Find adjoint of A where A={bmatrix}1 & 0 & 7 \\ 2 & 2 & 5 \\ 0 & 3 & 6{bmatrix}

Correct Answer is : [−321−14−12696−32]{bmatrix}-3 & 21 & -14 \\ -12 & 6 & 9 \\ 6 & -3 & 2{bmatrix}−3−126216−3−1492

{bmatrix}-3 & 21 & -14 \\ -12 & 6 & 9 \\ 6 & -3 & 2{bmatrix}

{bmatrix}-3 & -12 & 6 \\ 21 & 6 & -3 \\ -14 & 9 & 2{bmatrix}

{bmatrix}1 & 0 & 7 \\ 2 & 2 & 5 \\ 0 & 3 & 6{bmatrix}

Correct Answer is : [−321−14−12696−32]{bmatrix}-3 & 21 & -14 \\ -12 & 6 & 9 \\ 6 & -3 & 2{bmatrix}−3−126216−3−1492

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Matrices

Question : 14 of 36

Marks: +1, -0

Find adjoint of

A

A=\begin{bmatrix}1 & 0 & 7 \\ 2 & 2 & 5 \\ 0 & 3 & 6\end{bmatrix}

Solution:

Concept:

The adjoint of a matrix is the transpose of its cofactor matrix.

Explanation:

Given

A = \begin{bmatrix}1 & 0 & 7 \\ 2 & 2 & 5 \\ 0 & 3 & 6\end{bmatrix}

Compute each cofactor

C_{ij} = (-1)^{i+j} M_{ij}

where

M_{ij}

is the minor.

M_{11} = \begin{vmatrix}2 & 5 \\ 3 & 6\end{vmatrix} = 12-15 = -3

, so

C_{11} = -3

M_{12} = \begin{vmatrix}2 & 5 \\ 0 & 6\end{vmatrix} = 12-0 = 12

, so

C_{12} = -12

M_{13} = \begin{vmatrix}2 & 2 \\ 0 & 3\end{vmatrix} = 6-0 = 6

, so

C_{13} = 6

M_{21} = \begin{vmatrix}0 & 7 \\ 3 & 6\end{vmatrix} = 0-21 = -21

, so

C_{21} = 21

M_{22} = \begin{vmatrix}1 & 7 \\ 0 & 6\end{vmatrix} = 6-0 = 6

, so

C_{22} = 6

M_{23} = \begin{vmatrix}1 & 0 \\ 0 & 3\end{vmatrix} = 3-0 = 3

, so

C_{23} = -3

M_{31} = \begin{vmatrix}0 & 7 \\ 2 & 5\end{vmatrix} = 0-14 = -14

, so

C_{31} = -14

M_{32} = \begin{vmatrix}1 & 7 \\ 2 & 5\end{vmatrix} = 5-14 = -9

, so

C_{32} = 9

M_{33} = \begin{vmatrix}1 & 0 \\ 2 & 2\end{vmatrix} = 2-0 = 2

, so

C_{33} = 2

Thus cofactor matrix

C = \begin{bmatrix}-3 & -12 & 6 \\ 21 & 6 & -3 \\ -14 & 9 & 2\end{bmatrix}

Adjoint is transpose:

\operatorname{adj}(A) = C^T = \begin{bmatrix}-3 & 21 & -14 \\ -12 & 6 & 9 \\ 6 & -3 & 2\end{bmatrix}

This matches option A.

Answer:

Option A:

\begin{bmatrix}-3 & 21 & -14 \\ -12 & 6 & 9 \\ 6 & -3 & 2\end{bmatrix}

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