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JEE Main 202117 MarchMorning Shift
Exam
JEE Main
Year
2021
Date
17 March
Shift
Morning Shift
Subject
Mathematics
Chapter
Matrices

JEE Main 2021 · 17 March · Morning Shift · Mathematics · Matrices

If A = [ cases 2 & 3 0 & - 1 cases ], then the value of det ( A ^ 4...

If A = \left[\begin{cases}2 & 3 \\ 0 & - 1\end{cases}\right], then the value of det(A4)+det(A10−(Adj(2A))10)det \left({A}^{4}\right) + det \left({A}^{10} - ( Adj ( 2A ) {)}^{10}\right) is equal to ________.

Correct answer

0

Step-by-step solution

Given, A = \left[\begin{cases}2 & 3 \\ 0 & - 1\end{cases}\right]

⇒ {A}^{2} = \left[\begin{cases}2 & 3 \\ 0 & - 1\end{cases}\right]\left[\begin{cases}2 & 3 \\ 0 & - 1\end{cases}\right] = \left[\begin{cases}4 & 3 \\ 0 & 1\end{cases}\right]

⇒ {A}^{4} = \left[\begin{cases}4 & 3 \\ 0 & 1\end{cases}\right]\left[\begin{cases}4 & 3 \\ 0 & 1\end{cases}\right] = \left[\begin{cases}16 & 15 \\ 0 & 1\end{cases}\right]

⇒ {A}^{5} = \left[\begin{cases}16 & 15 \\ 0 & 1\end{cases}\right]\left[\begin{cases}2 & 3 \\ 0 & - 1\end{cases}\right] = \left[\begin{cases}32 & 33 \\ 0 & - 1\end{cases}\right]

⇒ {A}^{10} = \left[\begin{cases}32 & 33 \\ 0 & - 1\end{cases}\right]\left[\begin{cases}32 & 33 \\ 0 & - 1\end{cases}\right] = \left[\begin{cases}1024 & 1023 \\ 0 & 1\end{cases}\right] . . . . \left(i\right)

Now, 2A = \left[\begin{cases}4 & 6 \\ 0 & - 2\end{cases}\right]

⇒ Adj\left(2A\right) = {\left[\begin{cases}- 2 & 0 \\ - 6 & 4\end{cases}\right]}^{T}

⇒ Adj\left(2A\right) = \left[\begin{cases}- 2 & - 6 \\ 0 & 4\end{cases}\right]

⇒ {\left(Adj\left(2A\right)\right)}^{2} = \left[\begin{cases}- 2 & - 6 \\ 0 & 4\end{cases}\right]\left[\begin{cases}- 2 & - 6 \\ 0 & 4\end{cases}\right] = \left[\begin{cases}4 & - 12 \\ 0 & 16\end{cases}\right]

⇒ {\left(Adj\left(2A\right)\right)}^{4} = \left[\begin{cases}4 & - 12 \\ 0 & 16\end{cases}\right]\left[\begin{cases}4 & - 12 \\ 0 & 16\end{cases}\right] = \left[\begin{cases}16 & - 240 \\ 0 & 256\end{cases}\right]

⇒ {\left(Adj\left(2A\right)\right)}^{5} = \left[\begin{cases}16 & - 240 \\ 0 & 256\end{cases}\right]\left[\begin{cases}- 2 & - 6 \\ 0 & 4\end{cases}\right] = \left[\begin{cases}- 32 & - 1056 \\ 0 & 1024\end{cases}\right]

⇒ {\left(Adj\left(2A\right)\right)}^{10} = \left[\begin{cases}- 32 & - 1056 \\ 0 & 1024\end{cases}\right]\left[\begin{cases}- 32 & - 1056 \\ 0 & 1024\end{cases}\right] = \left[\begin{cases}1024 & - 1047552 \\ 0 & 1048576\end{cases}\right] . . . . . \left(ii\right)

From equation \left(i\right) \text{ and } \left(ii\right)</p><p></p><p>⇒ {A}^{10} - {\left(Adj\left(2A\right)\right)}^{10} = \left[{1024102301\begin{cases}1024 & 1023 \\ 0 & 1\end{cases}\right] - \left[{1024−104755201048576\begin{cases}1024 & - 1047552 \\ 0 & 1048576\end{cases}\right] = \left[{010485750−1048575\begin{cases}0 & 1048575 \\ 0 & - 1048575\end{cases}\right]</p><p></p><p>⇒ det\left({A}^{10} - {\left(Adj\left(2A\right)\right)}^{10}\right) = \left|{010485750−1048575\begin{cases}0 & 1048575 \\ 0 & - 1048575\end{cases}\right| = 0</p><p>Let</p><p>LetE = det \left({A}^{4}\right) + det \left({A}^{10} - ( Adj ( 2A ) {)}^{10}\right)</p><p></p><p>⇒ E = {\left(det \left(A\right)\right)}^{4} + 0</p><p></p><p>⇒ E = {\left(- 2 - 0\right)}^{4}</p><p></p><p>⇒ E = 16$

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