The Given Scalar Multiplication Problem

Oct 13, 2025 by magicalceylontours.com 65 views

Introduction

Scalar multiplication is a fundamental concept in linear algebra, where a scalar is multiplied with each element of a matrix. In this article, we will explore the process of finding the values of $x$ and $y$ in the given scalar multiplication problem. We will use the properties of matrix multiplication and scalar multiplication to solve for the unknown variables.

The Given Scalar Multiplication Problem

The given scalar multiplication problem is:

{ \frac{1}{5} \cdot \begin{bmatrix} 0 & x \\ 20 & y \end{bmatrix} = \begin{bmatrix} 0 & 6 \\ 4 & -3 \end{bmatrix} \}

Understanding Scalar Multiplication

Scalar multiplication is a process where a scalar is multiplied with each element of a matrix. In this case, the scalar is $\frac{1}{5}$ , and the matrix is $\begin{bmatrix} 0 & x \\ 20 & y \end{bmatrix}$ . The result of the scalar multiplication is the matrix $\begin{bmatrix} 0 & 6 \\ 4 & -3 \end{bmatrix}$ .

Properties of Matrix Multiplication

To solve for the values of $x$ and $y$ , we need to use the properties of matrix multiplication. Specifically, we will use the property that the product of a scalar and a matrix is equal to the scalar multiplied by each element of the matrix.

Solving for $x$

To solve for $x$ , we can start by multiplying the scalar $\frac{1}{5}$ with the element $x$ in the matrix. This gives us:

{ \frac{1}{5} \cdot x = 6 \}

To solve for $x$ , we can multiply both sides of the equation by 5:

{ x = 6 \cdot 5 \}

This gives us:

{ x = 30 \}

Solving for $y$

To solve for $y$ , we can start by multiplying the scalar $\frac{1}{5}$ with the element $y$ in the matrix. This gives us:

{ \frac{1}{5} \cdot y = -3 \}

To solve for $y$ , we can multiply both sides of the equation by 5:

{ y = -3 \cdot 5 \}

This gives us:

{ y = -15 \}

Conclusion

In this article, we have solved for the values of $x$ and $y$ in the given scalar multiplication problem. We have used the properties of matrix multiplication and scalar multiplication to find the values of the unknown variables. The final values of $x$ and $y$ are $x = 30$ and $y = -15$ .

Discussion

The problem of finding the values of $x$ and $y$ in the scalar multiplication is a classic example of a linear algebra problem. The solution involves using the properties of matrix multiplication and scalar multiplication to solve for the unknown variables. This type of problem is commonly encountered in linear algebra and is an important concept to understand.

Final Answer

The final answer is:

$x = 30$ $y = -15$

References

[1] "Linear Algebra and Its Applications" by Gilbert Strang
[2] "Matrix Algebra" by James E. Gentle
[3] "Linear Algebra" by David C. Lay

Note: The references provided are for informational purposes only and are not directly related to the problem at hand.

Introduction

In our previous article, we solved for the values of $x$ and $y$ in the given scalar multiplication problem. In this article, we will provide a Q&A section to help clarify any doubts or questions that readers may have.

Q: What is scalar multiplication?

A: Scalar multiplication is a process where a scalar is multiplied with each element of a matrix. In this case, the scalar is $\frac{1}{5}$ , and the matrix is $\begin{bmatrix} 0 & x \\ 20 & y \end{bmatrix}$ .

Q: How do you multiply a scalar with a matrix?

A: To multiply a scalar with a matrix, you multiply the scalar with each element of the matrix. In this case, we multiplied the scalar $\frac{1}{5}$ with each element of the matrix $\begin{bmatrix} 0 & x \\ 20 & y \end{bmatrix}$ .

Q: What is the result of the scalar multiplication?

A: The result of the scalar multiplication is the matrix $\begin{bmatrix} 0 & 6 \\ 4 & -3 \end{bmatrix}$ .

Q: How did you solve for $x$ ?

A: To solve for $x$ , we multiplied the scalar $\frac{1}{5}$ with the element $x$ in the matrix. This gave us $\frac{1}{5} \cdot x = 6$ . We then multiplied both sides of the equation by 5 to get $x = 6 \cdot 5$ .

Q: How did you solve for $y$ ?

A: To solve for $y$ , we multiplied the scalar $\frac{1}{5}$ with the element $y$ in the matrix. This gave us $\frac{1}{5} \cdot y = -3$ . We then multiplied both sides of the equation by 5 to get $y = -3 \cdot 5$ .

Q: What are the final values of $x$ and $y$ ?

A: The final values of $x$ and $y$ are $x = 30$ and $y = -15$ .

Q: What is the importance of scalar multiplication in linear algebra?

A: Scalar multiplication is an important concept in linear algebra as it allows us to scale matrices by a scalar value. This is useful in a variety of applications, including solving systems of linear equations and finding the inverse of a matrix.

Q: Can you provide more examples of scalar multiplication?

A: Yes, here are a few more examples of scalar multiplication:

$\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} \cdot 2 = \begin{bmatrix} 4 & 6 \\ 8 & 10 \end{bmatrix}$
$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \cdot 3 = \begin{bmatrix} 3 & 6 \\ 9 & 12 \end{bmatrix}$

Q: How do you find the inverse of a matrix using scalar multiplication?

A: To find the inverse of a matrix using scalar multiplication, you can multiply the matrix by the scalar value of $\frac{1}{\text{determinant}}$ . For example, if the determinant of the matrix is 5, you would multiply the matrix by $\frac{1}{5}$ to get the inverse.

Q: Can you provide more information on the properties of matrix multiplication?

A: Yes, here are a few more properties of matrix multiplication:

The product of two matrices is only defined if the number of columns in the first matrix is equal to the number of rows in the second matrix.
The product of two matrices is commutative, meaning that the order of the matrices does not matter.
The product of two matrices is associative, meaning that the order in which you multiply the matrices does not matter.