Question

using elementary transformation​

Answer 1

Answer:

$\tt{ A^{-1} = \left[\begin{array}{ccc}2/15&1/15&1/3\\-14/15&8/15&-1/3\\-1/3&1/3&-1/3\end{array}\right] }$

Step-by-step explanation:

Given: $\tt{ A = \left[\begin{array}{ccc}1&-2&3\\3&-1&4\\2&1&-2\end{array}\right] }$

To find: A⁻¹

Solution:

w.k.t., A = IA and I = A⁻¹ A

⇒ $\tt{ \left[\begin{array}{ccc}1&-2&3\\3&-1&4\\2&1&-2\end{array}\right] = \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] A}$

By R₂ → R₂ - 3R₁

⇒ $\tt{ \left[\begin{array}{ccc}1&-2&3\\0&5&-5\\2&1&-2\end{array}\right] = \left[\begin{array}{ccc}1&0&0\\-3&1&0\\0&0&1\end{array}\right] A}$

By R₃ → R₃ - 2R₁

⇒ $\tt{ \left[\begin{array}{ccc}1&-2&3\\0&5&-5\\0&5&-8\end{array}\right] = \left[\begin{array}{ccc}1&0&0\\-3&1&0\\-2&0&1\end{array}\right] A}$

By R₂ → R₂/5

⇒ $\tt{ \left[\begin{array}{ccc}1&-2&3\\0&1&-1\\0&5&-8\end{array}\right] = \left[\begin{array}{ccc}1&0&0\\-3/5&1/5&0\\-2&0&1\end{array}\right] A}$

By R₃ → R₃ - 5R₂

⇒ $\tt{ \left[\begin{array}{ccc}1&-2&3\\0&1&-1\\0&0&-3\end{array}\right] = \left[\begin{array}{ccc}1&0&0\\-3/5&1/5&0\\1&-1&1\end{array}\right] A}$

By R₃ → -R₃/3

⇒ $\tt{ \left[\begin{array}{ccc}1&-2&3\\0&1&-1\\0&0&1\end{array}\right] = \left[\begin{array}{ccc}1&0&0\\-3/5&1/5&0\\-1/3&1/3&-1/3\end{array}\right] A}$

By R₁ → R₁ + 2R₂

⇒ $\tt{ \left[\begin{array}{ccc}1&0&1\\0&1&-1\\0&0&1\end{array}\right] = \left[\begin{array}{ccc}-1/5&2/5&0\\-3/5&1/5&0\\-1/3&1/3&-1/3\end{array}\right] A}$

By R₁ → R₁ - R₃

⇒ $\tt{ \left[\begin{array}{ccc}1&0&0\\0&1&-1\\0&0&1\end{array}\right] = \left[\begin{array}{ccc}2/15&1/15&1/3\\-3/5&1/5&0\\-1/3&1/3&-1/3\end{array}\right] A}$

By R₂ → R₂ + R₃

⇒ $\tt{ \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] = \left[\begin{array}{ccc}2/15&1/15&1/3\\-14/15&8/15&-1/3\\-1/3&1/3&-1/3\end{array}\right] A}$

⇒ $\tt{ A^{-1} = \left[\begin{array}{ccc}2/15&1/15&1/3\\-14/15&8/15&-1/3\\-1/3&1/3&-1/3\end{array}\right] }$

Note :-

To solve the question, we have used Elementary Row Transformation.