Find a point of matrix {bmatrix} 5 & -6 & 4 \\ 7 & 4 & -3 \\ 2 & 1 & 6 {bmatrix}

Correct Answer is : [27402−482243−1−1762]{bmatrix} 27 & 40 & 2 \\ -48 & 22 & 43 \\ -1 & -17 & 62 {bmatrix}27−48−14022−1724362

{bmatrix} 2 & -2 & -4 \\ 0 & -1 & 1 \\ -2 & 1 & 3 {bmatrix}

{bmatrix} 27 & 40 & 2 \\ -48 & 22 & 43 \\ -1 & -17 & 62 {bmatrix}

{bmatrix} 27 & -48 & -1 \\ 40 & 22 & -17 \\ 2 & 43 & 62 {bmatrix}

Correct Answer is : [27402−482243−1−1762]{bmatrix} 27 & 40 & 2 \\ -48 & 22 & 43 \\ -1 & -17 & 62 {bmatrix}27−48−14022−1724362

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Question : 11 of 36

Marks: +1, -0

Find a point of matrix

\begin{bmatrix} 5 & -6 & 4 \\ 7 & 4 & -3 \\ 2 & 1 & 6 \end{bmatrix}

Solution:

Concept:

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

Explanation:

Given matrix

A = \begin{bmatrix} 5 & -6 & 4 \\ 7 & 4 & -3 \\ 2 & 1 & 6 \end{bmatrix}

Compute cofactors:

C_{11} = +\begin{vmatrix}4&-3\\1&6\end{vmatrix} = 27

C_{12} = -\begin{vmatrix}7&-3\\2&6\end{vmatrix} = -48

C_{13} = +\begin{vmatrix}7&4\\2&1\end{vmatrix} = -1

C_{21} = -\begin{vmatrix}-6&4\\1&6\end{vmatrix} = 40

C_{22} = +\begin{vmatrix}5&4\\2&6\end{vmatrix} = 22

C_{23} = -\begin{vmatrix}5&-6\\2&1\end{vmatrix} = -17

C_{31} = +\begin{vmatrix}-6&4\\4&-3\end{vmatrix} = 2

C_{32} = -\begin{vmatrix}5&4\\7&-3\end{vmatrix} = 43

C_{33} = +\begin{vmatrix}5&-6\\7&4\end{vmatrix} = 62

Cofactor matrix =

\begin{bmatrix}27&-48&-1\\40&22&-17\\2&43&62\end{bmatrix}

Transpose to get adjoint:

\begin{bmatrix}27&40&2\\-48&22&43\\-1&-17&62\end{bmatrix}

, which matches option B.

Answer:

Option B:

\begin{bmatrix}27&40&2\\-48&22&43\\-1&-17&62\end{bmatrix}

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