If a,b, and c are real numbers and determinant \(\Delta = \begin{vmatrix} b+c &c+a &a+b \\ c+a&a+b &b+c \\ a+b&b+c &c+a \end{vmatrix}\)
Show that either a+b+c=0 or a=b=c.
\(\Delta = \begin{vmatrix} b+c &c+a &a+b \\ c+a&a+b &b+c \\ a+b&b+c &c+a \end{vmatrix}\)
Applying R1\(\rightarrow\)R1+R2+R3, we have,
\(\Delta = \begin{vmatrix} 2(a+b+c) &2(a+b+c) &2(a+b+c) \\ c+a&a+b &b+c \\ a+b&b+c &c+a \end{vmatrix}\)
=2(a+b+c)\(\begin{vmatrix} 1&1 &1 \\ c+a&a+b &b+c \\ a+b&b+c &c+a \end{vmatrix}\)
Applying C2\(\rightarrow\)C2-C1 and C3\(\rightarrow\)C3-C1,we have,
Δ=2(a+b+c)\(\begin{vmatrix} 1&1 &1 \\ c+a&a+b &b+c \\ a+b&b+c &c+a \end{vmatrix}\)
Expanding along R1,we have:
Δ=2(a+b+c)(1)[(b-c)(c-b)-(b-a)(c-a)]
=2(a+b+c)[-b2-c2+2bc-bc+ba+ac-a2]
=2(a+b+c)[ab+bc+ca-a2-b2-c2]
It is given that Δ=0.
(a+b+c)[ab+bc+ca-a2-b2-c2]=0
⇒ Either a+b+c=0,or ab+bc+ca-a2-b2-c2=0.
Now,
ab+bc+ca-a2-b2-c2=0.
⇒ -2ab-2bc-2ca+2a2+2b2+2c2=0
⇒ (a-b)2+(b-c)2+(c-a)2=0
⇒ (a-b)2=(b-c)2=(c-a)2=0 [(a-b)2,(b-c)2,(c-a)2 are non-negative]
⇒ (a-b)=(b-c)=(c-a)=0
⇒ a=b=c
Hence,if ∆=0, then either a+b+c=0 or a=b=c.
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:
\(\begin{vmatrix} 3a& -a+b & -a+c\\ -b+a & 3b & -b+c \\-c+a&-c+b&3c\end{vmatrix}\)\(=3(a+b+c)(ab+bc+ca)\)