Question

Please solve the question no. 2 in an attached image

Answer 1

Question:

$\bf If\:A=\left[\begin{array}{ccc}1&2&1\\0&1&-1\\3&-1&1\end{array}\right],prove\:that\:A^{3} -3\:A^{2} -A+9\: I=0$

Answer:

Step-by-step explanation:

$\Large{\underline{\bf{Given:}}}$

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

$\Large{\underline{\bf{To\:Prove:}}}$

A³ - 3A² - A + 8I = 0

$\Large{\underline{\bf{Proof:}}}$

First finding A²

A² = A × A

Hence,

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

$A^{2} =\left[\begin{array}{ccc}1+0+3&2+2-1&1-2+1\\0+0-3&0+1+1&0-1-1\\3+0+3&6-1-1&3+1+1\end{array}\right]$

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

Now finding 3 A²

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

$3A^{2} =\left[\begin{array}{ccc}12&9&0\\-9&6&-6\\18&12&15\end{array}\right]$

Now finding A³

A³ = A² × A

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

$A^{3} =\left[\begin{array}{ccc}4+0+0&8+3+0&4-3+0\\-3+0-6&-6+2+2&-3-2-2\\6+0+15&12+4-5&6-4+5\end{array}\right]$

$A^{3} =\left[\begin{array}{ccc}4&11&1\\-9&-2&-7\\21&11&7\end{array}\right]$

Now we know that,

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

$9I= \left[\begin{array}{ccc}9&0&0\\0&9&0\\0&0&9\end{array}\right]$

First finding,

A³ - 3A²

$A^{3} -3A^{2} =\left[\begin{array}{ccc}4&11&1\\-9&-2&-7\\21&11&7\end{array}\right]-\left[\begin{array}{ccc}12&9&0\\-9&6&-6\\18&12&15\end{array}\right]$

$A^{3} -3A^{2} =\left[\begin{array}{ccc}-8&2&1\\0&-8&-1\\3&-1&-8\end{array}\right]$

Now finding,

A³-3A²-A

$A^{3} -3A^{2} -A=\left[\begin{array}{ccc}-8&2&1\\0&-8&-1\\3&-1&-8\end{array}\right]-\left[\begin{array}{ccc}1&2&1\\0&1&-1\\3&-1&1\end{array}\right]$

$A^{3} -3A^{2}-A=\left[\begin{array}{ccc}-9&0&0\\0&-9&0\\0&0&-9\end{array}\right]$

Now adding 9 I

$A^{3} -3A^{2}-A+9I=\left[\begin{array}{ccc}-9&0&0\\0&-9&0\\0&0&-9\end{array}\right]+ \left[\begin{array}{ccc}9&0&0\\0&9&0\\0&0&9\end{array}\right]$

$A^{3} -3A^{2}-A+9I=\left[\begin{array}{ccc}0&0&0\\0&0&0\\0&0&0\end{array}\right]$

= O

Hence proved.