Given The Matrices $A$ And $B$ Shown Below, Find $-\frac{1}{6} A - 3B$.$A = \begin{bmatrix} 0 & 30 & -18 \\ -36 & -24 & 12 \end{bmatrix}$B = \begin{bmatrix} -8 & 6 & 8 \\ -8 & 5 & -1

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Introduction to Matrix Operations

Matrix operations are a fundamental concept in linear algebra, and they play a crucial role in various fields such as physics, engineering, and computer science. In this article, we will focus on finding the result of the expression −16A−3B-\frac{1}{6} A - 3B, where AA and BB are given matrices.

Matrix A and Matrix B

The matrices AA and BB are given as follows:

A=[030−18−36−2412]A = \begin{bmatrix} 0 & 30 & -18 \\ -36 & -24 & 12 \end{bmatrix}

B=[−868−85−1]B = \begin{bmatrix} -8 & 6 & 8 \\ -8 & 5 & -1 \end{bmatrix}

Understanding Matrix Operations

To find the result of the expression −16A−3B-\frac{1}{6} A - 3B, we need to understand the rules of matrix operations. When we multiply a matrix by a scalar, we multiply each element of the matrix by that scalar. When we add or subtract two matrices, we add or subtract corresponding elements of the two matrices.

Finding −16A-\frac{1}{6} A

To find −16A-\frac{1}{6} A, we need to multiply each element of matrix AA by −16-\frac{1}{6}. This can be done as follows:

−16A=−16[030−18−36−2412]-\frac{1}{6} A = -\frac{1}{6} \begin{bmatrix} 0 & 30 & -18 \\ -36 & -24 & 12 \end{bmatrix}

=[0−5364−2]= \begin{bmatrix} 0 & -5 & 3 \\ 6 & 4 & -2 \end{bmatrix}

Finding −3B-3B

To find −3B-3B, we need to multiply each element of matrix BB by −3-3. This can be done as follows:

−3B=−3[−868−85−1]-3B = -3 \begin{bmatrix} -8 & 6 & 8 \\ -8 & 5 & -1 \end{bmatrix}

=[24−18−2424−153]= \begin{bmatrix} 24 & -18 & -24 \\ 24 & -15 & 3 \end{bmatrix}

Finding −16A−3B-\frac{1}{6} A - 3B

Now that we have found −16A-\frac{1}{6} A and −3B-3B, we can find the result of the expression −16A−3B-\frac{1}{6} A - 3B by adding the two matrices.

−16A−3B=[0−5364−2]+[24−18−2424−153]-\frac{1}{6} A - 3B = \begin{bmatrix} 0 & -5 & 3 \\ 6 & 4 & -2 \end{bmatrix} + \begin{bmatrix} 24 & -18 & -24 \\ 24 & -15 & 3 \end{bmatrix}

=[24−23−2130−111]= \begin{bmatrix} 24 & -23 & -21 \\ 30 & -11 & 1 \end{bmatrix}

Conclusion

In this article, we have found the result of the expression −16A−3B-\frac{1}{6} A - 3B, where AA and BB are given matrices. We have used the rules of matrix operations to multiply each element of the matrices by the given scalars and then added the two resulting matrices. The final result is a new matrix that represents the result of the given expression.

Final Answer

The final answer is [24−23−2130−111]\boxed{\begin{bmatrix} 24 & -23 & -21 \\ 30 & -11 & 1 \end{bmatrix}}.

Further Reading

For more information on matrix operations and linear algebra, we recommend the following resources:

  • Linear Algebra and Its Applications by Gilbert Strang
  • Matrix Algebra by James E. Gentle
  • Linear Algebra by David C. Lay

These resources provide a comprehensive introduction to linear algebra and matrix operations, and they are suitable for students and professionals alike.

Introduction

Matrix operations are a fundamental concept in linear algebra, and they play a crucial role in various fields such as physics, engineering, and computer science. In this article, we will provide a Q&A guide to help you understand matrix operations and how to apply them in different scenarios.

Q: What is a matrix?

A: A matrix is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns. Matrices are used to represent systems of linear equations, and they are a fundamental concept in linear algebra.

Q: What are the different types of matrix operations?

