Solution Found!
According to Proposition 4.10, if A is an m n matrix, then for each b C(A), there is a
Chapter 3, Problem 10(choose chapter or problem)
According to Proposition 4.10, if A is an \(m \times n\) matrix, then for each \(\mathbf{b} \in \mathbf{C}(\Lambda)\), there is a unique \(\mathbf{x} \in \mathbf{R}(\Lambda)\) with \(A \mathbf{x}=\mathbf{b}\). In each case, give a formula for that \(\mathbf{x}\).
a. \(A=\left[\begin{array}{lll}1 & 2 & 3 \\ 1 & 2 & 3\end{array}\right]\)
b. \(A=\left[\begin{array}{rrr}1 & 1 & 1 \\ 0 & 1 & -1\end{array}\right]\)
Questions & Answers
QUESTION:
According to Proposition 4.10, if A is an \(m \times n\) matrix, then for each \(\mathbf{b} \in \mathbf{C}(\Lambda)\), there is a unique \(\mathbf{x} \in \mathbf{R}(\Lambda)\) with \(A \mathbf{x}=\mathbf{b}\). In each case, give a formula for that \(\mathbf{x}\).
a. \(A=\left[\begin{array}{lll}1 & 2 & 3 \\ 1 & 2 & 3\end{array}\right]\)
b. \(A=\left[\begin{array}{rrr}1 & 1 & 1 \\ 0 & 1 & -1\end{array}\right]\)
ANSWER:Step 1 of 2
According to the proposition:
For each
Therefore, it is easy to see that a basis for 
That is, any element of 
Also, a basis for 
Thus, for any
If we set,
Then x is the unique element of 
