Question

Let \(\mathbf{u}=\left[\begin{array}{r}3 \\ -1 \\ 0\end{array}\right], \mathbf{v}=\left[\begin{array}{l}4 \\ 0 \\ 1\end{array}\right],\) and \(\mathbf{w}=\left[\begin{array}{c}-1 \\ 1 \\ 5\end{array}\right] .\) In each case, find \(\mathbf{x}\) such that: a. \(3(2 \mathbf{u}+\mathbf{x})+\mathbf{w}=2 \mathbf{x}-\mathbf{v}\) b. \(2(3 \mathbf{v}-\mathbf{x})=5 \mathbf{w}+\mathbf{u}-3 \mathbf{x}\)

Short answer

a. \(\mathbf{x} = \left[\begin{array}{r}-21 \\ 5 \\ -6\end{array}\right]\); b. \(\mathbf{x} = \left[\begin{array}{r}26 \\ -4 \\ 19\end{array}\right]\)."

Step-by-step solution

Simplify the Given Equation for Part (a)

We start by expanding and simplifying the equation given:- The left side is: \(3(2 \mathbf{u} + \mathbf{x}) + \mathbf{w} = 6 \mathbf{u} + 3\mathbf{x} + \mathbf{w}\).- The right side is: \(2 \mathbf{x} - \mathbf{v}\).The equation becomes:\[6 \mathbf{u} + 3\mathbf{x} + \mathbf{w} = 2 \mathbf{x} - \mathbf{v}\].

Substitute Known Values

Substituting the vectors \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\) into the equation:- \(6 \mathbf{u} = 6\left[\begin{array}{r}3 \ -1 \ 0\end{array}\right] = \left[\begin{array}{r}18 \ -6 \ 0\end{array}\right]\)- \(3 \mathbf{x} = \left[\begin{array}{r}3x_1 \ 3x_2 \ 3x_3\end{array}\right]\)- \(\mathbf{w} = \left[\begin{array}{c}-1 \ 1 \ 5\end{array}\right]\)- \(- \mathbf{v} = -\left[\begin{array}{l}4 \ 0 \ 1\end{array}\right] = \left[\begin{array}{r}-4 \ 0 \ -1\end{array}\right]\)The equation now is:\[\left[\begin{array}{r}18 \ -6 \ 0\end{array}\right] + \left[\begin{array}{r}3x_1 \ 3x_2 \ 3x_3\end{array}\right] + \left[\begin{array}{c}-1 \ 1 \ 5\end{array}\right] = \left[\begin{array}{r}2x_1 \ 2x_2 \ 2x_3\end{array}\right] - \left[\begin{array}{r}4 \ 0 \ 1\end{array}\right]\].

Balance the Equation

Combine terms on each side:- Left side: \(\left[\begin{array}{r}17 + 3x_1 \ -5 + 3x_2 \ 5 + 3x_3\end{array}\right]\)- Right side: \(\left[\begin{array}{r}2x_1 - 4 \ 2x_2 \ \ 2x_3 - 1\end{array}\right]\)Equating both sides gives:- \(17 + 3x_1 = 2x_1 - 4\)- \(-5 + 3x_2 = 2x_2\)- \(5 + 3x_3 = 2x_3 - 1\).

Solve for \(\mathbf{x}\) in Part (a)

Solve each equation for \(x_i\):- Solve \(17 + 3x_1 = 2x_1 - 4\) to find \(x_1 = -21\).- Solve \(-5 + 3x_2 = 2x_2\) to find \(x_2 = 5\).- Solve \(5 + 3x_3 = 2x_3 - 1\) to find \(x_3 = -6\).Thus, \(\mathbf{x} = \left[\begin{array}{r}-21 \ 5 \ -6\end{array}\right]\).

Simplify the Given Equation for Part (b)

Expand and simplify the equation:- The left side is: \(2(3 \mathbf{v} - \mathbf{x}) = 6 \mathbf{v} - 2\mathbf{x}\).- The right side is: \(5 \mathbf{w} + \mathbf{u} - 3 \mathbf{x}\).The equation becomes:\[6 \mathbf{v} - 2\mathbf{x} = 5 \mathbf{w} + \mathbf{u} - 3 \mathbf{x}\].

Substitute Known Values

Substituting the vectors \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\) into the equation:- \(6 \mathbf{v} = 6\left[\begin{array}{l}4 \ 0 \ 1\end{array}\right] = \left[\begin{array}{r}24 \ 0 \ 6\end{array}\right]\)- \(- 2\mathbf{x} = \left[\begin{array}{r}-2x_1 \ -2x_2 \ -2x_3\end{array}\right]\)- \(5 \mathbf{w} = 5\left[\begin{array}{c}-1 \ 1 \ 5\end{array}\right] = \left[\begin{array}{r}-5 \ 5 \ 25\end{array}\right]\)- \(\mathbf{u} = \left[\begin{array}{r}3 \ -1 \ 0\end{array}\right]\)- \(-3\mathbf{x} = \left[\begin{array}{r}-3x_1 \ -3x_2 \ -3x_3\end{array}\right]\)The equation now is:\[\left[\begin{array}{r}24 \ 0 \ 6\end{array}\right] - \left[\begin{array}{r}2x_1 \ 2x_2 \ 2x_3\end{array}\right] = \left[\begin{array}{r}-5 \ 5 \ 25\end{array}\right] + \left[\begin{array}{r}3 \ -1 \ 0\end{array}\right] - \left[\begin{array}{r}3x_1 \ 3x_2 \ 3x_3\end{array}\right]\].

