Let X=[ beginmatrix x1 x2 x3 endmatrix ],A=[ beginmatrix 1 -1 2 2 0 1 3 2 1 endmatrix ] and B=[ beginmatrix 3 1 4 endmatrix ] . If AX=B, then X is equal to:

Question

Let X=[ beginmatrix x1 x2 x3 endmatrix ],A=[ beginmatrix 1 -1 2 2 0 1 3 2 1 endmatrix ] and B=[ beginmatrix 3 1 4 endmatrix ] . If AX=B, then X is equal to: (A)  [ beginmatrix 1 2 3 endmatrix ]  (B)  [ beginmatrix -1 -2 3 endmatrix ]  (C)  [ beginmatrix -1 -2 -3 endmatrix ]  (D)  [ beginmatrix -1 2 3 endmatrix ]

Tardigrade Admin · Accepted Answer

∵ A=[ beginmatrix 1 -1 2 2 0 1 3 2 1 endmatrix ] ∴ A-1=(1/5)[ beginmatrix -2 5 -1 1 -5 3 4 -5 2 endmatrix ] Now, A-1B=(1/5)[ beginmatrix -2 5 -1 1 -5 3 4 -5 2 endmatrix ][ beginmatrix 3 1 4 endmatrix ] [ beginmatrix x1 x2 x3 endmatrix ]=[ beginmatrix -1 2 3 endmatrix ]

Q. Let $X = ⎣ ⎡ x_{1} x_{2} x_{3} ⎦ ⎤, A = ⎣ ⎡ 123 - 1 02 211 ⎦ ⎤$ and $B = ⎣ ⎡ 314 ⎦ ⎤$ . If $A X = B,$ then $X$ is equal to:

$⎣ ⎡ 123 ⎦ ⎤$

$⎣ ⎡ - 1 - 2 3 ⎦ ⎤$

$⎣ ⎡ - 1 - 2 - 3 ⎦ ⎤$

$⎣ ⎡ - 1 23 ⎦ ⎤$

$⎣ ⎡ 021 ⎦ ⎤$

$∵$ $A = ⎣ ⎡ 123 - 1 02 211 ⎦ ⎤$ $∴$ $A^{- 1} = \frac{1}{5} ⎣ ⎡ - 2 14 5 - 5 - 5 - 1 32 ⎦ ⎤$ Now, $A^{- 1} B = \frac{1}{5} ⎣ ⎡ - 2 14 5 - 5 - 5 - 1 32 ⎦ ⎤ ⎣ ⎡ 314 ⎦ ⎤$ $⎣ ⎡ x_{1} x_{2} x_{3} ⎦ ⎤ = ⎣ ⎡ - 1 23 ⎦ ⎤$

Q. Let X=⎣⎡​x1​x2​x3​​⎦⎤​,A=⎣⎡​123​−102​211​⎦⎤​ and B=⎣⎡​314​⎦⎤​ . If AX=B, then X is equal to:

⎣⎡​123​⎦⎤​

⎣⎡​−1−23​⎦⎤​

⎣⎡​−1−2−3​⎦⎤​

⎣⎡​−123​⎦⎤​

⎣⎡​021​⎦⎤​

∵ A=⎣⎡​123​−102​211​⎦⎤​ ∴ A−1=51​⎣⎡​−214​5−5−5​−132​⎦⎤​ Now, A−1B=51​⎣⎡​−214​5−5−5​−132​⎦⎤​⎣⎡​314​⎦⎤​ ⎣⎡​x1​x2​x3​​⎦⎤​=⎣⎡​−123​⎦⎤​