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Matrices and Determinants question

2017 · Shift 0 · Q29
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Matrices and Determinants question

2017 · Shift 0 · Q29

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If A=[2−3−41]A = \left[ {\begin{matrix} 2 & { - 3} \\ { - 4} & 1 \\ \end{matrix} } \right]A=[2−4​−31​], then adj(3A2 + 12A) is equal to
  1. A
    [51638472]\left[ {\begin{matrix} {51} & {63} \\ {84} & {72} \\ \end{matrix} } \right][5184​6372​]
  2. B
    [51846372]\left[ {\begin{matrix} {51} & {84} \\ {63} & {72} \\ \end{matrix} } \right][5163​8472​]
  3. C
    [72−63−8451]\left[ {\begin{matrix} {72} & {-63} \\ {-84} & {51} \\ \end{matrix} } \right][72−84​−6351​]
  4. D
    [72−84−6351]\left[ {\begin{matrix} {72} & {-84} \\ {-63} & {51} \\ \end{matrix} } \right][72−63​−8451​]
View written solutionFree

Correct answer: A

  1. Interpret the expression

Given

A=[2−3−41]A=\begin{bmatrix}2 & -3\\ -4 & 1\end{bmatrix}A=[2−4​−31​]

and we need to find

adj⁡(3A2+12A).\operatorname{adj}(3A^2+12A).adj(3A2+12A).
  1. Compute A2A^2A2
A2=A⋅A=[2−3−41][2−3−41]A^2=A\cdot A= \begin{bmatrix}2 & -3\\ -4 & 1\end{bmatrix} \begin{bmatrix}2 & -3\\ -4 & 1\end{bmatrix}A2=A⋅A=[2−4​−31​][2−4​−31​]

Now multiply:

  • First row, first column:
2⋅2+(−3)⋅(−4)=4+12=162\cdot 2+(-3)\cdot(-4)=4+12=162⋅2+(−3)⋅(−4)=4+12=16
  • First row, second column:
2⋅(−3)+(−3)⋅1=−6−3=−92\cdot(-3)+(-3)\cdot 1=-6-3=-92⋅(−3)+(−3)⋅1=−6−3=−9
  • Second row, first column:
(−4)⋅2+1⋅(−4)=−8−4=−12(-4)\cdot 2+1\cdot(-4)=-8-4=-12(−4)⋅2+1⋅(−4)=−8−4=−12
  • Second row, second column:
(−4)⋅(−3)+1⋅1=12+1=13(-4)\cdot(-3)+1\cdot 1=12+1=13(−4)⋅(−3)+1⋅1=12+1=13

So,

A2=[16−9−1213]A^2=\begin{bmatrix}16 & -9\\ -12 & 13\end{bmatrix}A2=[16−12​−913​]
  1. Compute 3A23A^23A2 and 12A12A12A
3A2=3[16−9−1213]=[48−27−3639]3A^2=3\begin{bmatrix}16 & -9\\ -12 & 13\end{bmatrix} =\begin{bmatrix}48 & -27\\ -36 & 39\end{bmatrix}3A2=3[16−12​−913​]=[48−36​−2739​] 12A=12[2−3−41]=[24−36−4812]12A=12\begin{bmatrix}2 & -3\\ -4 & 1\end{bmatrix} =\begin{bmatrix}24 & -36\\ -48 & 12\end{bmatrix}12A=12[2−4​−31​]=[24−48​−3612​]
  1. Add them
3A2+12A=[48−27−3639]+[24−36−4812]=[72−63−8451]3A^2+12A= \begin{bmatrix}48 & -27\\ -36 & 39\end{bmatrix} + \begin{bmatrix}24 & -36\\ -48 & 12\end{bmatrix} = \begin{bmatrix}72 & -63\\ -84 & 51\end{bmatrix}3A2+12A=[48−36​−2739​]+[24−48​−3612​]=[72−84​−6351​]

Let

M=[72−63−8451]M=\begin{bmatrix}72 & -63\\ -84 & 51\end{bmatrix}M=[72−84​−6351​]
  1. Find the adjoint of a 2×22\times 22×2 matrix

For

[abcd],\begin{bmatrix}a & b\\ c & d\end{bmatrix},[ac​bd​],

we have

adj⁡[abcd]=[d−b−ca]\operatorname{adj}\begin{bmatrix}a & b\\ c & d\end{bmatrix} = \begin{bmatrix}d & -b\\ -c & a\end{bmatrix}adj[ac​bd​]=[d−c​−ba​]

Thus,

adj⁡(M)=[51638472]\operatorname{adj}(M)= \begin{bmatrix}51 & 63\\ 84 & 72\end{bmatrix}adj(M)=[5184​6372​]
  1. Match with the options

This matches Option A.

[51638472]\boxed{\begin{bmatrix}51 & 63\\ 84 & 72\end{bmatrix}}[5184​6372​]​
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