|x+a x x x x+a x x x x+a|=0
Applying R1→R1+R2+R3,we get:
|3x+a 3x+a 3x+a x x+a x x x x+a|=0
⇒(3x+a)|111 x x+a x x x x+a|=0
Applying C2→C2-C1 and C3→C3-C1,we have:
(3x+a)|100 xa0 x0a|=0
Expanding along R1,we have:
(3x+a)[1xa2]=0
⇒a2(3x+a)=0
But a≠0,
Therefore we have
3x+a=0
⇒x=-a/3.
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:
\(\begin{vmatrix} 3a& -a+b & -a+c\\ -b+a & 3b & -b+c \\-c+a&-c+b&3c\end{vmatrix}\)\(=3(a+b+c)(ab+bc+ca)\)