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Chapter 3: Problem 9

Prove that the system of linear equations \(A x=b\) has a solution if and only if \(b \in \mathrm{R}\left(\mathrm{L}_{A}\right)\). Visit goo.gl/JfwjBa for a solution.

Short Answer

Expert verified
The system of linear equations \(Ax = b\) has a solution if and only if \(b \in R(L_A)\). We proved this by showing that if the system has a solution, there exists a vector \(x\) such that \(Ax = b\), making \(b\) in the range of \(L_A\). Conversely, if \(b\) is in the range of \(L_A\), there exists a vector \(x\) such that \(L_A(x) = Ax = b\), indicating the system has a solution.

Step by step solution

01

Part 1: If the system of linear equations \(Ax = b\) has a solution, then \(b\) is in the range of \(L_A\).

Let's assume that the system of linear equations \(Ax = b\) has a solution. That means there exists a vector \(x\) such that when multiplied by the matrix \(A\), we get the vector \(b\). Now, we know that the range of a linear transformation \(L_A\) is defined as the set of all possible outputs when applying the transformation to a vector. In this case, the transformation is given by \(L_A(x) = Ax\). Since there exists a vector \(x\) such that \(Ax = b\), we can say that the vector \(b\) is in the range of \(L_A\). Thus, if the system of linear equations \(Ax = b\) has a solution, then \(b\) is in the range of \(L_A\).
02

Part 2: If \(b\) is in the range of \(L_A\), then the system of linear equations \(Ax = b\) has a solution.

Let's assume that \(b \in R(L_A)\). This means that there exists a vector \(x\) such that \(L_A(x) = Ax = b\). Therefore, if \(b\) is in the range of \(L_A\), then the system of linear equations \(Ax = b\) has a solution. Having shown both parts, we conclude that the system of linear equations \(Ax = b\) has a solution if and only if \(b \in R(L_A)\).

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Most popular questions from this chapter

Let $$ A=\left(\begin{array}{rrr} 1 & 2 & 3 \\ 1 & 0 & 1 \\ 1 & -1 & 1 \end{array}\right), B=\left(\begin{array}{rrr} 1 & 0 & 3 \\ 1 & -2 & 1 \\ 1 & -3 & 1 \end{array}\right), \text { and } C=\left(\begin{array}{rrr} 1 & 0 & 3 \\ 0 & -2 & -2 \\ 1 & -3 & 1 \end{array}\right) \text {. } $$ Find an elementary operation that transforms \(A\) into \(B\) and an elementary operation that transforms \(B\) into \(C\). By means of several additional operations, transform \(C\) into \(I_{3}\).

Determine which of the following systems of linear equations has a solution. \(x_{1}+x_{2}-x_{3}+2 x_{4}=2\) (a) \(x_{1}+x_{2}+2 x_{3}=1\) (b) \(x_{1}+x_{2}-x_{3}=1\) \(2 x_{1}+2 x_{2}+x_{3}+2 x_{4}=4\) (b) \(x_{1}+x_{2}-x_{3}=1\) \(2 x_{1}+x_{2}+3 x_{3}=2\) \(x_{1}+2 x_{2}+3 x_{3}=1\) (d) \(x_{1}+x_{2}+3 x_{3}-x_{4}=0\) \(x_{1}+x_{2}+x_{3}+x_{4}=1\) $x_{1}-2 x_{2}+x_{3}-x_{4}=1$ (c) \(x_{1}+x_{2}-x_{3}=0\) \(x_{1}+2 x_{2}+x_{3}=3\) \(4 x_{1}+x_{2}+8 x_{3}-x_{4}=0\) (e) \(2 x_{1}+x_{2}+2 x_{3}=3\) \(x_{1}-4 x_{2}+7 x_{3}=4\)

Show that every elementary matrix \(E\) is invertible, and its inverse is an elementary matrix.

Prove or give a counterexample to the following statement: If the coefficient matrix of a system of \(m\) linear equations in \(n\) unknowns has rank \(m\), then the system has a solution.

Row reduce each of the following matrices to echelon form: (a) \(A=\left[\begin{array}{cccc}1 & 2 & -3 & 0 \\ 2 & 4 & -2 & 2 \\ 3 & 6 & -4 & 3\end{array}\right]\) (b) \(B=\left[\begin{array}{rrr}-4 & 1 & -6 \\ 1 & 2 & -5 \\ 6 & 3 & -4\end{array}\right]\)
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