Consider a 3x3 matrix denoted by A with first row [-1 3 2], second row [0 4 1] and third row [1 2 5], and another 3x3 matrix, B, with first row [6 2 -1], second row [3 5 2] and third row [-2 -3 0]. The matrix 2A-5B is:a.First row [-3 -2 -5], second row [-1 -1 -8] and third row [2 1 0]b.First row [2 -4 0], second row [5 7 -2] and third row [1 1 0]c.First row [-32 -4 9], second row [-15 -17 -8] and third row [12 19 10]d.First row [11 -3 5], second row [-1 6 -7] and third row [1 0 -2]
Question
Consider a 3x3 matrix denoted by A with first row [-1 3 2], second row [0 4 1] and third row [1 2 5], and another 3x3 matrix, B, with first row [6 2 -1], second row [3 5 2] and third row [-2 -3 0]. The matrix 2A-5B is:a.First row [-3 -2 -5], second row [-1 -1 -8] and third row [2 1 0]b.First row [2 -4 0], second row [5 7 -2] and third row [1 1 0]c.First row [-32 -4 9], second row [-15 -17 -8] and third row [12 19 10]d.First row [11 -3 5], second row [-1 6 -7] and third row [1 0 -2]
Solution
The matrix 2A-5B can be calculated by performing the following steps:
-
Multiply matrix A by 2:
- First row: [-12, 32, 2*2] = [-2, 6, 4]
- Second row: [02, 42, 1*2] = [0, 8, 2]
- Third row: [12, 22, 5*2] = [2, 4, 10]
-
Multiply matrix B by 5:
- First row: [65, 25, -1*5] = [30, 10, -5]
- Second row: [35, 55, 2*5] = [15, 25, 10]
- Third row: [-25, -35, 0*5] = [-10, -15, 0]
-
Subtract the result of step 2 from the result of step 1:
- First row: [-2-30, 6-10, 4-(-5)] = [-32, -4, 9]
- Second row: [0-15, 8-25, 2-10] = [-15, -17, -8]
- Third row: [2-(-10), 4-(-15), 10-0] = [12, 19, 10]
Therefore, the matrix 2A-5B is represented by the third option: First row [-32 -4 9], second row [-15 -17 -8], and third row [12 19 10].
Similar Questions
The matrix A transpose is:a.First row [-1 3 2], second row [0 4 1] and third row [1 2 5]b.First row [1 0 1], second row [0 1 0] and third row [0 0 1]c.First row [1 0 1], second row [0 -3 5] and third row [2 4 -5]d.First row [-1 0 1], second row [3 4 2] and third row [2 1
The matrix BA is:a.First row [0 7 -7], second row [1 4 7] and third row [2 -1 3]b.First row [-1 7 7], second row [10 17 8] and third row [2 -3 3]c.First row [-7 24 9], second row [-1 33 21] and third row [2 -18 -7]d.First row [-5 26 5], second row [-1 3 2] and third row [0 -18 -5]
The matrix AB is:a.First row [0 7 -7], second row [1 4 7] and third row [2 -1 3]b.First row [-1 7 7], second row [10 17 8] and third row [2 -3 3]c.First row [-1 0 1], second row [0 1 -8] and third row [-4 -3 3]d.First row [4 0 1], second row [1 1 8] and third row [2 -3 3]
Use the fact that matrices A and B are row-equivalent.A = −2 −5 8 0 −17 1 3 −5 1 5−5 −9 13 7 −671 7 −13 5 −3B = 1 0 1 0 1 0 1 −2 0 30 0 0 1 −50 0 0 0 0
Matrix A and B are given below:A =1 −3 4 −1 9−2 6 −6 −1 −10−3 9 −6 −6 −33 −9 4 9 0 and B =1 −3 0 5 −70 0 2 −3 80 0 0 0 50 0 0 0 0Assume that the matrix A is row equivalent to B. Find(a) (3 pts) rank A.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.