Let A=\(\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}\)
We know that A = IA
\(\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}\)=\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)A
Applying \(R_1\rightarrow R_1-\frac{1}{2}R_2\), we have:
\(\begin{bmatrix} 0 & 0 \\ 4 & 2 \end{bmatrix}\)=\(\begin{bmatrix} 1 & -\frac12 \\ 0 & 1 \end{bmatrix}\)A
Now, in the above equation, we can see all the zeros in the first row of the matrix on the L.H.S. Therefore, A−1 does not exist.
Let $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}$. If for some $\theta \in (0, \pi)$, $A^2 = A^T$, then the sum of the diagonal elements of the matrix $(A + I)^3 + (A - I)^3 - 6A$ is equal to
Let $ A $ be a $ 3 \times 3 $ matrix such that $ | \text{adj} (\text{adj} A) | = 81.
$ If $ S = \left\{ n \in \mathbb{Z}: \left| \text{adj} (\text{adj} A) \right|^{\frac{(n - 1)^2}{2}} = |A|^{(3n^2 - 5n - 4)} \right\}, $ then the value of $ \sum_{n \in S} |A| (n^2 + n) $ is:
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is:
A matrix for which matrix inversion operation exists, given that it satisfies the requisite conditions is known as an invertible matrix. Any given square matrix A of order n × n is called invertible if and only if there exists, another n × n square matrix B such that, AB = BA = In, where In is an identity matrix of order n × n.
It can be observed that the determinant of the following matrices is non-zero.