Matrix

Question

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

satisfies the following polynomial $A^{5}-4A^{4}-7A^{3}+11A^{2}-2A+kI=0$

Then the value of k is ______________

Answer 1

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

Answer 2

Comments

Well explained,,,diagonalization of matrix is part of gate syllabus??

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I think diagonalisation is in the gate syllabus. ( Not sure about it)

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Only eigen value and eigen vectors are mentioned.

Answer 3

Here's my approach

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

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

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

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

∴ $\color{green}{K =-15}$

Comments

Whenever a theorem can be applied, it is always better to use it because Profs. make questions for that purpose only. Brute force method will waste time and also completely useless for interviews. It can be used by those who do not prepare well in exam.

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Ok, will try to solve questions by using theorems from now on :)

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One thing, whenever writing SIR it is automatically changes to ***