If 3A + 4B' = [7&-10&17 0&6&31] and 2B - 3A' [-1&18 4&0 -5&-7] then B =

Question

If 3A + 4B' = [7&-10&17 0&6&31] and 2B - 3A' [-1&18 4&0 -5&-7] then B = (A) [-1&-18 4&-16 -5&-7]  (B) [1&3 -1&1 2&4]  (C) [1&3 -1&1 2&-4]  (D)  [1&-3 -1&1 2&4]

Tardigrade Admin · Accepted Answer

3A+4B' =[ beginmatrix7&-10&17 0&6&31 endmatrix] dots(1) (2B-3A')'=(2B)'-(3A')'=2B'-3A ⇒ 2B' -3A=[ beginmatrix-1&4&-5 18&0&-7 endmatrix] dots(2) Adding 1 and 2, we get 6B'=[ beginmatrix6&-6&12 18&6&24 endmatrix] ⇒ B' =[ beginmatrix1&-1&2 3&1&4 endmatrix] ∴ B=[ beginmatrix1&3 -1&1 2&4 endmatrix]

Q. If $3 A + 4 B^{'} =$ $[70 - 10 6 1731]$ and $2 B - 3 A^{'}$ $⎣ ⎡ - 1 4 - 5 180 - 7 ⎦ ⎤$ then $B =$

$⎣ ⎡ - 1 4 - 5 - 18 - 16 - 7 ⎦ ⎤$

$⎣ ⎡ 1 - 1 2 314 ⎦ ⎤$

$⎣ ⎡ 1 - 1 2 31 - 4 ⎦ ⎤$

$⎣ ⎡ 1 - 1 2 - 3 14 ⎦ ⎤$

$3 A + 4 B^{'} = [70 - 10 6 1731] \dots (1)$
$(2 B - 3 A^{'})^{'} = (2 B)^{'} - (3 A^{'})^{'} = 2 B^{'} - 3 A$
$\Rightarrow 2 B^{'} - 3 A = [- 1 18 40 - 5 - 7] \dots (2)$
Adding 1 and 2, we get $6 B^{'} = [618 - 6 6 1224]$
$\Rightarrow B^{'} = [13 - 1 1 24] ∴ B = ⎣ ⎡ 1 - 1 2 314 ⎦ ⎤$

Q. If 3A+4B′=[70​−106​1731​] and 2B−3A′ ⎣⎡​−14−5​180−7​⎦⎤​ then B=

⎣⎡​−14−5​−18−16−7​⎦⎤​

⎣⎡​1−12​314​⎦⎤​

⎣⎡​1−12​31−4​⎦⎤​

⎣⎡​1−12​−314​⎦⎤​