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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants5 Marks
Question
If $ A = \left[ {\begin{array}{*{20}{c}} 3&1 \\ { - 1}&2 \end{array}} \right]$ Show that $A^2 – 5A + 7I = O.$ Hence find $A^{-1}$
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Answer

We have, $A^2=A$
$=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}=\begin{bmatrix}8&5\\-5&2\end{bmatrix}$
$|A| = (3)(2) - (1)(-1)$
$= 6 + 1$
$= 7   \neq0$
$\Rightarrow A$ 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,
$A^2 – 5A + 7I = O \ ($given$)$
$A^2 – 5A = -7I$
Post multiplying by $A^{-1},$ we get,
$A^2A^{-1}-5AA^{-1} = -7IA^{-1}$
$AAA^{-1} – 5AA^{-1} = -7IA^{-1}$
$A – 5I = -7A^{-1} [AA^{-1} = I]$
$7A^{-1} = 5I – A$
$ = 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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