Math, asked by ashwinrangani, 4 months ago

Please solve the question no. 2 in an attached image

Attachments:

Answers

Answered by TheValkyrie
10

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.

Similar questions