By using properties of determinants, show that: \(\begin{vmatrix}-a^2&ab&ac\\ba&-b^2&bc\\ca&cb&-c^2\end{vmatrix}\)=4a2b2c2
△=\(\begin{vmatrix}-a^2&ab&ac\\ba&-b^2&bc\\ca&cb&-c^2\end{vmatrix}\)
=\(abc\begin{vmatrix}-a&b&c\\a&-b&c\\a&b&-c\end{vmatrix}\) [Taking out factors a,b,c from R1,R2and R3]
=a2b2c2 \(\begin{vmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{vmatrix}\) [Taking out factors a,b,c from C1,C2and C3]
Applying R2 → R2 + R1 and R3 → R3 + R1, we have:
△=a2b2c2\(\begin{vmatrix}-1&1&1\\0&0&2\\0&2&0\end{vmatrix}\)
=a2b2c2(-1)\(\begin{vmatrix}0&2\\2&0\end{vmatrix}\)
=-a2b2c2(0-4)
=4a2b2c2
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