Balance the Equation

Combine terms on each side:- Left side: \(\left[\begin{array}{r}24 - 2x_1 \ -2x_2 \ 6 - 2x_3\end{array}\right]\)- Right side: \(\left[\begin{array}{r}-2 - 3x_1 \ 4 - 3x_2 \ 25 - 3x_3\end{array}\right]\)Equating both sides gives:- \(24 - 2x_1 = -2 - 3x_1\)- \(-2x_2 = 4 - 3x_2\)- \(6 - 2x_3 = 25 - 3x_3\).

Solve for \(\mathbf{x}\) in Part (b)

Solve each equation for \(x_i\):- Solve \(24 - 2x_1 = -2 - 3x_1\) to find \(x_1 = 26\).- Solve \(-2x_2 = 4 - 3x_2\) to find \(x_2 = -4\).- Solve \(6 - 2x_3 = 25 - 3x_3\) to find \(x_3 = 19\).Thus, \(\mathbf{x} = \left[\begin{array}{r}26 \ -4 \ 19\end{array}\right]\).

Key concepts

Vector Operations

Vectors are fundamental objects in linear algebra used to represent quantities that have both magnitude and direction. Understanding vector operations is key to grasping more complex concepts in linear algebra.

• Vector Addition: To add two vectors, simply add their corresponding components. For vectors \(\mathbf{a} = \begin{bmatrix} a_1 \ a_2 \ a_3 \end{bmatrix}\) and \(\mathbf{b} = \begin{bmatrix} b_1 \ b_2 \ b_3 \end{bmatrix}\), their sum is \(\mathbf{a} + \mathbf{b} = \begin{bmatrix} a_1 + b_1 \ a_2 + b_2 \ a_3 + b_3 \end{bmatrix}\).


• Scalar Multiplication: Multiplying a vector by a scalar means scaling each component by the scalar. For \(c \cdot \mathbf{a} = \begin{bmatrix} c \cdot a_1 \ c \cdot a_2 \ c \cdot a_3 \end{bmatrix}\).


• Subtraction: Subtracting vector \(\mathbf{b}\) from \(\mathbf{a}\) is the same as adding \(\mathbf{a}\) to the negative of \(\mathbf{b}\), or \(\mathbf{a} - \mathbf{b} = \begin{bmatrix} a_1 - b_1 \ a_2 - b_2 \ a_3 - b_3 \end{bmatrix}\).

These basic operations are crucial when solving equations incorporating vectors. Making sure each step is performed correctly ensures accuracy when solving more complex problems.

Equation Solving

Solving equations in linear algebra often involves manipulating algebraic expressions to find unknown vector elements. Each vector component must satisfy the equation independently of others. Here are essential tips for solving vector equations efficiently:

• Isolate the Variable: Aim to have the vector or variable you are solving for on one side of the equation. Processes like combining like terms and using properties of equality are useful strategies. For instance, the equation \( 3(2\mathbf{u}+\mathbf{x})+\mathbf{w}=2\mathbf{x}-\mathbf{v} \) can be simplified by multiplying through and matching like terms.


• Substitution: Replace known vectors with their corresponding elements for clarity. This helps transform the problem into more manageable linear equations you can solve with basic algebra. In our example, replace \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) with their respective values.


• Solve Component-wise: When reduced to basic linear forms, solve the equations for each component separately. Each resulting equation corresponds to one component of the unknown vector, leading to efficient problem-solving.

By following these methods, you can systematically solve vector equations, no matter their complexity.

Matrix Manipulation

Matrix manipulation involves performing operations like addition, multiplication, and inversion that apply directly to matrix objects. Understanding these operations helps with solving systems of equations and performing transformations in linear algebra.

• Matrix Addition: Similar to vectors, add matrices by adding their corresponding elements. This only works if the matrices have the same dimensions.


• Matrix Multiplication: Multiply a matrix by another matrix, or vector, by utilizing the dot product of rows and columns. For matrices \(A\) of dimension \(m \times n\) and \(B\) of dimension \(n \times p\), their product \(AB\) is an \(m \times p\) matrix.


• Inversion and Transposition: These operations are crucial for solving system equations: finding inverses allows for direct solutions of matrix equations like \(AX = B\), while transposing swaps the matrix's rows and columns.

Mastery of matrix manipulation enables one to approach complex algebraic problems with different strategies and interpret broader application themes.