Let A - begin bmatrix 1 -2 - 1-2 - 3 - 1111 - 1 - 5lend bmatrix A -1-1- A

|A^ 1|= 14/13 ( 4/169 9/169 )+11/13 ( 11/169 15/169 )+5/13 ( 33/169+20/169 ) = 14/13 ( 13/169 )+11/13 ( 26/169 )+5/13 ( 13/169 ) =14/169 22/169 5/169= 1 ...

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Byju's Answer

Standard XII

Mathematics

Proof by mathematical induction

Let A = begin...

Question

Let $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 2 & 1 - 2 & 3 & 1 1 & 1 & 5 \end{matrix} ⎤ ⎥ ⎦$

$(A^{- 1})^{- 1} = A$

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Solution

$| A^{- 1} | = - \frac{14}{13} (- \frac{4}{169} - \frac{9}{169}) + \frac{11}{13} (- \frac{11}{169} - \frac{15}{169}) + \frac{5}{13} (- \frac{33}{169} + \frac{20}{169}) = - \frac{14}{13} (- \frac{13}{169}) + \frac{11}{13} (- \frac{26}{169}) + \frac{5}{13} (- \frac{13}{169}) = \frac{14}{169} - \frac{22}{169} - \frac{5}{169} = - \frac{13}{169} = - \frac{1}{13}$
and $a d j (A^{- 1}) = \frac{1}{13} ⎡ ⎢ ⎣ \begin{matrix} - 1 & 2 & - 1 2 & - 3 & - 1 - 1 & - 1 & - 5 \end{matrix} ⎤ ⎥ ⎦$
$∴ (A^{- 1}) = \frac{1}{| A^{- 1} |} (a d j A^{- 1}) = \frac{1}{- \frac{1}{13}} \times \frac{1}{13} ⎡ ⎢ ⎣ \begin{matrix} - 1 & 2 & - 1 2 & - 3 & - 1 - 1 & - 1 & - 5 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 2 & 1 - 2 & 3 & 1 1 & 1 & 5 \end{matrix} ⎤ ⎥ ⎦ = A$

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The expression which is not defined is:

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Let A=⎡⎢⎣1−21−231115⎤⎥⎦ (A−1)−1=A

Let $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 2 & 1 - 2 & 3 & 1 1 & 1 & 5 \end{matrix} ⎤ ⎥ ⎦$

$(A^{- 1})^{- 1} = A$