Using the property of determinants and without expanding, prove that: \(\begin{vmatrix}a-b&b-c&c-a\\b-c&c-a&a-b\\c-a&a-b&b-c\end{vmatrix}\)=0
\(\triangle\)= \(\begin{vmatrix}a-b&b-c&c-a\\b-c&c-a&a-b\\c-a&a-b&b-c\end{vmatrix}\)
Applying R1\(\to\) R1+R2,we have
\(\triangle\)= \(\begin{vmatrix}a-c&b-a&c-b\\b-c&c-a&a-b\\-(a-c)&-(b-a)&-(c-b)\end{vmatrix}\)
= \(-\begin{vmatrix}a-c&b-a&c-b\\b-c&c-a&a-b\\(a-c)&(b-a)&(c-b)\end{vmatrix}\)
Here, the two rows R1 and R3 are identical.
\(\triangle\) = 0.
Given :
\(\begin{vmatrix}a-b&b-c&c-a\\b-c&c-a&a-b\\c-a&a-b&b-c\end{vmatrix}\)
So, \(\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix}-\begin{vmatrix}b&c&a\\c&a&b\\a&b&c\end{vmatrix}\)
Therefore, by finding the determinant of each we get :
= a(cb - a2) - b(b2 - ca) + c(ba - c2)
−b(ac - b2) + c(c2 - ab) - a(cb - a2)
= abc - a3 - b3 + abc + abc - c3 - abc + b3 + c3 - abc - abc + a3
= 3abc - 3abc
= 0
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