For the matrix A -11112-32-13 - Show that A -3-6 A 2-5 A -11 I 3- O- Hence- find A-1-

A=1 #160; #160; #160; #160; #160;1 #160; #160; #160; #160; #160; #160;11 #160; #160; #160; #160; #160;2 #160; #160; #160; ...

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Byju's Answer

Standard XII

Mathematics

Integration by Partial Fractions

For the matri...

Question

For the matrix $A = [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{matrix}]$ . Show that $A^{- 3} - 6 A^{2} + 5 A + 11 I_{3} = O$ . Hence, find A⁻¹.

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Solution

$A = [1 1 1 1 2 - 3 2 - 1 3] \Rightarrow |A| = |1 1 1 1 2 - 3 2 - 1 3| = (1 \times 3) - (1 \times 9) + (1 \times - 5) = 3 - 9 - 5 = - 11 Since, |A| \neq 0 Hence, A^{- 1} exists . Now, A^{2} = [1 1 1 1 2 - 3 2 - 1 3] [1 1 1 1 2 - 3 2 - 1 3] = [1 + 1 + 2 1 + 2 - 1 1 - 3 + 3 1 + 2 - 6 1 + 4 + 3 1 - 6 - 9 2 - 1 + 6 2 - 2 - 3 2 + 3 + 9] = [4 2 1 - 3 8 - 14 7 - 3 14] and A^{3} = A^{2} . A = [4 2 1 - 3 8 - 14 7 - 3 14] [1 1 1 1 2 - 3 2 - 1 3] = [4 + 2 + 2 4 + 4 - 1 4 - 6 + 3 - 3 + 8 - 28 - 3 + 16 + 14 - 3 - 24 - 42 7 - 3 + 28 7 - 6 - 14 7 + 9 + 42] = [8 7 1 - 23 27 - 69 32 - 13 58] Now, A^{3} - 6 A^{2} + 5 A + 11 I_{3} = [8 7 1 - 23 27 - 69 32 - 13 58] - 6 [4 2 1 - 3 8 - 14 7 - 3 14] + 5 [1 1 1 1 2 - 3 2 - 1 3] + 11 [1 0 0 0 1 0 0 0 1] = [8 - 24 + 5 + 11 7 - 12 + 5 + 0 1 - 6 + 5 + 0 - 23 + 18 + 5 + 0 27 - 48 + 10 + 11 - 69 + 84 - 15 + 0 32 - 42 + 10 + 0 - 13 + 18 - 5 + 0 58 - 84 + 15 + 11] = [0 0 0 0 0 0 0 0 0] = O (Null matrix) Again, A^{3} - 6 A^{2} + 5 A + 11 I_{3} = O \Rightarrow A^{- 1} \times (A^{3} - 6 A^{2} + 5 A + 11 I_{3}) = A^{- 1} \times O (Pre - multiplying both sides because A^{- 1} exists) \Rightarrow (A^{2} - 6 A + 5 I_{3} + 11 A^{- 1}) = 0 \Rightarrow [4 2 1 - 3 8 - 14 7 - 3 14] - 6 [1 1 1 1 2 - 3 2 - 1 3] + 5 [1 0 0 0 1 0 0 0 1] = - 11 A^{- 1} \Rightarrow [4 - 6 + 5 2 - 6 + 0 1 - 6 + 0 - 3 - 6 + 0 8 - 12 + 5 - 14 + 18 + 0 7 - 12 + 0 - 3 + 6 + 0 14 - 18 + 5] = - 11 A^{- 1} \Rightarrow A^{- 1} = - \frac{1}{11} [3 - 4 - 5 - 9 1 4 - 5 3 1]$

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Similar questions

Q. For the matrix show that A 3 − 6 A 2 + 5 A + 11 I = O. Hence, find A −1.

Q. For the matrix

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 1 & 2 & - 3 2 & - 1 & 3 \end{matrix} ⎤ ⎥ ⎦

, show that

A^{3} - {6 A}^{2} + 5 A + 11 I = 0

. Hence find

A^{- 1}

For the matrixshow that A³ − 6A² + 5A + 11 I = O. Hence, find A^−1.

Q. If

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 1 & 2 & - 3 2 & - 1 & 3 \end{matrix} ⎤ ⎥ ⎦

show that

A^{3} - 6 A^{2} + 5 A + 11 I = 0

. Hence find

A^{- 1}

For the matrix $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 1 & 2 & - 3 2 & - 1 & 3 \end{matrix} ⎤ ⎥ ⎦$ . Show that $A^{3} - 6 A^{2} + 5 A + 11 I = 0$ . Hence, find $A^{- 1}$

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For the matrix A=11112-32-13. Show that A-3-6A2+5A+11 I3=O. Hence, find A−1.

For the matrix $A = [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{matrix}]$ . Show that $A^{- 3} - 6 A^{2} + 5 A + 11 I_{3} = O$ . Hence, find A⁻¹.