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

Short Answer

Expert verified
The result is \( \begin{bmatrix} 9 & -7 \\ 11 & -6 \end{bmatrix} \).

Step by step solution

01

Matrix Subtraction

Subtract matrix \(B\) from matrix \(A\). Compute each element of the matrix \(A-B\) by subtracting the corresponding elements of \(B\) from \(A\): \[ A-B = \begin{bmatrix} 1-(-3) & -1-2 \ 7-5 & 4-9 \end{bmatrix} = \begin{bmatrix} 4 & -3 \ 2 & -5 \end{bmatrix} \]
02

Scalar Multiplication

Multiply the resulting matrix \(A-B\) by the scalar \(3\). Multiply each element of the matrix by \(3\): \[ 3(A-B) = 3 \cdot \begin{bmatrix} 4 & -3 \ 2 & -5 \end{bmatrix} = \begin{bmatrix} 12 & -9 \ 6 & -15 \end{bmatrix} \]
03

Add Matrices

Add the matrix \(3(A-B)\) to \(B\). Compute each element of the resulting matrix \(3(A-B) + B\) by adding the corresponding elements of \(B\) to \(3(A-B)\): \[ 3(A-B) + B = \begin{bmatrix} 12 & -9 \ 6 & -15 \end{bmatrix} + \begin{bmatrix} -3 & 2 \ 5 & 9 \end{bmatrix} = \begin{bmatrix} 9 & -7 \ 11 & -6 \end{bmatrix} \]
04

Result Matrix

The result of the expression \(3(A-B) + B\) is the matrix: \( \begin{bmatrix} 9 & -7 \ 11 & -6 \end{bmatrix} \).

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Key Concepts

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

Matrix Subtraction
Matrix Subtraction is a fundamental concept in matrix algebra. It involves subtracting elements of one matrix from the corresponding elements of another matrix of the same size. In the exercise above, we subtract matrix \( B \) from matrix \( A \). This means for each position in the matrix, we subtract the element of matrix \( B \) from the element of matrix \( A \).

For instance, we calculate \( A[1,1] - B[1,1] \), that is, \( 1 - (-3) = 4 \). Similarly, for the position \( [1,2] \), we have \( -1 - 2 = -3 \).
  • Remember: The two matrices in subtraction must be of the same dimensions.
  • The operation is element-wise, so you handle each entry individually.
  • Matrix subtraction is straightforward but requires careful attention to negative signs and arithmetic details.
Scalar Multiplication
Scalar Multiplication involves multiplying every element of a matrix by the same scalar value. This scalar could be any number, positive or negative. In the given exercise, after determining \( A-B \), we multiply each element of the resulting matrix by the scalar \( 3 \).

This means each element like \( 4 \) becomes \( 4 \times 3 = 12 \), or \( -3 \) becomes \( -3 \times 3 = -9 \).
  • Key Concepts: Scalar multiplication impacts all elements uniformly.
  • Multiplying by a positive scalar retains the sign of the elements, while a negative scalar flips them.
  • Great for scaling transformations in graphics and physics applications.

It's essential to ensure each multiplication is accurate to avoid errors in further matrix operations.
Matrix Addition
Matrix Addition is one of the basic operations where you add corresponding elements of two matrices. For this method, just like subtraction, the matrices must be of the same size. In the final step of the exercise, we add matrix \( 3(A-B) \) with matrix \( B \).
For example, you calculate \( 12 + (-3) = 9 \) and \( -9 + 2 = -7 \), moving sequentially through each element.
  • Essential Points: Remember to match entries to ensure corresponding positions are added.
  • Addition is done element by element, maintaining matrix dimensions.
  • Matrix addition is intuitive and mirrors the algebraic addition you use with single numbers.

Ensuring proper alignment during addition helps you avoid mismatched elements, which could lead to errors in calculations.

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Most popular questions from this chapter

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 \(A \vec{u}=A \vec{v}=\vec{b}\). $$ \begin{array}{l} A=\left[\begin{array}{cc} 1 & -2 \\ -3 & 6 \end{array}\right] \\ \vec{b}=\left[\begin{array}{c} 2 \\ -6 \end{array}\right], \vec{u}=\left[\begin{array}{c} 0 \\ -1 \end{array}\right], \vec{v}=\left[\begin{array}{l} 2 \\ 0 \end{array}\right] \end{array} $$

Matrices \(A\) and \(B\) are given. Solve the matrix equation \(A X=B\). $$ \begin{array}{l} A=\left[\begin{array}{cc} -3 & -6 \\ 4 & 0 \end{array}\right] \\ B=\left[\begin{array}{cc} 48 & -30 \\ 0 & -8 \end{array}\right] \end{array} $$

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

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 \(A \vec{u}=A \vec{v}=\vec{b}\). $$ \begin{array}{l} A=\left[\begin{array}{cccc} 0 & -3 & -1 & -3 \\ -4 & 2 & -3 & 5 \end{array}\right] \\ \vec{b}=\left[\begin{array}{c} 48 \\ 36 \end{array}\right], \vec{u}=\left[\begin{array}{c} -17 \\ -16 \\ 0 \\ 0 \end{array}\right] \\ \vec{v}=\left[\begin{array}{c} -8 \\ -28 \\ 0 \\ 12 \end{array}\right] \end{array} $$

A matrix \(A\) and vector \(\vec{b}\) are given. (a) Solve the equation \(A \vec{x}=\vec{O}\) (b) Solve the equation \(A \vec{x}=\vec{b}\). In each of the above, be sure to write your answer in vector format. Also, when possible, give 2 particular solutions to each equation. $$ A=\left[\begin{array}{ll} -4 & -1 \\ -3 & -2 \end{array}\right], \vec{b}=\left[\begin{array}{l} 1 \\ 4 \end{array}\right] $$

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