Question

Find values for the scalars \(a\) and \(b\) that satisfy the given equation. $$ a\left[\begin{array}{c} 4 \\ -2 \end{array}\right]+b\left[\begin{array}{c} -6 \\ 3 \end{array}\right]=\left[\begin{array}{c} 10 \\ -5 \end{array}\right] $$

Short answer

One solution is \(a = 2.5\) and \(b = 0\). Any \(b\) would work, but \(a = 2.5 + 1.5b\).

Step-by-step solution

Write Out the Equation in Vector Form

The given problem states that the linear combination of two vectors should equal another vector. We need to express that algebraically as follows:\[ a \begin{bmatrix} 4 \ -2 \end{bmatrix} + b \begin{bmatrix} -6 \ 3 \end{bmatrix} = \begin{bmatrix} 10 \ -5 \end{bmatrix} \]

Separate into System of Linear Equations

Since each component of the vectors must be equal, we set up two separate equations:1. First component: \( 4a - 6b = 10 \)2. Second component: \( -2a + 3b = -5 \)

Solve the First Equation for 'a'

We will solve the first equation for \(a\):\[ 4a - 6b = 10 \]Rearranging gives:\[ 4a = 10 + 6b \]\[ a = \frac{10 + 6b}{4} \]\[ a = 2.5 + 1.5b \]

Substitute 'a' in the Second Equation

Substitute \(a = 2.5 + 1.5b\) back into the second equation:\[ -2(2.5 + 1.5b) + 3b = -5 \]Expanding:\[ -5 - 3b + 3b = -5 \]This shows that any value of \(b\) satisfies the equation.

Solve for 'a' in Terms of 'b'

Since the expression for \(b\) allows multiple values, let's use \(b = 0\) as a specific instance to find \(a\):If \(b = 0\), then \(a = 2.5 + 1.5(0) = 2.5\).Thus, one particular solution is \(a = 2.5\) and \(b = 0\).

Simplification and Conclusion

Since Step 4 shows that any value of \(b\) is valid, another approach is not needed. The clean solution for any scalable outcome is to choose a simple value for \(b\), such as 0:- a possible solution is \( (a, b) = (2.5, 0) \).This satisfies both vector equations.

Key concepts

Linear Combinations

A linear combination involves adding together multiples of vectors to create another vector. Vectors are like arrows with both a direction and length, and they are defined by numbers organized in a specific order. In mathematics, you can take any set of vectors and combine them by multiplying each vector by a number (called a scalar) and then adding them together.
This process is expressed as: \( extbf{v} = a \textbf{u} + b \textbf{w} \), where \( \textbf{v} \) is the resulting vector, \( \textbf{u} \) and \( \textbf{w} \) are the initial vectors, and \( a \) and \( b \) are the scalars.

• To find if a vector is a linear combination of others, break it into components.
• Ensure scalars exist that make the combination equal to your target vector.

Understanding linear combinations is like seeing how vectors mix to form new directions and magnitudes. It's a core concept in vector algebra.

Systems of Equations

When working with vectors, you'll often need to solve systems of equations to find specific values for your scalars. This means setting up equations where each component of the vector sum must match those of a target vector.
In the given exercise, we derived two equations: one for each component of the resulting vector:

• \( 4a - 6b = 10 \) for the first component.
• \( -2a + 3b = -5 \) for the second component.

Each equation is based on the corresponding parts of the vector. Solving systems of equations means finding the values for \( a \) and \( b \) that make both conditions true at once.
Techniques include substitution (using one equation to solve for a variable and plugging it into another) and elimination (adding equations to cancel out a variable). These strategies are powerful for solving problems in linear algebra.

Scalar Multiplication

Scalar multiplication is the process of multiplying a vector by a scalar, a single number, which scales or changes its magnitude (size) without altering its direction, unless it's negative which reverses the direction. It's like stretching or shrinking a vector.
For example, multiplying a vector \( \begin{bmatrix} 4 \ -2 \end{bmatrix} \) by a scalar \( a \), results in \( a \begin{bmatrix} 4 \ -2 \end{bmatrix} = \begin{bmatrix} 4a \ -2a \end{bmatrix} \). In the exercise given, the vector multiplication translates directly into expressions for each element.

• Scalar multiplication is performed element-wise.
• One equation results from the sum of all first elements and another from all second elements.

Scaling is a foundational vector operation allowing you to adjust vectors to fit certain conditions, crucial in making vectors part of something bigger, like solving equations or analyzing data.