Verify A(adj A)=(adj A)A=IAII. \(\begin{bmatrix}2&3\\-4&-6\end{bmatrix}\)
A=\(\begin{bmatrix}2&3\\-4&-6\end{bmatrix}\)
we have IAI=-12-(-12)=-12+12=0
so IAII=0\(\begin{vmatrix}1&0\\0&1\end{vmatrix}\)=\(\begin{vmatrix}0&0\\0&0\end{vmatrix}\)
Now A11=-6, A12=4, A21=-3, A22=2
so adj A=\(\begin{vmatrix}-6&3\\4&2\end{vmatrix}\)
Now A(adj A)=\(\begin{bmatrix}2&3\\-4&-6\end{bmatrix}\)\(\begin{vmatrix}-6&3\\4&2\end{vmatrix}\)
=\(\begin{vmatrix}-12+12&-6+6\\24-24&12-12\end{vmatrix}\)=\(\begin{vmatrix}0&0\\0&0\end{vmatrix}\)
Also (adj A)A=\(\begin{vmatrix}-6&3\\4&2\end{vmatrix}\)\(\begin{bmatrix}2&3\\-4&-6\end{bmatrix}\)
=\(\begin{vmatrix}-12+12&-18+18\\8-8&12-12\end{vmatrix}\)
=\(\begin{vmatrix}0&0\\0&0\end{vmatrix}\)
Hence A(adj A)=(adj A)A=IAII.
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:
\(\begin{vmatrix} 3a& -a+b & -a+c\\ -b+a & 3b & -b+c \\-c+a&-c+b&3c\end{vmatrix}\)\(=3(a+b+c)(ab+bc+ca)\)