If A= [ cc 3 & 1 -1 & 2 ] Show that A ^2-5 A +7 I = O. Hence find A ^ -1

We have, l A^2=AA= 3&1 -1&2 3&1 -1&2 = 8&5 -5&2 |A| = (3)(2) - (1)(-1) = 6 + 1 = 7 0 A is non singular and hence A-1 exists. A^2 - 5A + 7I = [ * 20 c 8&5 - 5 &2 ] - 5 [ * 20 c 3&1 - 1 &2 ] + 7 [ * 20 c 1&0 0&1 ] = [ * 20 c 8 - 15 + 7 & 5 - 5 + 0 - 5 + 50 & 3 - 10 + 7 ] = [ * 20 c 0&0 0&0 ]= 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 [ * 20 c 1&0 0&1 ] - [ * 20 c 3&1 - 1 &2 ] = [ * 20 c 2&-1 1&3 ] A^ - 1 = 1 7 [ * 20 c 2& - 1 1&3 ]

If A= [ cc 3 & 1 -1 & 2 ] Show that A ^2-5 A +7 I = O. Hence find…

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Question

If $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$ Show that $A ^2-5 A +7 I = O$. Hence find $A ^{-1}$

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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]$