By using properties of determinants, show that: \(\begin{vmatrix}1&x&x^2\\x^2&1&x\\x&x^2&1\end{vmatrix}\)=(1-x3)2
△=\(\begin{vmatrix}1&x&x^2\\x^2&1&x\\x&x^2&1\end{vmatrix}\)
Applying R1 → R1 + R2 + R3, we have:
△=\(\begin{vmatrix}1+x+x^2&1+x+x^2&1+x+x^2\\x^2&1&x\\x&x^2&1\end{vmatrix}\)
=(1+x+x2)\(\begin{vmatrix}1&1&1\\x^2&1&x\\x&x^2&1\end{vmatrix}\)
Applying C2 → C2 − C1 and C3 → C3 − C1, we have:
△=(1+x+x2)\(\begin{vmatrix}1&0&0\\x^2&1-x^2&x-x^2\\-x&x^2-x&1-x\end{vmatrix}\)
=(1+x+x2)(1-x)(1-x)I100 x2 1+x x x -x 1I
=(1-x3)(1-x)I100 x2 1+x x x -x 1I
Expanding along R1, we have:
△=(1-x3)(1-x)(1)I1+x x -x 1I
=(1-x3)(1-x)(1+x+x2)
=(1-x3)(1-x3)
=(1-x3)2
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:
\(\begin{vmatrix} 3a& -a+b & -a+c\\ -b+a & 3b & -b+c \\-c+a&-c+b&3c\end{vmatrix}\)\(=3(a+b+c)(ab+bc+ca)\)
Read More: Properties of Determinants