Find –5A + 4B
matrices
A equals
start 3 by 2 matrix
1st row
1st column 6
2nd column 1
2nd row
1st column negative 4
2nd column negative 6
3rd row
1st column 7
2nd column negative 7
end matrix
B equals
start 3 by 2 matrix
1st row
1st column negative 5
2nd column negative 1
2nd row
1st column negative 3
2nd column negative 8
3rd row
1st column 6
2nd column 8
end matrix
(1 point)
Responses
matrix
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To find -5A + 4B, we need to subtract 5 times matrix A from 4 times matrix B.
First, let's find -5A:
-5A = -5 * A
= -5 * [6 1; -4 -6; 7 -7]
= [-30 -5; 20 30; -35 35]
Next, let's find 4B:
4B = 4 * B
= 4 * [-5 -1; -3 -8; 6 8]
= [-20 -4; -12 -32; 24 32]
Now, let's add -5A and 4B:
-5A + 4B = [-30 -5; 20 30; -35 35] + [-20 -4; -12 -32; 24 32]
= [-30 -5; 20 30; -35 35] + [-20 -4; -12 -32; 24 32]
= [-50 -9; 8 -2; -11 67]
So, -5A + 4B is the matrix:
[-50 -9;
8 -2;
-11 67]
To find -5A + 4B, we need to perform scalar multiplication on matrix A and matrix B, and then subtract the resulting matrices.
First, let's perform scalar multiplication on matrix A:
-5 * A =
-5 * (start 3 by 2 matrix
1st row
1st column 6
2nd column 1
2nd row
1st column -4
2nd column -6
3rd row
1st column 7
2nd column -7
end matrix) =
(start 3 by 2 matrix
1st row
1st column -30
2nd column -5
2nd row
1st column 20
2nd column 30
3rd row
1st column -35
2nd column 35
end matrix)
Next, let's perform scalar multiplication on matrix B:
4 * B =
4 * (start 3 by 2 matrix
1st row
1st column -5
2nd column -1
2nd row
1st column -3
2nd column -8
3rd row
1st column 6
2nd column 8
end matrix) =
(start 3 by 2 matrix
1st row
1st column -20
2nd column -4
2nd row
1st column -12
2nd column -32
3rd row
1st column 24
2nd column 32
end matrix)
Finally, let's subtract the resulting matrices:
(start 3 by 2 matrix
1st row
1st column -30
2nd column -5
2nd row
1st column 20
2nd column 30
3rd row
1st column -35
2nd column 35
end matrix) - (start 3 by 2 matrix
1st row
1st column -20
2nd column -4
2nd row
1st column -12
2nd column -32
3rd row
1st column 24
2nd column 32
end matrix) =
(start 3 by 2 matrix
1st row
1st column -10
2nd column -1
2nd row
1st column 32
2nd column 62
3rd row
1st column -59
2nd column 3
end matrix)
Therefore, -5A + 4B equals:
start 3 by 2 matrix
1st row
1st column -10
2nd column -1
2nd row
1st column 32
2nd column 62
3rd row
1st column -59
2nd column 3
end matrix
To find -5A + 4B, you need to perform scalar multiplication and matrix addition.
First, let's calculate -5A:
-5A equals:
start 3 by 2 matrix
1st row
1st column -5 * 6 = -30
2nd column -5 * 1 = -5
2nd row
1st column -5 * (-4) = 20
2nd column -5 * (-6) = 30
3rd row
1st column -5 * 7 = -35
2nd column -5 * (-7) = 35
end matrix
Next, let's calculate 4B:
4B equals:
start 3 by 2 matrix
1st row
1st column 4 * (-5) = -20
2nd column 4 * (-1) = -4
2nd row
1st column 4 * (-3) = -12
2nd column 4 * (-8) = -32
3rd row
1st column 4 * 6 = 24
2nd column 4 * 8 = 32
end matrix
Now, let's perform the matrix addition of -5A and 4B:
-5A + 4B equals:
start 3 by 2 matrix
1st row
1st column -30 + (-20) = -50
2nd column -5 + (-4) = -9
2nd row
1st column 20 + (-12) = 8
2nd column 30 + (-32) = -2
3rd row
1st column -35 + 24 = -11
2nd column 35 + 32 = 67
end matrix
Therefore, -5A + 4B is equal to:
start 3 by 2 matrix
1st row
1st column -50
2nd column -9
2nd row
1st column 8
2nd column -2
3rd row
1st column -11
2nd column 67
end matrix