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Exercise B · Q8

Q.For A=[1233−21421]A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix} show that A3−23A−40I=OA^3 - 23A - 40I = O, where II is an identity matrix of order 3, and OO is zero matrix.

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Computing A2A^2, then A3A^3, gives A3−23A=40IA^3-23A=40I, so A3−23A−40I=OA^3-23A-40I=O.

Evaluate the matrix polynomial by successive multiplication: A2=A⋅AA^2=A\cdot A, A3=A2⋅AA^3=A^2\cdot A, then combine with 23A23A and 40I40I.

Given A=[1233−21421]A=\begin{bmatrix}1&2&3\\3&-2&1\\4&2&1\end{bmatrix}.

  1. Compute A2=A⋅AA^2=A\cdot A:

A2=[1+6+122−4+63+2+33−6+46+4+29−2+14+6+48−4+212+2+1]=[1948112814615]A^2=\begin{bmatrix}1+6+12&2-4+6&3+2+3\\3-6+4&6+4+2&9-2+1\\4+6+4&8-4+2&12+2+1\end{bmatrix}=\begin{bmatrix}19&4&8\\1&12&8\\14&6&15\end{bmatrix}

  1. Compute A3=A2⋅AA^3=A^2\cdot A. Row 1: [19+12+32, 38−8+16, 57+4+8]=[63,46,69][19+12+32,\ 38-8+16,\ 57+4+8]=[63,46,69]; Row 2: [1+36+32, 2−24+16, 3+12+8]=[69,−6,23][1+36+32,\ 2-24+16,\ 3+12+8]=[69,-6,23]; Row 3: [14+18+60, 28−12+30, 42+6+15]=[92,46,63][14+18+60,\ 28-12+30,\ 42+6+15]=[92,46,63]: A3=[63466969−623924663]A^3=\begin{bmatrix}63&46&69\\69&-6&23\\92&46&63\end{bmatrix} …

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