If A -102021203- then show that A is a root of the polynomial fx-x3-6 x2-7 x-2

Given: #8201; #160;fx=x3 6x2+7x+2fA=A3 6A2+7A+2I3Now, #160; #160;A2=AA #8658;A2=102021203102021203 #8658;A2=1+0+40+0+02+0+60+0+20+4+00+2+32+0+60+0+0 ...

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Standard XII

Mathematics

Inequalities Involving Modulus Function

If A =1020212...

Question

If $A = [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}]$ , then show that A is a root of the polynomial f (x) = x³ − 6x² + 7x + 2.

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Solution

$Given : f (x) = x^{3} - 6 x^{2} + 7 x + 2 f (A) = A^{3} - 6 A^{2} + 7 A + 2 I_{3} Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 + 0 + 4 & 0 + 0 + 0 & 2 + 0 + 6 \\ 0 + 0 + 2 & 0 + 4 + 0 & 0 + 2 + 3 \\ 2 + 0 + 6 & 0 + 0 + 0 & 4 + 0 + 9 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix}] A^{3} = A^{2} A \Rightarrow A^{3} = [\begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix}] [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] \Rightarrow A^{3} = [\begin{matrix} 5 + 0 + 16 & 0 + 0 + 0 & 10 + 0 + 24 \\ 2 + 0 + 10 & 0 + 8 + 0 & 4 + 4 + 15 \\ 8 + 0 + 26 & 0 + 0 + 0 & 16 + 0 + 39 \end{matrix}] \Rightarrow A^{3} = [\begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix}] A^{3} - 6 A^{2} + 7 A + 2 I_{3} \Rightarrow A^{3} - 6 A^{2} + 7 A + 2 I_{3} = [\begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix}] - 6 [\begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix}] + 7 [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] + 2 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow A^{3} - 6 A^{2} + 7 A + 2 I_{3} = [\begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix}] - [\begin{matrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{matrix}] + [\begin{matrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{matrix}] + [\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}] \Rightarrow A^{3} - 6 A^{2} + 7 A + 2 I_{3} = [\begin{matrix} 21 - 30 + 7 + 2 & 0 - 0 + 0 + 0 & 34 - 48 + 14 + 0 \\ 12 - 12 + 0 + 0 & 8 - 24 + 14 + 2 & 23 - 30 + 7 + 0 \\ 34 - 48 + 14 + 0 & 0 - 0 + 0 + 0 & 55 - 78 + 21 + 2 \end{matrix}] \Rightarrow A^{3} - 6 A^{2} + 7 A + 2 I_{3} = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] = 0 Since f (A) = 0, A is the root of f (x) = x^{3} - 6 x^{2} + 7 x + 2 .$

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If A=102021203, then show that A is a root of the polynomial f (x) = x3 − 6x2 + 7x + 2.

If $A = [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}]$ , then show that A is a root of the polynomial f (x) = x³ − 6x² + 7x + 2.