Find adjoint of each of the matrices \(\begin{bmatrix}1&-1&2\\2&3&5\\-2&0&1\end{bmatrix}\)
Let A=\(\begin{bmatrix}1&-1&2\\2&3&5\\-2&0&1\end{bmatrix}\)
we have
A11=\(\begin{vmatrix}3&5\\0&1\end{vmatrix}\)=3-0=3
A12=-\(\begin{vmatrix}2&5\\-2&1\end{vmatrix}\)=-12
A13=\(\begin{vmatrix}2&3\\-2&0\end{vmatrix}\)=0+6=6
A21=-\(\begin{vmatrix}-1&2\\0&1\end{vmatrix}\)=-(-1-0)=1
A22=\(\begin{vmatrix}1&2\\-2&1\end{vmatrix}\)=1+4=5
A23=-\(\begin{vmatrix}1&-1\\-2&0\end{vmatrix}\)=-(0-2)=2
A31=\(\begin{vmatrix}-1&2\\3&5\end{vmatrix}\)=-5-6=-11
A32=-\(\begin{vmatrix}1&2\\2&5\end{vmatrix}\)=-(5-4)=-1
A33=\(\begin{vmatrix}1&-1\\2&3\end{vmatrix}\)=3+2=5
Hence adj A=\(\begin{bmatrix}A_{11}&A_{21}&A_{31}\\A_{12}&A_{22}&A_{23}\\A_{13}&A_{23}&A_{33}\end{bmatrix}\)=\(\begin{bmatrix}3&1&-11\\ -12&5&-1\\ 6&2&5\end{bmatrix}\)
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:
\(\begin{vmatrix} 3a& -a+b & -a+c\\ -b+a & 3b & -b+c \\-c+a&-c+b&3c\end{vmatrix}\)\(=3(a+b+c)(ab+bc+ca)\)