Problem 14
Question
Operations with Matrices Find, if possible, \((a) A+B,(b) A-B,(c) 3 A,\) and \((d) 3 A-2 B.\) Use the matrix capabilities of a graphing utility to verify your results. $$A=\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right], \quad B=\left[\begin{array}{rr} -3 & -2 \\ 4 & 2 \end{array}\right]$$
Step-by-Step Solution
Verified Answer
The result for \(A+B\) is \(\left[\begin{array}{ll} -2 & 0 \ 6 & 3 \ \end{array}\right]\), \(A-B\) is \(\left[\begin{array}{ll} 4 & 4 \ -2 & -1 \ \end{array}\right]\), \(3A\) is \(\left[\begin{array}{ll} 3 & 6 \ 6 & 3 \ \end{array}\right]\), and \(3A-2B\) is \(\left[\begin{array}{ll} 9 & 10 \ 2 & -1 \ \end{array}\right]\). Compare with the results from a graphing utility to verify.
1Step 1: (a) Addition of Matrices
To add two matrices, their dimensions must be identical. Then add corresponding elements from each matrix to get the sum. Given: \(A=\left[\begin{array}{ll} 1 & 2 \ 2 & 1 \ \end{array}\right], \quad B=\left[\begin{array}{rr} -3 & -2 \ 4 & 2 \ \end{array}\right] \), The sum of A and B is, \(A+B = \left[\begin{array}{ll} 1+(-3) & 2+(-2) \ 2+4 & 1+2 \ \end{array}\right] = \left[\begin{array}{ll} -2 & 0 \ 6 & 3 \ \end{array}\right]\)
2Step 2: (b) Subtraction of Matrices
For subtracting two matrices, similar to addition, their dimensions should be the same. Subtract corresponding elements from each matrix. The difference of A and B is, \(A-B = \left[\begin{array}{ll} 1-(-3) & 2-(-2) \ 2-4 & 1-2 \ \end{array}\right] = \left[\begin{array}{ll} 4 & 4 \ -2 & -1 \ \end{array}\right]\)
3Step 3: (c) Multiplication of Matrix by a Scalar
The multiplication of a scalar with a matrix means that each element of the matrix is multiplied by the scalar. The multiplication of A by 3 is, \(3A = 3*\left[\begin{array}{ll} 1 & 2 \ 2 & 1 \ \end{array}\right] = \left[\begin{array}{ll} 3*1 & 3*2 \ 3*2 & 3*1 \ \end{array}\right] = \left[\begin{array}{ll} 3 & 6 \ 6 & 3 \ \end{array}\right]\)
4Step 4: (d) Combination of Matrix Operations
Combining matrix operations require the same approach. Multiply the matrices A and B by scalars 3 and 2 respectively, then subtract. \(3A-2B = 3*\left[\begin{array}{ll} 1 & 2 \ 2 & 1 \ \end{array}\right] - 2*\left[\begin{array}{rr} -3 & -2 \ 4 & 2 \ \end{array}\right] = \left[\begin{array}{ll} 3*1-2*(-3) & 3*2-2*(-2) \ 3*2-2*4 & 3*1-2*2 \ \end{array}\right] = \left[\begin{array}{ll} 9 & 10 \ 2 & -1 \ \end{array}\right]\)
Key Concepts
Matrix AdditionMatrix SubtractionScalar Multiplication of MatricesCombining Matrix Operations
Matrix Addition
Matrix addition is the simplest binary operation that can be performed with two matrices. To add two matrices, they must be of the same size, meaning they have the same number of rows and columns. The process then involves adding each corresponding element from the two matrices to find the sum.
For example, given two matrices:
A = \(\left[\begin{array}{cc} a & b \ c & d \end{array}\right]\) and B = \(\left[\begin{array}{cc} e & f \ g & h \end{array}\right]\),the matrix sum A + B is given by \(\left[\begin{array}{cc} a+e & b+f \ c+g & d+h \end{array}\right]\).
Adding matrices is straightforward when you focus on each element at a time, always matching the row and column positions. It is often helpful to visualize this operation as overlaying one matrix on top of another and then combining the numbers.
For example, given two matrices:
A = \(\left[\begin{array}{cc} a & b \ c & d \end{array}\right]\) and B = \(\left[\begin{array}{cc} e & f \ g & h \end{array}\right]\),the matrix sum A + B is given by \(\left[\begin{array}{cc} a+e & b+f \ c+g & d+h \end{array}\right]\).
