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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Matrices

NDA 2017 · Mathematics · Matrices

The adjoint of the matrix A=[ array lll 1 & 0 & 2 2 & 1 & 0 0 & 3 &...

The adjoint of the matrix A=\left[102210031\begin{array}{lll}1 & 0 & 2 \\ 2 & 1 & 0 \\ 0 & 3 & 1\end{array}\right] is

Options

  1. A\left[−162−21−4631\begin{array}{lrr}-1 & 6 & 2 \\ -2 & 1 & -4 \\ 6 & 3 & 1\end{array}\right]
  2. B\left[16−2−2146−31\begin{array}{lcr}1 & 6 & -2 \\ -2 & 1 & 4 \\ 6 & -3 & 1\end{array}\right]
  3. C\left[6124−1263−1\begin{array}{rrr}6 & 1 & 2 \\ 4 & -1 & 2 \\ 6 & 3 & -1\end{array}\right]
  4. D\left[−6214−2131−6\begin{array}{lll}-6 & 2 & 1 \\ 4 & -2 & 1 \\ 3 & 1 & -6\end{array}\right]

Correct answer

B. \left[16−2−2146−31\begin{array}{lcr}1 & 6 & -2 \\ -2 & 1 & 4 \\ 6 & -3 & 1\end{array}\right]

Step-by-step solution

A=[102210031]A11=1−0=1;A12=−(2−0)=−2;A13=6−0=6A21=−(0−6)=6;A22=1−0=1;A23=−3−0=−3A31=0−2=−2;A32=−(0−4)=4;A33=1−0=1\begin{aligned} & A=\left[\begin{array}{lll}1 & 0 & 2 \\ 2 & 1 & 0 \\ 0 & 3 & 1\end{array}\right] \\ & A_{11}=1-0=1 ; A_{12}=-(2-0)=-2 ; A_{13}=6-0=6 \\ & A_{21}=-(0-6)=6 ; A_{22}=1-0=1 ; A_{23}=-3-0=-3 \\ & A_{31}=0-2=-2 ; A_{32}=-(0-4)=4 ; A_{33}=1-0=1\end{aligned}\therefore \operatorname{Adj} A=\left[16−2−2146−31\begin{array}{ccc}1 & 6 & -2 \\ -2 & 1 & 4 \\ 6 & -3 & 1\end{array}$\right]
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