Form a 4 by 3 matrix, B, such that a) bij= i*j b)=(-1)^i+j
a) To form a 4 by 3 matrix B such that B[i][j] = i*j, you can follow these steps:
Step 1: Define a 4 by 3 matrix B.
Step 2: Assign the value of i*j to each element B[i][j], where i ranges from 0 to 3 and j ranges from 0 to 2.
Step 3: Repeat the process for all elements of B until the matrix is filled.
Here is how the matrix B will look like:
B = | 0 0 0 |
| 0 1 2 |
| 0 2 4 |
| 0 3 6 |
b) To form a 4 by 3 matrix B such that B[i][j] = (-1)^(i+j), you can follow these steps:
Step 1: Define a 4 by 3 matrix B.
Step 2: Assign the value of (-1)^(i+j) to each element B[i][j], where i ranges from 0 to 3 and j ranges from 0 to 2.
Step 3: Repeat the process for all elements of B until the matrix is filled.
Here is how the matrix B will look like:
B = | 1 -1 1 |
|-1 1 -1 |
| 1 -1 1 |
|-1 1 -1 |
To form a 4 by 3 matrix B with the given conditions, we need to assign values to each element in the matrix according to the given formulas.
a) For the condition bij = i * j, where i represents the row number and j represents the column number, we can fill in the elements as follows:
B =
| 1 * 1 1 * 2 1 * 3 |
| 2 * 1 2 * 2 2 * 3 |
| 3 * 1 3 * 2 3 * 3 |
| 4 * 1 4 * 2 4 * 3 |
Simplifying the expressions, we get:
B =
| 1 2 3 |
| 2 4 6 |
| 3 6 9 |
| 4 8 12 |
Therefore, the matrix B that satisfies the first condition is:
B =
| 1 2 3 |
| 2 4 6 |
| 3 6 9 |
| 4 8 12 |
b) For the condition (-1)^(i + j), we need to alternate the signs of each element based on the sum of the row number (i) and column number (j). We can fill in the elements as follows:
B =
| (-1)^(1+1) (-1)^(1+2) (-1)^(1+3) |
| (-1)^(2+1) (-1)^(2+2) (-1)^(2+3) |
| (-1)^(3+1) (-1)^(3+2) (-1)^(3+3) |
| (-1)^(4+1) (-1)^(4+2) (-1)^(4+3) |
Simplifying the expressions, we get:
B =
| 1 -1 1 |
| -1 1 -1 |
| 1 -1 1 |
| -1 1 -1 |
Therefore, the matrix B that satisfies the second condition is:
B =
| 1 -1 1 |
| -1 1 -1 |
| 1 -1 1 |
| -1 1 -1 |
a) To form a matrix B with bij = i*j, we simply need to multiply the row number by the column number for each element of the matrix.
B = [1*1 1*2 1*3
2*1 2*2 2*3
3*1 3*2 3*3
4*1 4*2 4*3]
B = [1 2 3
2 4 6
3 6 9
4 8 12]
b) To form a matrix B with bij = (-1)^(i+j), we need to alternate signs for each element of the matrix based on the sum of row number (i) and column number (j).
B = [(-1)^(1+1) (-1)^(1+2) (-1)^(1+3)
(-1)^(2+1) (-1)^(2+2) (-1)^(2+3)
(-1)^(3+1) (-1)^(3+2) (-1)^(3+3)
(-1)^(4+1) (-1)^(4+2) (-1)^(4+3)]
B = [1 -1 1
-1 1 -1
1 -1 1
-1 1 -1]
Note: (-1)^0 is equal to 1.