y+z)^{2}} & {x^{2}} & {x^{2}} \\ {y^{2}} & {(x+z)^{2}} & {y^{2}} \\ {z^{2}} & {z^{2}} & {(x+y)^{2}} \end{array}\right|\)
Applying C2 \(\rightarrow\) C2 – C1, C3 \(\rightarrow\) C3 – C1, we have
\(\Delta=\left|\begin{array}{ccc} {(y+z)^{2}} & {x^{2}-(y+z)^{2}} & {x^{2}-(y+z)^{2}} \\ {y^{2}} & {(x+z)^{2}-y^{2}} & {0} \\ {z^{2}} & {0} & {(x+y)^{2}-z^{2}} \end{array}\right|\)
Taking common factor (x + y + z) from C2 and C3 , we have
\(\Delta=(x+y+z)^{2}\left|\begin{array}{ccc} {(y+z)^{2}} & {x-(y+z)} & {x-(y+z)} \\ {y^{2}} & {(x+z)-y} & {0} \\ {z^{2}} & {0} & {(x+y)-z} \end{array}\right|\)
Applying R1 \(\rightarrow\) R1 – (R2 + R3 ), we have
\(\Delta=(x+y+z)^{2}\left|\begin{array}{ccc} {2 y z} & {-2 z} & {-2 y} \\ {y^{2}} & {x-y+z} & {0} \\ {z^{2}} & {0} & {x+y-z} \end{array}\right|\)
Applying C2 \(\rightarrow\) (C2 + \(\frac{1}{y}\) C1) and C3 \(\rightarrow\) C3 + \(\frac{1}{z}\)C1 we get
\(\Delta=(x+y+z)^{2}\left|\begin{array}{ccc} {2 y z} & {0} & {0} \\ {y^{2}} & {x+z} & {\frac{y^{2}}{z}} \\ {z^{2}} & {\frac{z^{2}}{y}} & {x+y} \end{array}\right|\)
Finally expanding along R1, we have
\(\Delta\) = (x + y + z)2 (2yz) [(x + z) (x + y) – yz] = (x + y + z)2 (2yz) (x 2 + xy + xz)
= (x + y + z)3 (2xyz)
\(\Delta=\frac{1}{x y z}\left|\begin{array}{ccc} {x(y+z)^{2}} & {x^{2} y} & {x^{2} z} \\ {x y^{2}} & {y(x+z)^{2}} & {y^{2} z} \\ {x z^{2}} & {y z^{2}} & {z(x+y)^{2}} \end{array}\right|\)
Taking common factors x, y, z from C1 C2 and C3, respectively, we get
\(\Delta=\frac{x y z}{x y z}\left|\begin{array}{ccc} {(y+z)^{2}} & {x^{2}} & {x^{2}} \\ {y^{2}} & {(x+z)^{2}} & {y^{2}} \\ {z^{2}} & {z^{2}} & {(x+y)^{2}} \end{array}\right|\)
Applying C2 \(\rightarrow\) C2 – C1, C3 \(\rightarrow\) C3 – C1, we have
\(\Delta=\left|\begin{array}{ccc} {(y+z)^{2}} & {x^{2}-(y+z)^{2}} & {x^{2}-(y+z)^{2}} \\ {y^{2}} & {(x+z)^{2}-y^{2}} & {0} \\ {z^{2}} & {0} & {(x+y)^{2}-z^{2}} \end{array}\right|\)
Taking common factor (x + y + z) from C2 and C3 , we have
\(\Delta=(x+y+z)^{2}\left|\begin{array}{ccc} {(y+z)^{2}} & {x-(y+z)} & {x-(y+z)} \\ {y^{2}} & {(x+z)-y} & {0} \\ {z^{2}} & {0} & {(x+y)-z} \end{array}\right|\)
Applying R1 \(\rightarrow\) R1 – (R2 + R3 ), we have
\(\Delta=(x+y+z)^{2}\left|\begin{array}{ccc} {2 y z} & {-2 z} & {-2 y} \\ {y^{2}} & {x-y+z} & {0} \\ {z^{2}} & {0} & {x+y-z} \end{array}\right|\)
Applying C2 \(\rightarrow\) (C2 + \(\frac{1}{y}\) C1) and C3 \(\rightarrow\) C3 + \(\frac{1}{z}\)C1 we get
\(\Delta=(x+y+z)^{2}\left|\begin{array}{ccc} {2 y z} & {0} & {0} \\ {y^{2}} & {x+z} & {\frac{y^{2}}{z}} \\ {z^{2}} & {\frac{z^{2}}{y}} & {x+y} \end{array}\right|\)
Finally expanding along R1, we have
\(\Delta\) = (x + y + z)2 (2yz) [(x + z) (x + y) – yz] = (x + y + z)2 (2yz) (x 2 + xy + xz)
= (x + y + z)3 (2xyz)
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