If B = ≤ft[ {{array}{ccc} 3&2&0\\ 2&4&0\\ 1&1&0 {array}} ], then what is adjoint of B equal to?

Correct Answer is : [000000−2−18]≤ft[ {{array}{ccc} 0&0&0\\ 0&0&0\\ { - 2}&{ - 1}&8 {array}} ]00−200−1008

≤ft[ {{array}{ccc} 0&0&0\\ 0&0&0\\ { - 2}&{ - 1}&8 {array}} ]

≤ft[ {{array}{ccc} 0&0&{ - 2}\\ 0&0&{ - 1}\\ 0&0&8 {array}} ]

≤ft[ {{array}{ccc} 0&0&2\\ 0&0&1\\ 0&0&0 {array}} ]

Correct Answer is : [000000−2−18]≤ft[ {{array}{ccc} 0&0&0\\ 0&0&0\\ { - 2}&{ - 1}&8 {array}} ]00−200−1008

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UPSC NDA I 2019 Mathematics Solved Paper

Question : 7 of 120

Marks: +1, -0

B = \left[ {\begin{array}{ccc} 3&2&0\\ 2&4&0\\ 1&1&0 \end{array}} \right]

, then what is adjoint of B equal to?

$\left[ {\begin{array}{ccc} 0&0&0\\ 0&0&0\\ { - 2}&{ - 1}&8 \end{array}} \right]$
$\left[ {\begin{array}{ccc} 0&0&{ - 2}\\ 0&0&{ - 1}\\ 0&0&8 \end{array}} \right]$
$\left[ {\begin{array}{ccc} 0&0&2\\ 0&0&1\\ 0&0&0 \end{array}} \right]$
It does not exist

Solution:

Concept:

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

Explanation:

Given matrix

B = \begin{bmatrix} 3 & 2 & 0 \\ 2 & 4 & 0 \\ 1 & 1 & 0 \end{bmatrix}

We first find the cofactor

B_{ij}

of each element

b_{ij}

B_{ij} = (-1)^{i+j} \times M_{ij}

, where

M_{ij}

is the minor (determinant after deleting row

i

and column

j

Compute all cofactors:

B_{11} = (-1)^{1+1} \cdot \begin{vmatrix}4 & 0 \\ 1 & 0\end{vmatrix} = 1 \cdot (4\cdot0 - 0\cdot1) = 0

B_{12} = (-1)^{1+2} \cdot \begin{vmatrix}2 & 0 \\ 1 & 0\end{vmatrix} = -1 \cdot (2\cdot0 - 0\cdot1) = 0

B_{13} = (-1)^{1+3} \cdot \begin{vmatrix}2 & 4 \\ 1 & 1\end{vmatrix} = 1 \cdot (2\cdot1 - 4\cdot1) = -2

B_{21} = (-1)^{2+1} \cdot \begin{vmatrix}2 & 0 \\ 1 & 0\end{vmatrix} = -1 \cdot (2\cdot0 - 0\cdot1) = 0

B_{22} = (-1)^{2+2} \cdot \begin{vmatrix}3 & 0 \\ 1 & 0\end{vmatrix} = 1 \cdot (3\cdot0 - 0\cdot1) = 0

B_{23} = (-1)^{2+3} \cdot \begin{vmatrix}3 & 2 \\ 1 & 1\end{vmatrix} = -1 \cdot (3\cdot1 - 2\cdot1) = -1

B_{31} = (-1)^{3+1} \cdot \begin{vmatrix}2 & 0 \\ 4 & 0\end{vmatrix} = 1 \cdot (2\cdot0 - 0\cdot4) = 0

B_{32} = (-1)^{3+2} \cdot \begin{vmatrix}3 & 0 \\ 2 & 0\end{vmatrix} = -1 \cdot (3\cdot0 - 0\cdot2) = 0

B_{33} = (-1)^{3+3} \cdot \begin{vmatrix}3 & 2 \\ 2 & 4\end{vmatrix} = 1 \cdot (3\cdot4 - 2\cdot2) = 12 - 4 = 8

Thus, the cofactor matrix

C = \begin{bmatrix} 0 & 0 & -2 \\ 0 & 0 & -1 \\ 0 & 0 & 8 \end{bmatrix}

The adjoint is the transpose:

\operatorname{adj}(B) = C^T = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ -2 & -1 & 8 \end{bmatrix}

Answer:

Option A:

\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ -2 & -1 & 8 \end{bmatrix}

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