Find the inverse of each of the matrices (if it exists). \(\begin{bmatrix}1&0&0\\3&3&0\\5&2&-1\end{bmatrix}\)
Let A=\(\begin{bmatrix}1&0&0\\3&3&0\\5&2&-1\end{bmatrix}\)
we have IAI=1(-3-0)-0+0=-3
now A11=-3-0=-3, A12=-(-3-0)=3, A13=6-15=9
A21=-(0-0)=0, A22=-1-0=-1,A23=-(2-0)=-2
A31=0-0=0,A32=-(0-0)=0,A33=3-0=3
therefore adj A=\(\begin{bmatrix}-3&0&0\\3&-1&0\\-9&-2&3\end{bmatrix}\)
so A-1=\(\frac{1}{\mid A \mid}\)adj A=\(-\frac{1}{3}\)\(\begin{bmatrix}-3&0&0\\3&-1&0\\-9&-2&3\end{bmatrix}\)
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:
\(\begin{vmatrix} 3a& -a+b & -a+c\\ -b+a & 3b & -b+c \\-c+a&-c+b&3c\end{vmatrix}\)\(=3(a+b+c)(ab+bc+ca)\)