Math  /  Algebra

QuestionLet A=[2142]A=\begin{bmatrix}2 & 1 \\ -4 & -2\end{bmatrix} and b=[b1b2]\mathbf{b}=\begin{bmatrix}b_{1} \\ b_{2}\end{bmatrix}. Show when Ax=bA\mathbf{x}=\mathbf{b} has no solution and describe valid b\mathbf{b}.

Studdy Solution
From the row reduced form of the augmented matrix, we can see that the system Ax=bA \mathbf{x}=\mathbf{b} has a solution if and only if 2b1b2=02b_{1} - b_{2} =0. This is the condition for the rightmost column not to be a pivot column.
Therefore, the set of all b\mathbf{b} for which Ax=bA \mathbf{x}=\mathbf{b} does have a solution is given by all vectors b=[b1b2]\mathbf{b} = \left[\begin{array}{l}b_{1} \\ b_{2}\end{array}\right] such that 2b1b2=02b_{1} - b_{2} =0.

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