Problem 37
Question
In Exercises \(37-44\), perform the indicated matrix operations given that \(A, B,\) and \(C\) are defined as follows If an operation is not defined, state the reason. $$A=\left[\begin{array}{rr}4 & 0 \\\\-3 & 5 \\\0 & 1\end{array}\right] \quad B=\left[\begin{array}{rr} 5 & 1 \\\\-2 & -2\end{array}\right] \quad C=\left[\begin{array}{rr}1 & -1 \\\\-1 & 1\end{array}\right]$$ $$4 B-3 C$$
Step-by-Step Solution
Verified Answer
The resulting matrix of the operation \(4B - 3C\) is \(\left[\begin{array}{rr}17 & 7 \\-5 & -11 \end{array}\right]\)
1Step 1: Multiply Matrix B by Scalar 4
The first operation is multiplying matrix B by 4. The result is a new matrix where every entry is four times the corresponding entry in B. Here’s how you do it: \[4B = 4*\left[\begin{array}{rr}5 & 1 \\-2 & -2 \end{array}\right] = \left[\begin{array}{rr}20 & 4 \\-8 & -8 \end{array}\right] \]
2Step 2: Multiply Matrix C by Scalar 3
The second operation is multiplying matrix C by 3. The result is a new matrix where every entry is three times the corresponding entry in C. Here’s how you do it: \[3C = 3*\left[\begin{array}{rr}1 & -1 \\-1 & 1 \end{array}\right] = \left[\begin{array}{rr}3 & -3 \\-3 & 3 \end{array}\right] \]
3Step 3: Subtract the resulting Matrix of Step 2 from Matrix obtained in Step 1
The final operation is subtracting the matrix resulting from Step 2 from the matrix resulting from Step 1. Remember that subtraction is done entry by entry. Here’s how you do it:\[4B - 3C = \left[\begin{array}{rr}20 & 4 \\-8 & -8 \end{array}\right] - \left[\begin{array}{rr}3 & -3 \\-3 & 3 \end{array}\right] = \left[\begin{array}{rr}17 & 7 \\-5 & -11 \end{array}\right] \]
Key Concepts
Scalar MultiplicationMatrix SubtractionMatrix Addition
Scalar Multiplication
Scalar multiplication is a basic matrix operation where each element of a matrix is multiplied by a real number called a scalar. This operation adjusts the magnitude of the matrix without altering its dimensions. Here's how it works. Suppose you have a matrix:\[B = \begin{bmatrix}5 & 1 \-2 & -2\end{bmatrix}\]If you want to multiply this matrix by a scalar, say 4, each entry in the matrix gets multiplied by 4. The operation would look like this:\[4B = \begin{bmatrix} 4*5 & 4*1 \4*(-2) & 4*(-2)\end{bmatrix} = \begin{bmatrix} 20 & 4 \-8 & -8\end{bmatrix}\]The key steps in scalar multiplication include:
- Identifying the scalar.
- Multiplying each entry of the matrix by this scalar.
- Retaining the matrix's original size, with only the values changed.
Matrix Subtraction
Matrix subtraction, like addition, is a straightforward matrix operation that requires both matrices to be of the same dimensions. It involves subtracting corresponding entries of one matrix from another. Consider the matrices:\[A = \begin{bmatrix}20 & 4 \-8 & -8\end{bmatrix}, \ 3C = \begin{bmatrix}3 & -3 \-3 & 3\end{bmatrix}\]To subtract matrix \(3C\) from matrix \(A\), you perform the subtraction entry-wise as follows:\[A - 3C = \begin{bmatrix}20 - 3 & 4 - (-3) \-8 - (-3) & -8 - 3\end{bmatrix} = \begin{bmatrix}17 & 7 \-5 & -11\end{bmatrix}\]Important points to consider include:
- Matrices must have the same dimensions.
- Subtract corresponding elements of the matrices.
- Produce a resultant matrix of the same dimensions.
Matrix Addition
Matrix addition is among the simplest matrix operations and requires matrices of the same size. In matrix addition, each element of one matrix is added to the corresponding element of another matrix. Suppose you have two matrices:\[C = \begin{bmatrix}1 & -1 \-1 & 1\end{bmatrix} \text{ and } D = \begin{bmatrix}2 & 3 \4 & -4\end{bmatrix}\]To add these matrices, perform entry-wise addition, which results in:\[C + D = \begin{bmatrix}1 + 2 & -1 + 3 \-1 + 4 & 1 + (-4)\end{bmatrix} = \begin{bmatrix}3 & 2 \3 & -3\end{bmatrix}\]Here are some essentials for matrix addition:
- The matrices should be of the same dimensions.
- Add corresponding elements from both matrices.
- The result is a matrix of the same dimensions as the original matrices.
Other exercises in this chapter
Problem 36
Write each matrix equation as a system of linear equations without matrices. $$\left[\begin{array}{rrr}-1 & 0 & 1 \\\0 & -1 & 0 \\\0 & 1 & 1\end{array}\right]\l
View solution Problem 36
solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{array}{c} w+x+y+z=
View solution Problem 37
Evaluate each determinant. $$\left|\begin{array}{rrrr}4 & 2 & 8 & -7 \\\\-2 & 0 & 4 & 1 \\\5 & 0 & 0 & 5 \\\4 & 0 & 0 & -1\end{array}\right|$$
View solution Problem 37
Describe what happens when Gaussian elimination is used to solve an inconsistent system.
View solution