retagged by
630 views
14 votes
14 votes

$A x=b$ has solutions $x_1=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)$ and $x_2=\left(\begin{array}{l}4 \\ 5 \\ 6\end{array}\right)$, and possibly other solutions, for some (real) matrix $A$ and right-hand side $b$.

Which of the following MUST be a solution for $\mathrm{A}x=0?$


  1. $\left(\begin{array}{l}
    1 \\
    1 \\
    1
    \end{array}\right)$
  2. $\left(\begin{array}{l}
    2 \\
    2 \\
    2
    \end{array}\right)$
  3. $\left(\begin{array}{l}
    3 \\
    3 \\
    3
    \end{array}\right)$
  4. $\left(\begin{array}{l}
    4 \\
    4 \\
    4\\
    \end{array} \right)$
retagged by

1 Answer

16 votes
16 votes
$Ax_{1} = b$ …...eq1

$Ax_{2} = b$ …...eq2

computing eq2 & eq1 we will get

$Ax_{1} = Ax_{2} $

$A(x_{2} –  x_{1}) = 0$

$Ay = 0$

where $(x_{2} – x_{1}) = y= \begin{bmatrix} 3\\ 3\\3 \end{bmatrix}$

clearly $ y \neq 0 → Ay = 0 $ has non-trival solution → columns of A are Linearly Dependent

so $ky$ is also solution for $Ax = 0$ where $k$ is a scaler

Hence options$ A,B,C,D$ all of them are solutions of $Ax= 0$.
Answer:

Related questions