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Ch. 6 - Matrices and Determinants
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Blitzer 8th EditionCh. 6 - Matrices and Determinants문제 37
7장, 문제 37

Perform the indicated matrix operations given that A, B and C are defined as follows. If an operation is not defined, state the reason.
A=[40−3501],B=[51−2−2],C=[1−1−11]A=\(\begin{bmatrix}\)4 & 0\\ -3 & 5\\ 0 & 1\(\end{bmatrix}\),B=\(\begin{bmatrix}\)5 & 1\\ -2 & -2\(\end{bmatrix}\),C=\(\begin{bmatrix}\)1 & -1\\ -1 & 1\(\end{bmatrix}\)
4B - 3C

검증된 단계별 안내
1
Step 1: Identify the matrices and the operation to be performed. We are asked to compute the expression \(4B - 3C\), where matrices \(B\) and \(C\) are given as: \(B = \begin{bmatrix} 5 & 1 \\ -2 & -2 \end{bmatrix}\) and \(C = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\).
Step 2: Multiply matrix \(B\) by the scalar 4. This means multiplying each element of \(B\) by 4: \(4B = 4 \times \begin{bmatrix} 5 & 1 \\ -2 & -2 \end{bmatrix} = \begin{bmatrix} 4 \times 5 & 4 \times 1 \\ 4 \times (-2) & 4 \times (-2) \end{bmatrix}\).
Step 3: Multiply matrix \(C\) by the scalar 3. This means multiplying each element of \(C\) by 3: \(3C = 3 \times \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 3 \times 1 & 3 \times (-1) \\ 3 \times (-1) & 3 \times 1 \end{bmatrix}\).
Step 4: Subtract the matrix \$3C\( from the matrix \)4B$. This involves subtracting corresponding elements: \(4B - 3C = \begin{bmatrix} (4B)_{11} - (3C)_{11} & (4B)_{12} - (3C)_{12} \\ (4B)_{21} - (3C)_{21} & (4B)_{22} - (3C)_{22} \end{bmatrix}\).
Step 5: Verify the dimensions of matrices \(B\) and \(C\) are the same before performing the subtraction. Since both are \(2 \times 2\) matrices, the subtraction is defined and can be performed element-wise.

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이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Matrix Multiplication

Matrix multiplication involves multiplying rows of the first matrix by columns of the second matrix and summing the products. It is only defined when the number of columns in the first matrix equals the number of rows in the second. This operation is not commutative, meaning AB does not necessarily equal BA.
추천 영상:
03:42
Finding Zeros & Their Multiplicity

Scalar Multiplication of Matrices

Scalar multiplication involves multiplying every element of a matrix by a constant (scalar). This operation changes the magnitude of the matrix elements but not the matrix dimensions. It is distributive over matrix addition and associative with scalar multiplication.
추천 영상:
4:35
Introduction to Matrices

Matrix Dimensions and Operation Validity

The dimensions of matrices determine which operations are defined. For addition or subtraction, matrices must have the same dimensions. For multiplication, the number of columns in the first matrix must equal the number of rows in the second. Understanding dimensions helps identify if an operation is possible.
추천 영상:
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Performing Row Operations on Matrices
관련 실천
교과서 질문

Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.

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교과서 질문

Write each matrix equation as a system of linear equations without matrices.

[20−1030110][xyz]=[695]\(\begin{bmatrix}\) 2 & 0 & -1 \\ 0 & 3 & 0 \\ 1 & 1 & 0 \(\end{bmatrix}\) \(\begin{bmatrix}\) x \\ y \\ z \(\end{bmatrix}\) = \(\begin{bmatrix}\) 6 \\ 9 \\ 5 \(\end{bmatrix}\)

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교과서 질문

In Exercises 31–36, use the alternative method for evaluating third-order determinants on here to evaluate each determinant. ∣0.5750.5390.513∣\(\begin{vmatrix}\) 0.5 & 7 & 5 \\ 0.5 & 3 & 9 \\ 0.5 & 1 & 3 \(\end{vmatrix}\)

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교과서 질문

In Exercises 37–44, use Cramer's Rule to solve each system. {x+y+z=02x−y+z=−1−x+3y−z=−8\(\begin{cases}\) x + y + z = 0 \\ 2x - y + z = -1 \\ -x + 3y - z = -8 \(\end{cases}\)

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교과서 질문

In Exercises 37–38, find the products and to determine whether B is the multiplicative inverse of A.

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교과서 질문

In Exercises 27 - 36, find (if possible) the following matrices: a. AB b. BA 1 2 2 - 3 1 - 1 - 1 1 A = B = 1 1 - 2 1 5 4 10 5

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