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.
If \[ A = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix} \] then find \( A^{-1} \). Hence, solve the system of linear equations: \[ x - 2y = 10, \] \[ 2x - y - z = 8, \] \[ -2y + z = 7. \]
Complete and balance the following chemical equations: (a) \[ 2MnO_4^-(aq) + 10I^-(aq) + 16H^+(aq) \rightarrow \] (b) \[ Cr_2O_7^{2-}(aq) + 6Fe^{2+}(aq) + 14H^+(aq) \rightarrow \]
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.