Adding matrices is straightforward when you focus on each element at a time, always matching the row and column positions. It is often helpful to visualize this operation as overlaying one matrix on top of another and then combining the numbers.
Matrix Subtraction
Matrix subtraction is quite similar to matrix addition and follows the same rules for size compatibility. Just as with addition, the two matrices in question must have identical dimensions, meaning they should have the same number of rows and columns.
For instance, if we consider the same matrices A and B as before, the result of matrix subtraction, A - B, will be formed by subtracting the corresponding elements:
\( A - B = \left[\begin{array}{cc} a-e & b-f \ c-g & d-h \end{array}\right]\).
The key here is to take each element pair and simply perform a subtraction, element by element. Subtraction often helps in representing the difference or change between two sets of data points reflected by the matrices.
For instance, if we consider the same matrices A and B as before, the result of matrix subtraction, A - B, will be formed by subtracting the corresponding elements:
\( A - B = \left[\begin{array}{cc} a-e & b-f \ c-g & d-h \end{array}\right]\).
The key here is to take each element pair and simply perform a subtraction, element by element. Subtraction often helps in representing the difference or change between two sets of data points reflected by the matrices.
Scalar Multiplication of Matrices
Scalar multiplication involves the multiplication of a matrix by a single number, known as a scalar. In this operation, each element of the matrix is multiplied by the scalar to yield a new matrix.
For example, if we have a scalar k and matrix A as before, the result of the scalar multiplication kA is obtained by multiplying each element of A by k:
\( kA = k\cdot\left[\begin{array}{cc} a & b \ c & d \end{array}\right] = \left[\begin{array}{cc} ka & kb \ kc & kd \end{array}\right]\).
This operation is useful in situations where we need to rescale a dataset or adjust the weighting of a matrix's elements. Remembering to multiply every single element by the scalar is critical in correctly performing this operation.
For example, if we have a scalar k and matrix A as before, the result of the scalar multiplication kA is obtained by multiplying each element of A by k:
\( kA = k\cdot\left[\begin{array}{cc} a & b \ c & d \end{array}\right] = \left[\begin{array}{cc} ka & kb \ kc & kd \end{array}\right]\).
This operation is useful in situations where we need to rescale a dataset or adjust the weighting of a matrix's elements. Remembering to multiply every single element by the scalar is critical in correctly performing this operation.
Combining Matrix Operations
Combining matrix operations often involves performing more than one type of operation sequentially. This could mean adding and then multiplying, or subtracting and then taking a scalar multiple, among other combinations. It's essential to follow the rules of each operation and to conduct them in the given order.
For instance, if you are given the operation 3A - 2B, you need to first perform scalar multiplication on both matrices A and B, respectively, and then subtract the resultant matrices. It's a step by step process.
Being attentive to the order of operations is crucial, as reversing the steps can lead to an entirely different result.
For instance, if you are given the operation 3A - 2B, you need to first perform scalar multiplication on both matrices A and B, respectively, and then subtract the resultant matrices. It's a step by step process.
- Multiply matrix A by 3.
- Multiply matrix B by 2.
- Subtract the matrix resulting from 2 from the matrix resulting from 1.
Being attentive to the order of operations is crucial, as reversing the steps can lead to an entirely different result.
Other exercises in this chapter
Problem 14
Solve the system by the method of elimination and check any solutions algebraically. $$\left\\{\begin{array}{l} 3 x-5 y=2 \\ 2 x+5 y=13 \end{array}\right.$$
View solution Problem 14
Solve the system by the method of substitution. Check your solution graphically. $$\left\\{\begin{aligned} -2 x+y &=-5 \\ x^{2}+y^{2} &=25 \end{aligned}\right.$
View solution Problem 14
Use back-substitution to solve the system of linear equations. $$\left\\{\begin{aligned} 4 x-3 y-2 z &=21 \\ 6 y-5 z &=-8 \\ z &=-2 \end{aligned}\right.$$
View solution Problem 15
Use the matrix capabilities of a graphing utility to find the determinant of the matrix. $$\left[\begin{array}{rrr} 1.3 & 0.2 & 3.2 \\ 0.2 & 6.2 & 0.2 \\ -0.4 &
View solution