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(ii) If matrix \( A=\left[\begin{array}{rrr}3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1\end{array}\right] \), then find \( A^{-1} \) using elementary transformations (row or column).(P.B. 2011) |
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Answer» [3 -4 3 -2. 3 -2 8. -12 9] is the inverse of the given matrix \(A = \begin{bmatrix}3&0&-1\\2&3&0\\0&4&1\end{bmatrix}\) A = IA ⇒ \( \begin{bmatrix}3&0&-1\\2&3&0\\0&4&1\end{bmatrix}\) = \( \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A\) R2 →3R2 - 2R1 \( \begin{bmatrix}3&0&-1\\0&9&2\\0&4&1\end{bmatrix}\)= \( \begin{bmatrix}1&0&0\\-2&3&0\\0&0&1\end{bmatrix}A\) R1 → \(\frac{R_1}3\) \( \begin{bmatrix}1&0&-1/3\\0&9&2\\0&4&1\end{bmatrix}\) = \( \begin{bmatrix}1/3&0&0\\-2&3&0\\0&0&1\end{bmatrix}A\) R3 → 9R3 - 4R2 \( \begin{bmatrix}1&0&-1/3\\0&9&2\\0&0&1\end{bmatrix}\) = \( \begin{bmatrix}1/3&0&0\\-2&3&0\\8&-12&9\end{bmatrix}A\) R1 → R1 + \(\frac13R_3\) R2 → R2 - 2R3 \( \begin{bmatrix}1&0&0\\0&9&0\\0&0&1\end{bmatrix}\) = \( \begin{bmatrix}3&-4&3\\-2&3&-2\\8&-12&9\end{bmatrix}A\) ∴ A-1 = \( \begin{bmatrix}3&-4&3\\-2&3&-2\\8&-12&9\end{bmatrix}\) |
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