Question

Question:-
༒ Bʀᴀɪɴʟʏ Sᴛᴀʀs

༒Mᴏᴅᴇʀᴀᴛᴇʀs

༒Other best user
$\sf \: For \: A \: Matrix \: \: A \: \longrightarrow\rm\sf \: \: - \begin{gathered}\sf \left | \begin{array}{ccc}0&0& 1\\2&1&1\\1&1&1\end{array}\right | \end{gathered} \: {A}^{ - 1} \: is \: given \: by \:$
$\ \textless \ br /\ \textgreater \ \sf 1) A^2+2A+3I \qquad \qquad \qquad 2) A-3I \\ \\\sf 3) A^2-2A \qquad \qquad \qquad4) A^2-2A-I$
No Spam. ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given matrix is

$\rm \: A = \begin{gathered}\sf \left [ \begin{array}{ccc}0&0& 1\\2&1&1\\1&1&1\end{array}\right ] \end{gathered}$

We know, Characteristic equation is given by

$\rm \: |A - k \: I| \: = \: 0$

So,

$\rm \: \begin{gathered}\sf \left | \begin{array}{ccc} - k&0& 1\\2&1 - k&1\\1&1&1 - k\end{array}\right | \end{gathered} = 0$

$\rm \: - k[ {(1 - k)}^{2} - 1] + 1(2 - 1 + k) = 0$

$\rm \: - k[1 + {k}^{2} - 2k - 1] + (1 + k) = 0$

$\rm \: - k[{k}^{2} - 2k] + (1 + k) = 0$

$\rm \: -{k}^{3} + 2 {k}^{2} + 1 + k= 0$

$\rm \: -({k}^{3} - 2 {k}^{2} - 1 - k)= 0$

$\rm \: {k}^{3} - 2 {k}^{2} - k - 1= 0$

Now, We know

Cayley Hamilton Theorem states that every square matrix satisfy its characteristic equation.

So, A must satisfy the above characteristic equation.

$\rm \: {A}^{3} - 2 {A}^{2} - A - I= 0$

$\rm \: Now\:premultiply \: by \: {A}^{ - 1} \: on \: both \: sides$

$\rm \: {A}^{ - 1} {A}^{3} - 2{A}^{ - 1} {A}^{2} - {A}^{ - 1} A - {A}^{ - 1} I= 0$

$\rm \: ({A}^{ - 1}A) {A}^{2} - 2({A}^{ - 1} {A})A - I - {A}^{ - 1} = 0$

$\rm \: I{A}^{2} - 2IA - I - {A}^{ - 1} = 0$

$\rm \: {A}^{2} - 2A - I - {A}^{ - 1} = 0$

$\rm \: \rm\implies \: \: {A}^{ - 1} \: = \: {A}^{2} - 2A - I$

So, Option (4) is correct

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

BASIC RESULT USED

$\rm \: {A}^{ - 1} A = I$

$\rm \: AI \: = \: IA \: = \: A$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Answer 2

Answer:

Answer is given in attachment