A: There are several types of matrix operations, including:

  • Matrix addition: Adding two matrices by adding corresponding elements.
  • Matrix subtraction: Subtracting one matrix from another by subtracting corresponding elements.
  • Matrix multiplication: Multiplying two matrices by multiplying corresponding elements and summing the products.
  • Scalar multiplication: Multiplying a matrix by a scalar by multiplying each element of the matrix by the scalar.

Q: How do I add two matrices?

A: To add two matrices, you need to add corresponding elements of the two matrices. For example, if you have two matrices:

A=[123456]A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}

B=[789101112]B = \begin{bmatrix} 7 & 8 & 9 \\ 10 & 11 & 12 \end{bmatrix}

You can add them as follows:

A+B=[1+72+83+94+105+116+12]A + B = \begin{bmatrix} 1+7 & 2+8 & 3+9 \\ 4+10 & 5+11 & 6+12 \end{bmatrix}

=[81012141618]= \begin{bmatrix} 8 & 10 & 12 \\ 14 & 16 & 18 \end{bmatrix}

Q: How do I subtract one matrix from another?

A: To subtract one matrix from another, you need to subtract corresponding elements of the two matrices. For example, if you have two matrices:

A=[123456]A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}

B=[789101112]B = \begin{bmatrix} 7 & 8 & 9 \\ 10 & 11 & 12 \end{bmatrix}

You can subtract them as follows:

A−B=[1−72−83−94−105−116−12]A - B = \begin{bmatrix} 1-7 & 2-8 & 3-9 \\ 4-10 & 5-11 & 6-12 \end{bmatrix}

=[−6−6−6−6−6−6]= \begin{bmatrix} -6 & -6 & -6 \\ -6 & -6 & -6 \end{bmatrix}

Q: How do I multiply two matrices?

A: To multiply two matrices, you need to multiply corresponding elements of the two matrices and sum the products. For example, if you have two matrices:

A=[123456]A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}

B=[7810111213]B = \begin{bmatrix} 7 & 8 \\ 10 & 11 \\ 12 & 13 \end{bmatrix}

You can multiply them as follows:

AB=[1∗7+2∗10+3∗121∗8+2∗11+3∗134∗7+5∗10+6∗124∗8+5∗11+6∗13]AB = \begin{bmatrix} 1*7+2*10+3*12 & 1*8+2*11+3*13 \\ 4*7+5*10+6*12 & 4*8+5*11+6*13 \end{bmatrix}

=[7+20+368+22+3928+50+7232+55+78]= \begin{bmatrix} 7+20+36 & 8+22+39 \\ 28+50+72 & 32+55+78 \end{bmatrix}

=[6369150165]= \begin{bmatrix} 63 & 69 \\ 150 & 165 \end{bmatrix}

Q: How do I multiply a matrix by a scalar?

A: To multiply a matrix by a scalar, you need to multiply each element of the matrix by the scalar. For example, if you have a matrix:

A=[123456]A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}

And you want to multiply it by a scalar 2, you can do it as follows:

2A=[2∗12∗22∗32∗42∗52∗6]2A = \begin{bmatrix} 2*1 & 2*2 & 2*3 \\ 2*4 & 2*5 & 2*6 \end{bmatrix}

=[24681012]= \begin{bmatrix} 2 & 4 & 6 \\ 8 & 10 & 12 \end{bmatrix}

Conclusion

In this article, we have provided a Q&A guide to help you understand matrix operations and how to apply them in different scenarios. We have covered matrix addition, subtraction, multiplication, and scalar multiplication, and we have provided examples to illustrate each concept.

Final Answer

The final answer is that matrix operations are a fundamental concept in linear algebra, and they play a crucial role in various fields such as physics, engineering, and computer science. By understanding matrix operations, you can solve systems of linear equations, find the inverse of a matrix, and perform other important tasks in linear algebra.

Further Reading

For more information on matrix operations and linear algebra, we recommend the following resources:

  • Linear Algebra and Its Applications by Gilbert Strang
  • Matrix Algebra by James E. Gentle
  • Linear Algebra by David C. Lay

These resources provide a comprehensive introduction to linear algebra and matrix operations, and they are suitable for students and professionals alike.