Problem 12
Question
Refer to the following matrices: \(A=\left[\begin{array}{rr}-1 & 2 \\ 3 & -2 \\ 4 & 0\end{array}\right] \quad B=\left[\begin{array}{rr}2 & 4 \\ 3 & 1 \\ -2 & 2\end{array}\right]\) \(C=\left[\begin{array}{rrr}3 & -1 & 0 \\ 2 & -2 & 3 \\ 4 & 6 & 2\end{array}\right] \quad D=\left[\begin{array}{rrr}2 & -2 & 4 \\ 3 & 6 & 2 \\\ -2 & 3 & 1\end{array}\right]\) Compute \(4 D-2 C\).
Step-by-Step Solution
Verified Answer
The result of \(4D - 2C\) is: \(\left[\begin{array}{rrr}14 & -10 & 16 \\\ 16 & 20 & 14 \\\ 0 & 24 & 8\end{array}\right]\).
1Step 1: Multiply the matrices with the scalars
First, we will multiply matrix \(C\) by the scalar \(-2\) and matrix \(D\) by the scalar \(4\).
For matrix \(C\):
\(-2 \times \left[\begin{array}{rrr}3 & -1 & 0 \\\ 2 & -2 & 3 \\\ 4 & 6 & 2\end{array}\right] =
\left[\begin{array}{rrr}-6 & 2 & 0 \\\ -4 & 4 & -6 \\\ -8 & -12 & -4\end{array}\right]\)
For matrix \(D\):
\(4 \times \left[\begin{array}{rrr}2 & -2 & 4 \\\ 3 & 6 & 2 \\\ -2 & 3 & 1\end{array}\right] =
\left[\begin{array}{rrr}8 & -8 & 16 \\\ 12 & 24 & 8 \\\ -8 & 12 & 4\end{array}\right]\)
2Step 2: Subtract the matrices
Now, subtract the matrix we obtained by multiplying \(-2C\) from the matrix we obtained by multiplying \(4D\).
\(\left[\begin{array}{rrr}8 & -8 & 16 \\\ 12 & 24 & 8 \\\ -8 & 12 & 4\end{array}\right] - \left[\begin{array}{rrr}-6 & 2 & 0 \\\ -4 & 4 & -6 \\\ -8 & -12 & -4\end{array}\right] = \left[\begin{array}{rrr}8-(-6) & -8-2 & 16-0 \\\ 12-(-4) & 24-4 & 8-(-6) \\\ -8-(-8) & 12-(-12) & 4-(-4)\end{array}\right]\)
3Step 3: Calculate the result
Now, perform the subtraction within the matrix.
\(\left[\begin{array}{rrr}14 & -10 & 16 \\\ 16 & 20 & 14 \\\ 0 & 24 & 8\end{array}\right]\)
So, the result of \(4D - 2C\) is: \(\left[\begin{array}{rrr}14 & -10 & 16 \\\ 16 & 20 & 14 \\\ 0 & 24 & 8\end{array}\right]\).
Key Concepts
Matrix MultiplicationScalar MultiplicationMatrix Subtraction
Matrix Multiplication
Matrix multiplication is a mathematical operation that takes two matrices and produces a third matrix. It is different from simple arithmetic multiplication and requires specific rules to be followed. When you multiply two matrices, you perform a series of dot products between the rows of the first matrix and the columns of the second matrix. This operation is only possible when the number of columns in the first matrix matches the number of rows in the second matrix.
To perform matrix multiplication, it's key to remember a few things:
To perform matrix multiplication, it's key to remember a few things:
- The resulting matrix's size will be determined by the number of rows from the first matrix and the number of columns from the second matrix.
- If you have matrix A (with dimensions m×n) and matrix B (with dimensions n×p), the resulting matrix AB will have dimensions m×p.
Scalar Multiplication
Scalar multiplication involves multiplying each entry of a matrix by a constant value called a scalar. This operation is quite straightforward and maintains the dimensionality of the matrix. The result is a new matrix where each element is the product of the corresponding element in the original matrix and the scalar.
Here's how scalar multiplication works:
Here's how scalar multiplication works:
- Given a scalar, say 3, and a matrix, you multiply each element of the matrix by 3.
- If matrix A is \[\begin{array}{cc}1 & 2 \3 & 4 \\end{array}\], multiplying it by 3 gives you \[\begin{array}{cc}3 & 6 \9 & 12 \\end{array}\].
Matrix Subtraction
Matrix subtraction is an operation that involves taking two matrices of the same dimensions and subtracting corresponding elements. The result is another matrix of the same size. For the subtraction to be possible, the two matrices must have identical dimensions.
Here is how you perform matrix subtraction:
Here is how you perform matrix subtraction:
- Ensure both matrices have the same number of rows and columns.
- Subtract the corresponding elements of the second matrix from the first matrix.
- For matrices A and B:\[\text{If } A = \begin{array}{cc}2 & 3 \4 & 5\end{array}, \text{and } B = \begin{array}{cc}1 & 1 \2 & 2\end{array}, \\text{then } A - B = \begin{array}{cc}1 & 2 \2 & 3\end{array}.\]
Other exercises in this chapter
Problem 12
Find the inverse of the matrix, if it exists. Verify your answer. \(\left[\begin{array}{rrr}1 & 2 & 0 \\ -3 & 4 & -2 \\ -5 & 0 & -2\end{array}\right]\)
View solution Problem 12
Compute the indicated products. \(\left[\begin{array}{rr}1 & 3 \\ -1 & 2\end{array}\right]\left[\begin{array}{lll}1 & 3 & 0 \\ 3 & 0 & 2\end{array}\right]\)
View solution Problem 12
Given that the augmented matrix in row-reduced form is equivalent to the augmented matrix of a system of linear equations, (a) determine whether the system has
View solution Problem 12
Indicate whether the matrix is in rowreduced form. \(\left[\begin{array}{ll|l}0 & 1 & 3 \\ 0 & 0 & 5\end{array}\right]\)
View solution