Find the inverse of each of the matrices(if it exists). \(\begin{bmatrix}2&1&3\\4&-1&0\\-7&2&1\end{bmatrix}\)
Let A=\(\begin{bmatrix}2&1&3\\4&-1&0\\-7&2&1\end{bmatrix}\)
We have IAI=2(-1-0)-1(4-0)+3(8-7)
=-2-4+3
=-3
Now A11=-1-0=-1, A12=-(4-0), A13=8-7=1
A21=-(1-6)=5, A22=2+21=23, A23=-(-(4+7)=-11
A31=0+3=3, A32=-(0-12)=12,A33=-2-4=-6
therefore adj A=\(\begin{bmatrix}-1&5&3\\-4&23&12\\1&-11&-6\end{bmatrix}\)
so A-1=\(\frac{1}{\mid A\mid}\)adj A=\(-\frac{1}{3}\) \(\begin{bmatrix}-1&5&3\\-4&23&12\\1&-11&-6\end{bmatrix}\)
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:
\(\begin{vmatrix} 3a& -a+b & -a+c\\ -b+a & 3b & -b+c \\-c+a&-c+b&3c\end{vmatrix}\)\(=3(a+b+c)(ab+bc+ca)\)