(i) △=\(\begin{vmatrix} a-b-c & 2a & 2a\\ 2b & b-c-a & 2b \\ 2c&2c&c-a-b\end{vmatrix}\)
Applying R1 → R1 + R2 + R3, we have:
△=I\(\begin{vmatrix} a+b+c & a+b+c & a+b+c\\ 2b & b-c-a & 2b \\ 2c&2c&c-a-b\end{vmatrix}\)
=(a+b+c)\(\begin{vmatrix} 1 & 1 & 1\\ 2b & b-c-a & 2b \\ 2c&2c&c-a-b\end{vmatrix}\)
Applying C2 → C2 − C1, C3 → C3 − C1, we have:
△=(a+b+c)\(\begin{vmatrix} 1 & 0 & 0\\ 2b & -a+b+c & 0 \\ 2c&0&-(a+b+c)\end{vmatrix}\)
=(a+b+c)3\(\begin{vmatrix} 1 & 0 & 0\\ 2b & -1 & 0 \\ 2c&0&-1\end{vmatrix}\)
Expanding along C3, we have:
△=(a+b+c)3(-1)(-1)=(a+b+c)3
Hence, the given result is proved.
(ii) △=\(\begin{vmatrix} x+y+2z & x & y\\ z & y+z+2x & y\\ z &x&z+x+2y \end{vmatrix}\)
Applying C1 → C1 + C2 + C3, we have:
△=\(\begin{vmatrix} 2(x+y+z) & x & y\\ 2(x+y+z) & y+z+2x & y\\ 2(x+y+z) &x&z+x+2y \end{vmatrix}\)
=2(x+y+z)\(\begin{vmatrix} 1 & x & y\\ 1 & y+z+2x & y\\ 1 &x&z+x+2y \end{vmatrix}\)
Applying R2 → R2 − R1 and R3 → R3 − R1, we have:
△=2(x+y+z)\(\begin{vmatrix} 1 & x & y\\ 0 &x+y+z & y\\ 0 &0&x+y+z \end{vmatrix}\)
△=2(x+y+z)3\(\begin{vmatrix} 1 & x & y\\ 0 &1 & 0\\ 0 &0&1\end{vmatrix}\)
Expanding along R3, we have:
△=2(x+y+z)3(1)(1-0)=2(x+y+z)3
Hence, the given result is proved.
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Evaluate the determinants in Exercises 1 and 2.
\(\begin{vmatrix}2&4\\-5&-1\end{vmatrix}\)
Evaluate the determinants in Exercises 1 and 2.
(i) \(\begin{vmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{vmatrix}\)
(ii) \(\begin{vmatrix}x^2&-x+1&x-1\\& x+1&x+1\end{vmatrix}\)
Using properties of determinants,prove that:
\(\begin{vmatrix} x & x^2 & 1+px^3\\ y & y^2 & 1+py^3\\z&z^2&1+pz^3 \end{vmatrix}\)\(=(1+pxyz)(x-y)(y-z)(z-x)\)
Using properties of determinants,prove that:
\(\begin{vmatrix} 3a& -a+b & -a+c\\ -b+a & 3b & -b+c \\-c+a&-c+b&3c\end{vmatrix}\)\(=3(a+b+c)(ab+bc+ca)\)
Read More: Properties of Determinants