MHT CET · Maths · Determinants
For the matrix \(\mathrm{A}=\left[\begin{array}{ccc}2 & 0 & -1 \\ 3 & 1 & 2 \\ -1 & 1 & 2\end{array}\right]\), the matrix of cofactors is
- A \(\left[\begin{array}{ccc}0 & 8 & -4 \\ -1 & 3 & 2 \\ 1 & -7 & 2\end{array}\right]\)
- B \(\left[\begin{array}{ccc}0 & -8 & 4 \\ -1 & 3 & -2 \\ 1 & -7 & 2\end{array}\right]\)
- C \(\left[\begin{array}{ccc}0 & 8 & -4 \\ 1 & -3 & 2 \\ -1 & 7 & -2\end{array}\right]\)
- D \(\left[\begin{array}{ccc}0 & -8 & 4 \\ -1 & 3 & 2 \\ -1 & -7 & 2\end{array}\right]\)
Answer & Solution
Correct Answer
(B) \(\left[\begin{array}{ccc}0 & -8 & 4 \\ -1 & 3 & -2 \\ 1 & -7 & 2\end{array}\right]\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{A}_{11}=(-1)^{1+1}\left|\begin{array}{ll}1 & 2 \\ 1 & 2\end{array}\right|=1(0)=0 \\ & \mathrm{~A}_{12}=(-1)^{1+2}\left|\begin{array}{cc}3 & 2 \\ -1 & 2\end{array}\right|=(-1)(8)=-8 \\ & \mathrm{~A}_{13}=(-1)^{1+3}\left|\begin{array}{cc}3 & 1 \\ -1 & 1\end{array}\right|=(1)(4)=4 \\ & \mathrm{~A}_{21}=(-1)^{2+1}\left|\begin{array}{cc}0 & -1 \\ 1 & 2\end{array}\right|=(-1)(1)=-1 \\ & \mathrm{~A}_{22}=(-1)^{2+2}\left|\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right|=(1)(3)=3 \\ & \mathrm{~A}_{23}=(-1)^{2+3}\left|\begin{array}{cc}2 & 0 \\ -1 & 1\end{array}\right|=(-1)(2)=-2 \\ & \mathrm{~A}_{31}=(-1)^{3+1}\left|\begin{array}{cc}0 & -1 \\ 1 & 2\end{array}\right|=(1)(1)=1 \\ & \mathrm{~A}_{32}=(-1)^{3+2}\left|\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right|=(-1)(7)=-7 \\ & \mathrm{~A}_{33}=(-1)^{3+3}\left|\begin{array}{ll}2 & 0 \\ 3 & 1\end{array}\right|=(1)(2)=2\end{aligned}\)
\(\therefore \) The matrix of the cofactors is \(\left[\begin{array}{ccc}0 & -8 & 4 \\ -1 & 3 & -2 \\ 1 & -7 & 2\end{array}\right]\) :
\(\therefore \) The matrix of the cofactors is \(\left[\begin{array}{ccc}0 & -8 & 4 \\ -1 & 3 & -2 \\ 1 & -7 & 2\end{array}\right]\) :
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