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=\(-\frac{1}{3}\)\(\begin{bmatrix}-3&0&0\\3&-1&0\\-9&-2&3\end{bmatrix}\)
Let I be the identity matrix of order 3 × 3 and for the matrix $ A = \begin{pmatrix} \lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{pmatrix} $, $ |A| = -1 $. Let B be the inverse of the matrix $ \text{adj}(A \cdot \text{adj}(A^2)) $. Then $ |(\lambda B + I)| $ is equal to _______
Let $A = \{ z \in \mathbb{C} : |z - 2 - i| = 3 \}$, $B = \{ z \in \mathbb{C} : \text{Re}(z - iz) = 2 \}$, and $S = A \cap B$. Then $\sum_{z \in S} |z|^2$ is equal to
If $ y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\27 & 28 & 27 \\1 & 1 & 1 \end{vmatrix} $, $ x \in \mathbb{R} $, then $ \frac{d^2y}{dx^2} + y $ is equal to
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is: