Given that A =(2 4) and B=( 11 3) find C such
3 6. 4 1
that B×C=A
To find C, we need to solve the equation B × C = A.
Let's call the elements of C as follows:
C = (a b)
(c d)
Then, the equation can be written as:
B × C = A
(11 3) × (a b) = (2 4)
(c d)
This gives us the following system of equations:
11a + 3c = 2
11b + 3d = 4
To solve this system, we can use the method of substitution or elimination.
Let's use the elimination method:
Multiply the first equation by 11 and the second equation by 3 to eliminate the variable a:
121a + 33c = 22
33a + 9c = 6
Subtract the second equation from the first:
88a = 16
a = 16/88
a = 2/11
Substitute this value of a into the first equation to solve for c:
11(2/11) + 3c = 2
2 + 3c = 2
3c = 0
c = 0/3
c = 0
Now, let's substitute the values of a and c into the second equation to solve for d:
11b + 3d = 4
11b + 3(0) = 4
11b + 0 = 4
11b = 4
b = 4/11
Finally, the values of a, b, c, and d that satisfy the equation B × C = A are:
a = 2/11
b = 4/11
c = 0
d = unknown (yet).
Therefore, C = (2/11 4/11)
(0 d)
Note: To find the exact value of d, we need additional information or equations.
To find matrix C such that B×C = A, we need to solve the equation B×C = A for C.
Let's denote the elements of matrix C as follows:
C = (a b)
(c d)
Now, we can write the given equation as:
B×C = A
(11 3) × (a b) = (2 4)
(c d) (3 6)
To multiply the matrices on the left side of the equation, we need to perform matrix multiplication using the row-column method:
Row 1 of B × Column 1 of C: (11 × a) + (3 × c) = 2
Row 1 of B × Column 2 of C: (11 × b) + (3 × d) = 4
Row 2 of B × Column 1 of C: (4 × a) + (1 × c) = 3
Row 2 of B × Column 2 of C: (4 × b) + (1 × d) = 6
Now, we can solve this system of linear equations to find the values of a, b, c, and d.
From equation 1: 11a + 3c = 2 --> Equation (1)
From equation 2: 11b + 3d = 4 --> Equation (2)
From equation 3: 4a + c = 3 --> Equation (3)
From equation 4: 4b + d = 6 --> Equation (4)
We can solve this system of linear equations by substitution or elimination method. Let's solve it using the elimination method:
From equation (1), we can express "a" in terms of "c":
11a = 2 - 3c
a = (2 - 3c)/11
Substituting this value of "a" in equation (3), we get:
4((2 - 3c)/11) + c = 3
Simplifying the equation, we get:
(8 - 12c)/11 + c = 3
Multiplying both sides of the equation by 11 to eliminate the fraction, we get:
8 - 12c + 11c = 33
Simplifying further, we have:
-c = 33 - 8
-c = 25
c = -25
Now, substituting this value of "c" back into equation (1), we get:
11a + 3(-25) = 2
Simplifying, we have:
11a - 75 = 2
11a = 77
a = 7
Similarly, substituting the values of "a" and "c" into equations (2) and (4), we can find the values of "b" and "d".
From equation (2): 11b + 3d = 4
11b + 3d = 4
11(7) + 3d = 4
77 + 3d = 4
3d = -73
d = -73/3
From equation (4): 4b + d = 6
4b + d = 6
4b + (-73/3) = 6
4b = 6 + 73/3
4b = (18 + 73)/3
4b = 91/3
b = (91/3) / 4
b = 91/12
So, the matrix C that satisfies B×C = A is:
C = (7 91/12)
(-25 -73/3)