%%

If $$A = \left[ {\begin{matrix} 2 & { - 3} \\ { - 4} & 1 \\ \end{matrix} } \right]$$, then adj(3A2 + 12A) is equal to

JEE · Math · previous-year question

  1. A.$$\left[ {\begin{matrix} {51} & {63} \\ {84} & {72} \\ \end{matrix} } \right]$$correct
  2. B.$$\left[ {\begin{matrix} {51} & {84} \\ {63} & {72} \\ \end{matrix} } \right]$$
  3. C.$$\left[ {\begin{matrix} {72} & {-63} \\ {-84} & {51} \\ \end{matrix} } \right]$$
  4. D.$$\left[ {\begin{matrix} {72} & {-84} \\ {-63} & {51} \\ \end{matrix} } \right]$$

Answer

A. $$\left[ {\begin{matrix} {51} & {63} \\ {84} & {72} \\ \end{matrix} } \right]$$

Explanation

We have, $$A = \left[ {\begin{matrix} 2 & { - 3} \\ { - 4} & 1 \\ \end{matrix} } \right]$$ $$ \therefore $$ A2 = A.A = $$\left[ {\begin{matrix} 2 & { - 3} \\ { - 4} & 1 \\ \end{matrix} } \right]\left[ {\begin{matrix} 2 & { - 3} \\ { - 4} & 1 \\ \end{matrix} } \right]$$ = $$\left[ {\begin{matrix} {4 + 12} & { - 6 - 3} \\ { - 8 - 4} & {12 + 1} \\ \end{matrix} } \right]$$ = $$\left[ {\begin{matrix} {16} & { - 9} \\ { - 12} & {13} \\ \end{matrix} } \right]$$ Now, 3A2 + 12A = $$3\left[ {\begin{matrix} {16} & { - 9} \\ { - 12} & {13} \\ \end{matrix} } \right] + 12\left[ {\begin{matrix} 2 & { - 3} \\ { - 4} & 1 \\ \end{matrix} } \right]$$ = $$\left[ {\begin{matrix} {48} & { - 27} \\ { - 36} & {39} \\ \end{matrix} } \right] + \left[ {\begin{matrix} {24} & { - 36} \\ { - 48} & {12} \\ \end{matrix} } \right]$$ = $$\left[ {\begin{matrix} {72} & { - 63} \\ { - 84} & {51} \\ \end{matrix} } \right]$$ $$ \therefore $$ adj(3A2 + 12A) = $$\left[ {\begin{matrix} {51} & {63} \\ {84} & {72} \\ \end{matrix} } \right]$$

Practice more JEE questions

Answer thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.

Practice JEE free →

More JEE Math questions