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Matrices \(A\) and \(B\) are given below. Simplify the given expression. $$ A=\left[\begin{array}{cc} 1 & -1 \\ 7 & 4 \end{array}\right] \quad B=\left[\begin{array}{cc} -3 & 2 \\ 5 & 9 \end{array}\right] $$ $$ 2 A-3 B $$

Short Answer

Expert verified
The simplified expression is \(\begin{bmatrix} 11 & -8 \\ -1 & -19 \end{bmatrix}\).

Step by step solution

01

Multiply Matrix A by Scalar 2

Multiply each element of matrix \(A\) by 2. Matrix \(A\) is: \[\begin{bmatrix} 1 & -1 \ 7 & 4 \end{bmatrix}\] After multiplying by 2, we get:\[2A = \begin{bmatrix} 2 \times 1 & 2 \times (-1) \ 2 \times 7 & 2 \times 4 \end{bmatrix} = \begin{bmatrix} 2 & -2 \ 14 & 8 \end{bmatrix}\]
02

Multiply Matrix B by Scalar 3

Multiply each element of matrix \(B\) by 3. Matrix \(B\) is: \[\begin{bmatrix} -3 & 2 \ 5 & 9 \end{bmatrix}\] After multiplying by 3, we get:\[3B = \begin{bmatrix} 3 \times (-3) & 3 \times 2 \ 3 \times 5 & 3 \times 9 \end{bmatrix} = \begin{bmatrix} -9 & 6 \ 15 & 27 \end{bmatrix}\]
03

Subtract Matrix 3B from Matrix 2A

Subtract the resulting matrix \(3B\) from matrix \(2A\): \[2A - 3B = \begin{bmatrix} 2 & -2 \ 14 & 8 \end{bmatrix} - \begin{bmatrix} -9 & 6 \ 15 & 27 \end{bmatrix} \] Subtract corresponding elements: \[= \begin{bmatrix} 2 - (-9) & -2 - 6 \ 14 - 15 & 8 - 27 \end{bmatrix} = \begin{bmatrix} 11 & -8 \ -1 & -19 \end{bmatrix}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Multiplication
Matrix multiplication is a process that involves taking two matrices and producing a third matrix. It’s crucial to note that the number of columns in the first matrix must match the number of rows in the second matrix to perform matrix multiplication. While this is not directly needed in the provided exercise, understanding this concept is essential when you move forward with more complex matrix operations.
  • Order of Multiplication Matters: Matrix multiplication is not commutative, meaning \( A \times B eq B \times A \). The order in which you multiply matrices changes the result.
  • Element Calculation: Each element of the resulting matrix is calculated by multiplying elements from the rows of the first matrix by the corresponding elements in the columns of the second matrix, and then summing those products.
  • Use of Identity Matrix: Multiplying any matrix by the identity matrix (a special matrix with 1s on the diagonal and 0s elsewhere) returns the original matrix.
Understanding how matrix multiplication works will greatly enhance your ability to manipulate matrices properly and is a fundamental skill in linear algebra.
Scalar Multiplication
Scalar multiplication involves multiplying each entry of a matrix by a single number, known as a scalar. This is a straightforward operation and is the first step in handling expressions like \( 2A - 3B \). For any matrix, scalar multiplication is performed element-wise throughout the matrix.
  • Simplicity of Calculation: Given matrix \(A\) and scalar \(k\), each element \(a_{ij}\) in matrix \(A\) becomes \(k \times a_{ij}\).
  • Applications: Scalar multiplication is often used in adjusting the magnitude of a matrix, particularly in scaling equations and in applying transformation matrices in physics and graphics.
For example, if \( A = \begin{bmatrix} 1 & -1 \ 7 & 4 \end{bmatrix} \), then \(2A\) results in \( \begin{bmatrix} 2 & -2 \ 14 & 8 \end{bmatrix} \). This can alter both the size and direction of the vectors within the matrix.
Matrix Subtraction
Matrix subtraction is the process of subtracting one matrix from another of the same dimensions by subtracting corresponding elements. This operation is similar to matrix addition and is used widely in solving matrix equations.
  • Dimensional Consistency: Both matrices must have the same dimensions for the operation to be valid. If \(A\) and \(B\) are two matrices of size \( m \times n \), their subtraction, \( A - B \), is defined as another \( m \times n \) matrix.
  • Element-Wise Operation: Every entry \( a_{ij} \) in matrix \( A \) is subtracted by the corresponding entry \( b_{ij} \) in matrix \( B \).
  • Result Interpretation: The result of matrix subtraction provides a matrix that signifies the difference between the original matrices, which can be interpreted as changes from \( B \) to \( A \).
For instance, in the expression \( 2A - 3B \), after scaling we subtract each element of \( 3B \) from \( 2A \), yielding \( \begin{bmatrix} 11 & -8 \ -1 & -19 \end{bmatrix} \). This operation can represent different transformations or shifts in the data being analyzed.

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Most popular questions from this chapter

A diagonal matrix \(D\) and a matrix \(A\) are given. Find the products \(D A\) and \(A D,\) where possible. $$ \begin{array}{l} D=\left[\begin{array}{ll} 4 & 0 \\ 0 & -3 \end{array}\right] \\ A=\left[\begin{array}{ll} 1 & 2 \\ 1 & 2 \end{array}\right] \end{array} $$

Matrices \(A\) and \(B\) are given. Compute \((A B)^{-1}\) and \(B^{-1} A^{-1}\). $$ A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right], \quad B=\left[\begin{array}{ll} 7 & 1 \\ 2 & 1 \end{array}\right] $$

Row and column vectors \(\vec{u}\) and \(\vec{v}\) are defined. Find the product \(\vec{u} \vec{v},\) where possible. $$ \vec{u}=\left[\begin{array}{lll} -3 & 6 & 1 \end{array}\right] \vec{v}=\left[\begin{array}{c} 1 \\ -1 \\ 1 \end{array}\right] $$

Row and column vectors \(\vec{u}\) and \(\vec{v}\) are defined. Find the product \(\vec{u} \vec{v},\) where possible. $$ \begin{array}{l} \vec{u}=\left[\begin{array}{cccc} 1 & 2 & 3 & 4 \end{array}\right] \\ \vec{v}=\left[\begin{array}{c} 1 \\ -1 \\ 1 \\ -1 \end{array}\right] \end{array} $$

A matrix \(A\) and vectors \(\vec{b}, \vec{u}\) and \(\vec{v}\) are given. Verify that \(\vec{u}\) and \(\vec{v}\) are both solutions to the equation \(A \vec{x}=\vec{b} ;\) that is, show that \(A \vec{u}=A \vec{v}=\vec{b}\). $$ \begin{array}{l} A=\left[\begin{array}{ll} 1 & 0 \\ 2 & 0 \end{array}\right] \\ \vec{b}=\left[\begin{array}{l} -3 \\ -6 \end{array}\right], \vec{u}=\left[\begin{array}{l} -3 \\ -1 \end{array}\right], \vec{v}=\left[\begin{array}{l} -3 \\ 59 \end{array}\right] \end{array} $$

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