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If \( A = \left[ {\begin{array}{*{20}{c}} 3&1 \\ { - 1}&2 \end{array}} \right]\) Show that A2 – 5A + 7I = O. Hence find A-1
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If \( A = \left[ {\begin{array}{*{20}{c}} 3&1 \\ { - 1}&2 \end{array}} \right]\) Show that A2 – 5A + 7I = O. Hence find A-1

Answer

we have, \( \begin{array}{l}A^2=AA=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}=\begin{bmatrix}8&5\\-5&2\end{bmatrix}\\\end{array}\)
|A| = (3)(2) - (1)(-1) = 6 + 1 = 7 \( \neq0\)
\(\Rightarrow\) A is non singular and hence A-1 exists.
\( {A^2} - 5A + 7I = \left[ {\begin{array}{*{20}{c}} 8&5 \\ { - 5}&2 \end{array}} \right] - 5\left[ {\begin{array}{*{20}{c}} 3&1 \\ { - 1}&2 \end{array}} \right] + 7\left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right]\)
\( = \left[ {\begin{array}{*{20}{c}} {8 - 15 + 7}&{5 - 5 + 0} \\ { - 5 + 50}&{3 - 10 + 7} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]\)= O
Now,
A2 – 5A + 7I = O(given)
A2 – 5A = -7I
Post multiplying by A-1, we get,
A2A-1 -5AA-1 = -7IA-1
AAA-1 – 5AA-1 = -7IA-1
A – 5I = -7A-1 [AA-1 = I]
7A-1 = 5I – A
\( = 5\left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 3&1 \\ { - 1}&2 \end{array}} \right]\)
\( = \left[ {\begin{array}{*{20}{c}} 2&-1\\ 1&3 \end{array}} \right]\)
\( {A^{ - 1}} = \frac{1}{7}\left[ {\begin{array}{*{20}{c}} 2&{ - 1} \\ 1&3 \end{array}} \right]\)
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