Let A=\(\begin{bmatrix} 2 & -3\\ -1 & 2 \end{bmatrix}\)
We know that A = IA
\(\begin{bmatrix} 2 & -3\\ -1 & 2 \end{bmatrix}\)=A\(\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\)
⇒ \(\begin{bmatrix} 1 & -1\\ -1 & 2 \end{bmatrix}\)= \(\begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}\)A \((R_1\rightarrow R_1+R_2)\)
⇒ \(\begin{bmatrix} 1 & -1\\ 0 & 1 \end{bmatrix}\)=\(\begin{bmatrix} 1 & 1\\ 1 & 2 \end{bmatrix}\)A \((R_2\rightarrow R_2+R_1)\)
⇒ \(\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\)= \(\begin{bmatrix} 2 & 3\\ 1 & 2 \end{bmatrix}\)A \((R_1\rightarrow R_1+R_2)\)
so A-1=\(\begin{bmatrix} 2 & 3\\ 1 & 2 \end{bmatrix}\)
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) \[ Cr_2O_7^{2-}(aq) + 6Fe^{2+}(aq) + 14H^+(aq) \rightarrow \]