Question

If a =[ 1 -2 1, 0 1 -1, 3 -1 1] then find the value of A3-3A2-A-3I

Answer 1

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

To Find : A³ - 3A²  - A - 3I

Solution:  

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

$A^2=\left[\begin{array}{ccc}4&-5&4\\-3&2&-2\\6&-8&5\end{array}\right]$

$3A^2=\left[\begin{array}{ccc}12&-15&12\\-9&6&-6\\18&-24&15\end{array}\right]$

$A^3=\left[\begin{array}{ccc}16&-17&13\\-9&10&-7\\21&-25&19\end{array}\right]$

$I=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$

$3I=\left[\begin{array}{ccc}3&0&0\\0&3&0\\0&0&3\end{array}\right]$

A³ - 3A²  - A - 3I  = $=\left[\begin{array}{ccc}0&0&0\\0&0&0\\0&0&0\end{array}\right]$

Value of A³ - 3A²  - A - 3I  = 0

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