Home/Class 12/Maths/

Question and Answer

Show that  \(\Delta=\left|\begin{array}{ccc} {(y+z)^{2}} & {x y} & {z x} \\ {x y} & {(x+z)^{2}} & {y z} \\ {x z} & {y z} & {(x+y)^{2}} \end{array}\right|=2 x y z(x+y+z)^{3}\)
loading
settings
Speed
00:00
11:15
fullscreen
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)
To Keep Reading This Answer, Download the App
4.6
star pngstar pngstar pngstar pngstar png
Review from Google Play
To Keep Reading This Answer, Download the App
4.6
star pngstar pngstar pngstar pngstar png
Review from Google Play
Correct9
Incorrect0
Watch More Related Solutions
Evaluate \(\left| {\begin{array}{*{20}{c}} 3&{ - 4}&5 \\ 1&1&{ - 2} \\ 2&3&1 \end{array}} \right|\)
If \(\Delta\)\(\left|\begin{array}{rrr} {2} & {-3} & {5} \\ {6} & {0} & {4} \\ {1} & {5} & {-7} \end{array}\right|\) and \(\Delta_{1}=\) \(\left|\begin{array}{rrr} {2} & {-3} & {5} \\ {1} & {5} & {-7}\\ {6} & {0} & {4} \end{array}\right|\)
then  \(\Delta=-\Delta_{1}\)
Let   A = \(\left[\begin{array}{ccc} {1} & {\sin \theta} & {1} \\ {-\sin \theta} & {1} & {\sin \theta} \\ {-1} & {-\sin \theta} & {1} \end{array}\right]\) where \(0 \leq \theta \leq 2 \pi\). Then
  • Det(A) \(\in\) (2, \(\infty\))
  • Det(A) = 0
  • Det(A) \(\in\) (2, 4)
  • Det(A) \(\in\) [2, 4]
For the matrix \(A = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&{ - 3} \\ 2&{ - 1}&3 \end{array}} \right]\), show that A3 - 6A2 + 5A + 11I = 0. Hence find A-1.
Which of the following is correct
  • Determinant is a number associated to a square matrix.
  • None of these
  • Determinant is a number associated to a matrix.
  • Determinant is a square matrix
If \(\left|\begin{array}{cc} {x} & {2} \\ {18} & {x} \end{array}\right|=\left|\begin{array}{cc} {6} & {2} \\ {18} & {6} \end{array}\right|\) , then x is equal to
  • \(\pm\)6
  • 0
  • -6
  • 6
Find minors and cofactors of all the elements of the determinant \(\left| {\begin{array}{*{20}{c}} 1&{ - 2} \\ 4&3 \end{array}} \right|\)
Find adjoint of the matrix \(\left[ {\begin{array}{*{20}{c}} 1&2 \\ 3&4 \end{array}} \right]\)
Evaluate \(\left|\begin{array}{cc} {x} & {x+1} \\ {x-1} & {x} \end{array}\right|\)
Evaluate: \(\left| {\begin{array}{*{20}{c}} x&y&{x + y} \\ y&{x + y}&x \\ {x+ y}&x&y \end{array}} \right|\)

Load More