Question

If A = $\left[\begin{array}{ccc}1&-2&1\\0&1&-1\\3&-1&1\end{array}\right],$ then find A³ - 3A² -A - 3I, where I is unit matrix of order 3.

Answer 1

Answer:

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

Step-by-step explanation:

if

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

so for

$A^{3} =A^{2}\times A \\ \\A^{2}=\left[\begin{array}{ccc}1&-2&1\\0&1&-1\\3&-1&1\end{array}\right]\times\left[\begin{array}{ccc}1&-2&1\\0&1&-1\\3&-1&1\end{array}\right]\\\\\\A^{2}=\left[\begin{array}{ccc}4&-5&4\\-3&2&-2\\6&-8&5\end{array}\right]$   ...eq1

for  $A^{3} =\left[\begin{array}{ccc}4&-5&4\\-3&2&-2\\6&-8&5\end{array}\right]\times\left[\begin{array}{ccc}1&-2&1\\0&1&-1\\3&-1&1\end{array}\right]\\\\\\=\left[\begin{array}{ccc}16&-17&13\\-9&10&-7\\21&-25&19\end{array}\right]$

....eq2

from eq1 and 2

A³ - 3A² -A - 3I

$=\left[\begin{array}{ccc}16&-17&13\\-9&10&-7\\21&-25&19\end{array}\right]-3\left[\begin{array}{ccc}4&-5&4\\-3&2&-2\\6&-8&5\end{array}\right]-\left[\begin{array}{ccc}1&-2&1\\0&1&-1\\3&-1&1\end{array}\right]-3\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$

$=\left[\begin{array}{ccc}16&-17&13\\-9&10&-7\\21&-25&19\end{array}\right]-\left[\begin{array}{ccc}12&-15&12\\-9&6&-6\\18&-24&15\end{array}\right]-\left[\begin{array}{ccc}1&-2&1\\0&1&-1\\3&-1&1\end{array}\right]-\left[\begin{array}{ccc}3&0&0\\0&3&0\\0&0&3\end{array}\right]$

$=\left[\begin{array}{ccc}16-12-1-3&-17+15+2&13-12-1\\-9+9-0-0&10-6-1-3&-7+6+1-0\\21-18-3&-25+24+1&19-15-1-3\end{array}\right]\\\\\\=\left[\begin{array}{ccc}0&0&0\\0&0&0\\0&0&0\end{array}\right]$