Choose the correct answer. Let A be a square matrix of order 3×3,then IkAI is equal to
\(K \mid A \mid\)
\(K^2\mid A\mid\)
\(K^3\mid A\mid\)
\(3K \mid A\mid\)
A is a square matrix of order 3 × 3.
Let A=\(\begin{bmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{bmatrix}\)
Then kA=\(\begin{bmatrix}ka_1&kb_1&kc_1\\ka_2&kb_2&kc_2\\ka_3&kb_3&kc_3\end{bmatrix}\)
so IkAI=\(\begin{vmatrix}ka_1&kb_1&kc_1\\ka_2&kb_2&kc_2\\ka_3&kb_3&kc_3\end{vmatrix}\)
=k3 \(\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}\)
=k3IAI
so IkAI=k3IAI
Hence, the correct answer is 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:
\(\begin{vmatrix} 3a& -a+b & -a+c\\ -b+a & 3b & -b+c \\-c+a&-c+b&3c\end{vmatrix}\)\(=3(a+b+c)(ab+bc+ca)\)
Read More: Properties of Determinants