Test Index

AP EAPCET 08-Jul-2022 Shift 2 Paper

Section: Mathematics

© examsnet.com

Question : 4 of 160

Marks: +1, -0

If

\begin{pmatrix} 5 & a & -7 \\ b & -7 & c \\ -7 & d & -1 \end{pmatrix}

is the adjoint of the matrix

\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{pmatrix}

a+b+c+d =

$8$
$10$
$0$
$2$

Solution:

C_{11} = (-1)^{1+1} (3 \times 2 - 1 \times 1) = 5

C_{12} = (-1)^{1+2} (2 \times 2 - 3 \times 1) = -1

C_{13} = (-1)^{1+3} (2 \times 1 - 3 \times 3) = -7

C_{21} = (-1)^{2+1} (4-3) = -1

C_{22} = (-1)^{2+2} (2-9) = -7

C_{23} = (-1)^{2+3} (1-6) = 5

C_{31} = (-1)^{3+1} (2-9) = -7

C_{32} = (-1)^{3+2} (1-6) = 5

C_{33} = (-1)^{3+3} (3-4) = -1

Co-factor of given matrix is given by-

\begin{bmatrix} 5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1 \end{bmatrix} = B\text{(say)}

Adjoin of given matrix-

⇒

B^T = \begin{bmatrix} 5 & -1 & -7 \\ -1 & (-7) & 5 \\ (-7) & 5 & -1 \end{bmatrix} = \begin{bmatrix} 5 & a & -7 \\ b & -7 & c \\ -7 & d & -1 \end{bmatrix}

After comparing we get,

a = b = -1

and

c = d = 5

Thus,

a+b+c+d = -1 -1 + 5 + 5 = 8

© examsnet.com

Go to Question:

Prev Question

Next Question