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    First published on Monday, Jan 6, 2025 and last modified on Wednesday, Jan 8, 2025

    Linear Algebra in the Euclidian Space: Section 4 test
    2022

    Fabienne Chaplais Mathedu SAS

    Keywords: Vectors

    1 Introduction

    Section 1.1

    In that test, you will solve vectorial equations of many kinds.

    In these equations, the unknown is a column vector with two real elements denoted \( X\) rather than the vectorial notation, but the two notations are strictly equivalent.

    2 Q. 1: Solve additive equations

    Use the equivalence :

    \[\begin{equation} \overrightarrow{w}=\overrightarrow{u}+\overrightarrow{v} \Leftrightarrow \overrightarrow{u}=\overrightarrow{w}-\overrightarrow{v} \end{equation}\]

    (1)

    to solve the following equations in the unlnown column vector \( X\) :

    1. \( X+\begin{bmatrix}3\\ 2\end{bmatrix}=\begin{bmatrix}5\\ 2\end{bmatrix}\)

    2. \( X+\begin{bmatrix}5\\ 2\end{bmatrix}=\begin{bmatrix}3\\ 2\end{bmatrix}\)

    3. \( X+\begin{bmatrix}-3\\ 2\end{bmatrix}=\begin{bmatrix}5\\ -2\end{bmatrix}\)

    3 Q. 2: Solve subtractive equations of type 1

    Use the equivalence :

    \[\begin{equation} \overrightarrow{w}=\overrightarrow{u}-\overrightarrow{v} \Leftrightarrow \overrightarrow{u}=\overrightarrow{w}+\overrightarrow{v} \end{equation}\]

    (2)

    to solve the following equations in the unlnown column vector \( X\) :

    1. \( X-\begin{bmatrix}3\\ 2\end{bmatrix}=\begin{bmatrix}5\\ 2\end{bmatrix}\)

    2. \( X-\begin{bmatrix}5\\ 2\end{bmatrix}=\begin{bmatrix}3\\ 2\end{bmatrix}\)

    3. \( X-\begin{bmatrix}-3\\ 2\end{bmatrix}=\begin{bmatrix}5\\ -2\end{bmatrix}\)

    4 Q. 3: Solve subtractive equations of type 2

    Use the equivalence :

    \[\begin{equation} \overrightarrow{w}=\overrightarrow{v}-\overrightarrow{u} \Leftrightarrow \overrightarrow{u}=\overrightarrow{v}-\overrightarrow{w} \end{equation}\]

    (3)

    to solve the following equations in the unlnown column vector \( X\) :

    1. \( \begin{bmatrix}3\\ 2\end{bmatrix}-X=\begin{bmatrix}5\\ 2\end{bmatrix}\)

    2. \( \begin{bmatrix}5\\ 2\end{bmatrix}-X=\begin{bmatrix}3\\ 2\end{bmatrix}\)

    3. \( \begin{bmatrix}-3\\ 2\end{bmatrix}-X=\begin{bmatrix}5\\ -2\end{bmatrix}\)

    5 Q. 4: Solve linear equations

    Use the equivalence :

    \[\begin{equation} \overrightarrow{v}=\lambda\overrightarrow{u} \Leftrightarrow \overrightarrow{u}=\frac{\overrightarrow{v}}{\lambda} \end{equation}\]

    (4)

    and the fact that:

    \[\begin{equation} \frac{\overrightarrow{v}}{\lambda}=\frac{1}{\lambda}\overrightarrow{v} \end{equation}\]

    (5)

    to solve the following equations in the unlnown column vector \( X\) :

    1. \( 2X=\begin{bmatrix}5\\ 2\end{bmatrix}\)

    2. \( -2X=\begin{bmatrix}5\\ 2\end{bmatrix}\)

    3. \( \frac{1}{2}X=\begin{bmatrix}-5\\ 2\end{bmatrix}\)

    4. \( \left(-\frac{1}{2}\right)X=\begin{bmatrix}-5\\ 2\end{bmatrix}\)

    6 Q. 5: Solve affine equations

    To solve the following equations in the unlnown column vector \( X\) , process in two steps:

    1. Solve an additive or subtractive equation.

    2. Then solve the resulting linear equation.

    Here are the equations:

    1. \( 2X+\begin{bmatrix}3\\ 2\end{bmatrix}=\begin{bmatrix}5\\ 2\end{bmatrix}\)

    2. \( -2X-\begin{bmatrix}3\\ 2\end{bmatrix}=\begin{bmatrix}5\\ 2\end{bmatrix}\)

    3. \( \frac{1}{2}X-\begin{bmatrix}3\\ 2\end{bmatrix}=\begin{bmatrix}-5\\ 2\end{bmatrix}\)

    4. \( \left(-\frac{1}{2}\right)X+\begin{bmatrix}3\\ 2\end{bmatrix}=\begin{bmatrix}-5\\ 2\end{bmatrix}\)