Let A=( 1 2 -1 2 5 -1 1 3 2 ). Given that 𝐱1^⋆=( 5

Name: Problem 13, Chapter 2: Applied Linear Algebra (Undergraduate Texts in Mathematics)
Uploaded: 2021-06-28T07:17:29-08:00
Duration: 1 min 45 s
Channel: Abhijith V
Description: Let A=( 1 2 -1 2 5 -1 1 3 2 ). Given that 𝐱1^⋆=( 5 -1 2 ) solves A 𝐱=𝐛1=( 1 3 6 ) and 𝐱2^⋆=( -11 5 -1 ) solves A 𝐱=𝐛2=( 0 4 2 ), find a solution to A 𝐱=2 𝐛1+𝐛2=( 2 10 14 ).

step 1 So here we are given a matrix, that is, matrix A is equal to 5, minus 3, minus 2, and 1. So we n

Question

Let $A=\left(\begin{array}{rrr}1 & 2 & -1 \\ 2 & 5 & -1 \\ 1 & 3 & 2\end{array}\right)$. Given that $\mathbf{x}_1^{\star}=\left(\begin{array}{r}5 \\ -1 \\ 2\end{array}\right)$ solves $A \mathbf{x}=\mathbf{b}_1=\left(\begin{array}{l}1 \\ 3 \\ 6\end{array}\right)$ and $\mathbf{x}_2^{\star}=\left(\begin{array}{r}-11 \\ 5 \\ -1\end{array}\right)$ solves $A \mathbf{x}=\mathbf{b}_2=\left(\begin{array}{l}0 \\ 4 \\ 2\end{array}\right)$, find a solution to $A \mathbf{x}=2 \mathbf{b}_1+\mathbf{b}_2=\left(\begin{array}{r}2 \\ 10 \\ 14\end{array}\right)$.

    Let $A=\left(\begin{array}{rrr}1 & 2 & -1 \\ 2 & 5 & -1 \\ 1 & 3 & 2\end{array}\right)$. Given that $\mathbf{x}_1^{\star}=\left(\begin{array}{r}5 \\ -1 \\ 2\end{array}\right)$ solves $A \mathbf{x}=\mathbf{b}_1=\left(\begin{array}{l}1 \\ 3 \\ 6\end{array}\right)$ and $\mathbf{x}_2^{\star}=\left(\begin{array}{r}-11 \\ 5 \\ -1\end{array}\right)$ solves $A \mathbf{x}=\mathbf{b}_2=\left(\begin{array}{l}0 \\ 4 \\ 2\end{array}\right)$, find a solution to $A \mathbf{x}=2 \mathbf{b}_1+\mathbf{b}_2=\left(\begin{array}{r}2 \\ 10 \\ 14\end{array}\right)$.

Applied Linear Algebra (Undergraduate Texts in Mathematics)

Peter J. Olver,… 2nd Edition

Chapter 2, Problem 13

Instant Answer

Solved by Expert Abhijith V

06/28/2021

Step 1

We are given a matrix $ A $ and two vectors $ \mathbf{x}_1^{\star} $ and $ \mathbf{x}_2^{\star} $ which are solutions to the equations $ A \mathbf{x} = \mathbf{b}_1 $ and $ A \mathbf{x} = \mathbf{b}_2 $ respectively. We need to find a solution \( Show more…

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Let $A=\left(\begin{array}{rrr}1 & 2 & -1 \\ 2 & 5 & -1 \\ 1 & 3 & 2\end{array}\right)$. Given that $\mathbf{x}_1^{\star}=\left(\begin{array}{r}5 \\ -1 \\ 2\end{array}\right)$ solves $A \mathbf{x}=\mathbf{b}_1=\left(\begin{array}{l}1 \\ 3 \\ 6\end{array}\right)$ and $\mathbf{x}_2^{\star}=\left(\begin{array}{r}-11 \\ 5 \\ -1\end{array}\right)$ solves $A \mathbf{x}=\mathbf{b}_2=\left(\begin{array}{l}0 \\ 4 \\ 2\end{array}\right)$, find a solution to $A \mathbf{x}=2 \mathbf{b}_1+\mathbf{b}_2=\left(\begin{array}{r}2 \\ 10 \\ 14\end{array}\right)$.

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Key Concepts

Linearity of Matrix Operations

Matrix multiplication is a linear operation, meaning that for any matrix A and any vectors x and y with corresponding scalars a and b, we have A(ax + by) = aA(x) + bA(y). This property allows one to combine known solutions of systems of equations to form new solutions.

Superposition Principle

In linear systems, the superposition principle states that the sum of solutions corresponding to different right-hand sides, appropriately weighted by scalars, is a solution of the system whose right-hand side is the corresponding linear combination of the original ones. This principle is crucial for solving complex systems by combining simpler, known solutions.

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