Prove that\(\begin{vmatrix} a^2&bc &ac+c^2 \\ a^2+ab&b^2 &ac\\ ab&b^2+bc &c^2 \end{vmatrix}=4a^2b^2c^2\)
\(\Delta = \begin{vmatrix} a^2&bc &ac+c^2 \\ a^2+ab&b^2 &ac\\ ab&b^2+bc &c^2 \end{vmatrix}\)
Taking out common factors a,b and c from C1,C2 and C3 we have,
\(\Delta=abc\begin{vmatrix} a&c &a+c \\ a+b&b &a \\ b&b+c &c \end{vmatrix}\)
Applying R2\(\rightarrow\)R2-R1 and R3\(\rightarrow\)R3-R1,we have:
\(\Delta=abc\begin{vmatrix} a&c &a+c \\ b&b-c &-c \\ b-a&b &-a \end{vmatrix}\)
Applying R2\(\rightarrow\)R2+R1,we heve:
\(\Delta=abc\begin{vmatrix} a&c &a+c \\ a+b&b &a \\ 2b&2b &0 \end{vmatrix}\)
Applying R3\(\rightarrow\)R3+R2,we heve:
\(\Delta=abc\begin{vmatrix} a&c &a+c \\ a+b&b &a \\ 2b&2b &0 \end{vmatrix}\)
\(\Delta=2a^2bc\begin{vmatrix} a&c &a+c \\ a+b&b &a \\ 2b&2b &0 \end{vmatrix}\)
ApplyingC2\(\rightarrow\)C2-C1, we have:
\(\Delta=2a^2bc\begin{vmatrix} a&c-a &a+c \\ a-b&-a &a \\ 0&0 &0 \end{vmatrix}\)
Expanding along R3,we have:
Δ=2ab2c[a(c-a)+a(a+c)]
=2ab2c[ac-a2+a2+ac]
=2ab2c(2ac)
=4a2b2c2
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:
\(\begin{vmatrix} 3a& -a+b & -a+c\\ -b+a & 3b & -b+c \\-c+a&-c+b&3c\end{vmatrix}\)\(=3(a+b+c)(ab+bc+ca)\)