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6 votes

According to Cayley-Hamilton Theorem ,

"Every Square Matrix satisfies its own characteristic equation"

Ref :-  https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem

So , for the given matrix :-

characteristic equation is :-   (1-λ)(3-λ) - 4*2 = 0 where  λ = eigen value

So,   λ2 - 4λ -5 = 0

Now , given square matrix will satisfy this characteristic equation.

So, A2 - 4A - 5I = 0  -------- equation (1)

now , here polynomial is :-

A5−4A4−7A3+11A2−2A+kI=0

A5 - 4A4 - 5A3 - 2A3 + 11A2 - 2A +kI = 0

A3(A2-4A - 5I) - 2A3 + 11A2 - 2A +kI = 0

-2A3 + 11A2 - 2A +kI = 0       (Since , according to equation (1) , A2 - 4A - 5I = 0 )

-2A3 + 8A2 +10A +3A2 -2A - 10A+kI = 0

-2A(A2 -4A - 5I) +3A2 -2A - 10A+kI = 0

3A2 -12A +kI = 0  (According to equation (1))

3A2 -12A -15I +15I +kI = 0

3(A2-4A - 5I) +15I +kI = 0

15I + kI = 0 (According to equation (1))

(15+k)I = 0

now put Identity matrix 'I' here and after solving it ,we will get ,  k =  -15

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Answer

1 votes
1 votes
Here's my approach

------------------------------------------------------------------

$A = \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix}$

∴ $\color{maroon}{-2A = \begin{bmatrix} -2 & -8 \\ -4 & -6 \end{bmatrix}}$

$A^2 = \begin{bmatrix} (1×1+4×2)=9 & (1×4+4×3)=16 \\ (2×1+3×2)=8 & (2×4+3×3)=17 \end{bmatrix}$

Or, $A^2 = \begin{bmatrix} 9 & 16 \\ 8 & 17 \end{bmatrix}$

Or, $11A^2 = \begin{bmatrix} (11×9) = 99 & (11 ×16) =176 \\ (11 ×8) =88 & (11 ×17) = 187 \end{bmatrix}$

Or, $\color{maroon}{11A^2 = \begin{bmatrix}  99 & 176 \\ 88 & 187 \end{bmatrix}}$

$A^3 = \begin{bmatrix} (9×1+16×2)=41 & (9×4+16×3)=84 \\ (8×1+17×2)=42 & (8×4+17×3)=83 \end{bmatrix}$

Or,$A^3 = \begin{bmatrix} 41 & 84 \\ 42 & 83 \end{bmatrix}$

Or, $-7A^3 = \begin{bmatrix} (-7×41) = -287 & (-7×84) = -588 \\ (-7×42)= -294 & (-7×83)= -581 \end{bmatrix}$

Or, $\color{maroon}{-7A^3 = \begin{bmatrix}  -287 &  -588 \\  -294 & -581 \end{bmatrix}}$

$A^4 = \begin{bmatrix} (41×1+84×2)=209 & (41×4+84×3)=416 \\ (42×1+83×2)=208 & (42×4+83×3)=417 \end{bmatrix}$

Or, $A^4 = \begin{bmatrix} 209 & 416 \\ 208 & 417 \end{bmatrix}$

Or, $-4A^4 = \begin{bmatrix} (-4×209)= -836 & (-4× 416)= -1664 \\ (-4×208)=-832 & (-4×417)=-1668 \end{bmatrix}$

Or,$\color{maroon}{-4A^4 = \begin{bmatrix} -836 &  -1664 \\ -832 & -1668 \end{bmatrix}}$

$A^5 = \begin{bmatrix} (209×1+416×2)=1041 & (209×4+416×3)=2084 \\ (208×1+417×2)=1042 & (208×4+417×3)=2083 \end{bmatrix}$

Or, $\color{maroon}{A^5 = \begin{bmatrix} 1041 & 2084 \\ 1042 & 2083 \end{bmatrix}}$

∴ $A^5-4A^4-7A^3+11A^2-2A = \begin{bmatrix} (1041-836-287+99-2)=15 & (2084-1664-588+176-8)= 0 \\ (1042-832-294+88-4)=0 & (2083-1668-581+187-6)=15 \end{bmatrix}$

Or, $\color{blue}{A^5-4A^4-7A^3+11A^2-2A = \begin{bmatrix} 15 & 0 \\ 0 & 15 \end{bmatrix}}$

Given that, $A^5-4A^4-7A^3+11A^2-2A+KI =0$

Or, $\color{teal}{\begin{bmatrix} 15 & 0 \\ 0 & 15 \end{bmatrix} + \begin{bmatrix} K & 0 \\ 0 & K \end{bmatrix}=0}$

Or, $\begin{bmatrix} K & 0 \\ 0 & K \end{bmatrix} = -\begin{bmatrix} 15 & 0 \\ 0 & 15 \end{bmatrix}$

Or, $\begin{bmatrix} K & 0 \\ 0 & K \end{bmatrix} = \begin{bmatrix} -15 & 0 \\ 0 & -15 \end{bmatrix}$

∴ $\color{green}{K =-15}$
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