Question
Download Solution PDFजर a, b, c या शून्य नसलेल्या वास्तविक संख्या असतील, तर सारणी \(A = \left[ {\begin{array}{*{20}{c}} a&0&0\\ 0&b&0\\ 0&0&c \end{array}} \right]\) चा व्यस्त काेणाच्या बरोबर आहे?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
सारणीचा व्यस्त: n × n सारणीचा व्यस्त द्वारे दिला जातो:
\({A^{ - 1}} = \frac{{adj\left( A \right)}}{{\left| A \right|}}\) जेथे adj(A) ला आयुग्मी सारणी म्हणतात.
आयुग्मी सारणी: जर Bn× nहे सारणी An× n चे सहघटक सारणी असेल तर An× n चे आयुग्मी सारणी adj(A) द्वारे दर्शविले जाते आणि BT म्हणून परिभाषित केले जाते. त्यामुळे, adj(A) = BT.
गणना:
दिलेले आहे: \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} a&0&0\\ 0&b&0\\ 0&0&c \end{array}} \right]\)
आपल्याला माहित आहे की, \({{\rm{A}}^{ - 1}} = \frac{{adj\left( A \right)}}{{\left| A \right|}}\)
⇒ |A| = [a (bc - 0) - 0(0) - 0(0)]
⇒ |A| = abc
\(\Rightarrow {\rm{adj}}\left( {\rm{A}} \right) = {\left[ {\begin{array}{*{20}{c}} {bc}&0&0\\ 0&{ac}&0\\ 0&0&{ab} \end{array}} \right]^T}\)
\(\Rightarrow {\rm{adj}}\left( {\rm{A}} \right) = \left[ {\begin{array}{*{20}{c}} {bc}&0&0\\ 0&{ac}&0\\ 0&0&{ab} \end{array}} \right]\)
\(\Rightarrow {A^{ - 1}} = \frac{{\left[ {\begin{array}{*{20}{c}} {bc}&0&0\\ 0&{ac}&0\\ 0&0&{ab} \end{array}} \right]}}{{abc}}\)
\(\Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {{a^{ - 1}}}&0&0\\ 0&{{b^{ - 1}}}&0\\ 0&0&{{c^{ - 1}}} \end{array}} \right]\)Last updated on Jun 18, 2025
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