Problem 16
Question
If possible, find each of the following. (a) \(A+B\) (b) \(3 A\) (c) \(2 A-3 B\) $$A=\left[\begin{array}{rrr}1 & -2 & 5 \\\3 & -4 & -1\end{array}\right]$$ $$B=\left[\begin{array}{rrr}0 & -1 & -5 \\\\-3 & 1 & 2\end{array}\right]$$
Step-by-Step Solution
Verified Answer
(a) \(\begin{bmatrix} 1 & -3 & 0 \\ 0 & -3 & 1 \end{bmatrix}\), (b) \(\begin{bmatrix} 3 & -6 & 15 \\ 9 & -12 & -3 \end{bmatrix}\), (c) \(\begin{bmatrix} 2 & -1 & 25 \\ 15 & -11 & -8 \end{bmatrix}\).
1Step 1: Addition of Matrices (A+B)
To find the sum of two matrices, add the corresponding elements from each matrix. Matrix \(A\) and \(B\) have the same dimensions, so we can perform the addition. \[A+B = \begin{bmatrix} 1 & -2 & 5 \ 3 & -4 & -1 \end{bmatrix} + \begin{bmatrix} 0 & -1 & -5 \ -3 & 1 & 2 \end{bmatrix} = \begin{bmatrix} (1+0) & (-2-1) & (5-5) \ (3-3) & (-4+1) & (-1+2) \end{bmatrix} = \begin{bmatrix} 1 & -3 & 0 \ 0 & -3 & 1 \end{bmatrix} \]The sum of matrices \(A\) and \(B\) is \(\begin{bmatrix} 1 & -3 & 0 \ 0 & -3 & 1 \end{bmatrix}\).
2Step 2: Scalar Multiplication of Matrix (3A)
To find \(3A\), multiply each element of matrix \(A\) by the scalar 3.\[3A = 3 \cdot \begin{bmatrix} 1 & -2 & 5 \ 3 & -4 & -1 \end{bmatrix} = \begin{bmatrix} 3 \cdot 1 & 3 \cdot (-2) & 3 \cdot 5 \ 3 \cdot 3 & 3 \cdot (-4) & 3 \cdot (-1) \end{bmatrix} = \begin{bmatrix} 3 & -6 & 15 \ 9 & -12 & -3 \end{bmatrix} \]The result of scalar multiplication is \(\begin{bmatrix} 3 & -6 & 15 \ 9 & -12 & -3 \end{bmatrix}\).
3Step 3: Linear Combination of Matrices (2A - 3B)
To find \(2A - 3B\), calculate \(2A\) by multiplying each element of matrix \(A\) by 2, and \(3B\) by multiplying each element of matrix \(B\) by 3. Then subtract \(3B\) from \(2A\).First calculate \(2A\):\[2A = 2 \cdot \begin{bmatrix} 1 & -2 & 5 \ 3 & -4 & -1 \end{bmatrix} = \begin{bmatrix} 2 & -4 & 10 \ 6 & -8 & -2 \end{bmatrix}\]Next calculate \(3B\):\[3B = 3 \cdot \begin{bmatrix} 0 & -1 & -5 \ -3 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 0 & -3 & -15 \ -9 & 3 & 6 \end{bmatrix}\]Now compute: \[2A - 3B = \begin{bmatrix} 2 & -4 & 10 \ 6 & -8 & -2 \end{bmatrix} - \begin{bmatrix} 0 & -3 & -15 \ -9 & 3 & 6 \end{bmatrix} = \begin{bmatrix} (2-0) & (-4+3) & (10+15) \ (6+9) & (-8-3) & (-2-6) \end{bmatrix} = \begin{bmatrix} 2 & -1 & 25 \ 15 & -11 & -8 \end{bmatrix}\]The result of the linear combination is \(\begin{bmatrix} 2 & -1 & 25 \ 15 & -11 & -8 \end{bmatrix}\).
Key Concepts
Matrix AdditionScalar MultiplicationLinear Combination
Matrix Addition
Matrix addition is a straightforward process that involves adding corresponding elements of the matrices involved. For the matrices to be added, they need to have the same dimensions, meaning they have the same number of rows and columns.
For instance, if we have two matrices, \( A \) and \( B \), both are \( 2 \times 3 \) matrices, you would add each element in \( A \) with each corresponding element in \( B \).
Consider:
For instance, if we have two matrices, \( A \) and \( B \), both are \( 2 \times 3 \) matrices, you would add each element in \( A \) with each corresponding element in \( B \).
Consider:
- The element in the first row, first column of \( A \) is added to the first row, first column of \( B \).
- This is repeated for elements in subsequent positions, such as second row, first column, and so forth.
Scalar Multiplication
Scalar multiplication involves multiplying each element of a matrix by a single number, known as a scalar. It's as though you were multiplying the entire matrix by that number.
For example, to compute \( 3A \) for a matrix \( A \), take each element in \( A \) and multiply it by the scalar \( 3 \). This process affects all elements uniformly:
For example, to compute \( 3A \) for a matrix \( A \), take each element in \( A \) and multiply it by the scalar \( 3 \). This process affects all elements uniformly:
- Suppose an element of \( A \) is \( 1 \). After multiplication by \( 3 \), it becomes \( 3 \times 1 = 3 \).
- All elements undergo the same operation, whether positive, negative, or zero.
Linear Combination
A linear combination of matrices involves two or more matrices and typically includes both scalar multiplication and addition or subtraction.
To obtain a linear combination like \( 2A - 3B \), follow these steps:
To obtain a linear combination like \( 2A - 3B \), follow these steps:
- First, apply scalar multiplication to multiply each element of matrix \( A \) by \( 2 \) and each element of matrix \( B \) by \( 3 \).
- You'll get new matrices for \( 2A \) and \( 3B \).
- Subtract the resultant \( 3B \) from \( 2A \), meaning subtract each element of \( 3B \) from the corresponding element of \( 2A \).
Other exercises in this chapter
Problem 15
Solve the equation for \(x\) and then solve it for \(y .\) $$ x-y^{2}=5 $$
View solution Problem 15
If possible, solve the system. $$ \begin{array}{r} x+3 y+z=6 \\ 3 x+y-z=6 \\ x-y-z=0 \end{array} $$
View solution Problem 16
Let \(A\) be the given matrix. Find det \(A\) by using the method of co factors. $$ \left[\begin{array}{rrr} 1 & 1 & 5 \\ -3 & -3 & 0 \\ 7 & 0 & 0 \end{array}\r
View solution Problem 16
( Refer to Examples 3-5.) LetA be the given matrix. Find \(A^{-1}\) without a calculator. $$ \left[\begin{array}{rr} 1 & 0 \\ 1 & -1 \end{array}\right] $$
View solution