Problem 33 - Algebra 2

Question

Use matrices \(A, B, C,\) and \(D\) to find the following. $$A=\left[\begin{array}{rr}{5} & {7} \\ {-1} & {6} \\ {3} & {-9}\end{array}\right] \quad B=\left[\begin{array}{ll}{8} & {3} \\ {5} & {1} \\\ {4} & {4}\end{array}\right] \quad C=\left[\begin{array}{rr}{0} & {4} \\\ {-2} & {5} \\ {7} & {-1}\end{array}\right] \quad D=\left[\begin{array}{ll}{6} & {2} \\ {9} & {0} \\ {-3} & {0}\end{array}\right]$$ \(3 C-4 A+B\)

Step-by-Step Solution

Answer

The result is \(\left[\begin{array}{rr}{-12} & {-13} \\ {3} & {-8} \\ {13} & {37}\end{array}\right]\).

Step 1: Multiply Matrix C by 3

To find \(3C\), multiply each element of matrix \(C\) by 3. Matrix \(C\) is:\[C=\left[\begin{array}{rr}{0} & {4} \{-2} & {5} \{7} & {-1}\end{array}\right]\]Thus, \(3C\) becomes:\[3C = \left[\begin{array}{rr}{3 \times 0} & {3 \times 4} \{3 \times (-2)} & {3 \times 5} \{3 \times 7} & {3 \times (-1)}\end{array}\right] = \left[\begin{array}{rr}{0} & {12} \{-6} & {15} \{21} & {-3}\end{array}\right]\]

Step 2: Multiply Matrix A by 4

To find \(4A\), multiply each element of matrix \(A\) by 4. Matrix \(A\) is:\[A=\left[\begin{array}{rr}{5} & {7} \{-1} & {6} \{3} & {-9}\end{array}\right]\]Thus, \(4A\) becomes:\[4A = \left[\begin{array}{rr}{4 \times 5} & {4 \times 7} \{4 \times (-1)} & {4 \times 6} \{4 \times 3} & {4 \times (-9)}\end{array}\right] = \left[\begin{array}{rr}{20} & {28} \{-4} & {24} \{12} & {-36}\end{array}\right]\]

Step 3: Calculate 3C - 4A

Subtract the elements of \(4A\) from \(3C\).\(3C\) is:\[\left[\begin{array}{rr}{0} & {12} \{-6} & {15} \{21} & {-3}\end{array}\right] \]\(4A\) is:\[\left[\begin{array}{rr}{20} & {28} \{-4} & {24} \{12} & {-36}\end{array}\right] \]Thus, \(3C - 4A\) becomes:\[\begin{array}{rr}{0-20} & {12-28} \{-6-(-4)} & {15-24} \{21-12} & {-3-(-36)} \end{array} = \left[\begin{array}{rr}{-20} & {-16} \{-2} & {-9} \{9} & {33}\end{array}\right]\]

Step 4: Add Matrix B

Add the matrix \(B\) to the result from Step 3.Matrix \(B\) is:\[B=\left[\begin{array}{ll}{8} & {3} \{5} & {1} \{4} & {4}\end{array}\right] \]Thus, \(3C - 4A + B\) becomes:\[\begin{array}{rr}{-20 + 8} & {-16 + 3} \{-2 + 5} & {-9 + 1} \{9 + 4} & {33 + 4} \end{array} = \left[\begin{array}{rr}{-12} & {-13} \{3} & {-8} \{13} & {37}\end{array}\right]\]

Step 5: Result

The resulting matrix from the expression \(3C - 4A + B\) is:\[\left[\begin{array}{rr}{-12} & {-13} \{3} & {-8} \{13} & {37}\end{array}\right]\]

Key Concepts

Scalar Multiplication

Scalar multiplication is a fundamental operation in linear algebra. It involves multiplying each element within a matrix by a single constant number, called a scalar.
This operation is straightforward and serves as a building block for more complex matrix manipulations. For example, if we have a matrix \( C \) and we want to multiply it by a scalar \( 3 \), we multiply each element of \( C \) by \( 3 \).
An important point to note is that scalar multiplication does not change the matrix dimensions; it only scales each entry.

• For a matrix \( C = \begin{bmatrix} 0 & 4 \ -2 & 5 \ 7 & -1 \end{bmatrix} \)
• Multiplying by scalar \( 3 \) gives: \[ 3C = \begin{bmatrix} 0 \times 3 & 4 \times 3 \ -2 \times 3 & 5 \times 3 \ 7 \times 3 & -1 \times 3 \end{bmatrix} = \begin{bmatrix} 0 & 12 \ -6 & 15 \ 21 & -3 \end{bmatrix} \]

This simple operation is widely used in various applications such as shifting the range of data or altering images in computer graphics.

Matrix Addition

Matrix addition involves adding two or more matrices by summing their corresponding elements. For matrices to be added together, they must have the same dimensions, i.e., the same number of rows and columns.
This additive operation maintains the original shape and size of the matrices.
Consider the matrices:

• \( A = \begin{bmatrix} 5 & 7 \ -1 & 6 \ 3 & -9 \end{bmatrix} \)
• \( B = \begin{bmatrix} 8 & 3 \ 5 & 1 \ 4 & 4 \end{bmatrix} \)

To add them, we perform the operation element-wise:

• The top left corner: \( 5 + 8 = 13 \)
• The top right corner: \( 7 + 3 = 10 \)
• And this continues cell by cell...
• Resulting in a new matrix : \( A + B = \begin{bmatrix} 13 & 10 \ 4 & 7 \ 7 & -5 \end{bmatrix} \)

Matrix addition is commutative and associative, meaning the order of addition does not matter and it can be grouped in any way.

Matrix Subtraction

Matrix subtraction follows a similar procedure to matrix addition, in which subtraction occurs element by element. The matrices require matching dimensions for subtraction to take place.
For instance, consider matrices \( P \) and \( Q \):

• \( P = \begin{bmatrix} 5 & 12 \ 8 & 20 \ 7 & 1 \end{bmatrix} \)
• \( Q = \begin{bmatrix} 1 & 3 \ 4 & 9 \ 6 & 8 \end{bmatrix} \)

Subtraction is executed as follows:

• The top left element: \( 5 - 1 = 4 \)
• The top right element: \( 12 - 3 = 9 \)
• And so on for each element...
• Resulting in the matrix: \( P - Q = \begin{bmatrix} 4 & 9 \ 4 & 11 \ 1 & -7 \end{bmatrix} \)

Matrix subtraction is not commutative, so the order of the matrices matters: \( P - Q eq Q - P \). This property distinguishes it from addition.

Linear Algebra

Linear algebra is the mathematical study of vectors, vector spaces, and linear transformations such as systems of linear equations.
This branch of mathematics is foundational to many aspects of science and engineering, particularly in areas involving multiple dimensions and systems of equations. Matrices are pivotal in linear algebra as they facilitate the handling of complex linear equations.
Some key applications include:

• Robot kinematics, where matrices model transformations in space.
• Computer graphics, using matrices for rotating and scaling images.
• Data analysis, where matrices represent data and transformations.
• Solving systems of linear equations common in physics and engineering.

The key operations—scalar multiplication, addition, and subtraction—form the basic tools required for these applications. Enhancing one's understanding of these concepts paves the way to mastering more advanced topics like eigenvalues and singular value decomposition.