1)Given that A=[2 4] and B=[11 3],find C such
[3 6]. [3 6 ]
that B×C=A
There is no unique solution, and the question is rather ill expressed, but
C =
[4 5]
[-14,-17]
works just fine
Apologies for the confusion, you are correct. There can be multiple solutions for C that satisfy the equation B*C = A.
By solving the equation, we get:
11c1 + 3c2 = 2 (equation 1)
11c1 + 3c2 = 4 (equation 2)
If we subtract equation 1 from equation 2, we get:
0 = 2
This tells us that the equations are inconsistent. However, if we modify the second equation to be:
11c1 + 3c2 = 6
Then, subtracting equation 1 from equation 2 gives:
0 = 4
Which is also inconsistent.
Thus, there is no unique solution for C that satisfies the equation B*C = A.
To find matrix C, we need to solve the equation B × C = A.
Given that B is a 2x2 matrix:
B = [11 3]
[3 6]
And A is a 2x1 matrix:
A = [2]
[4]
Let's denote the elements of matrix C as follows:
C = [a b]
[c d]
Now, we can multiply B by C:
B × C = [11 3] × [a b] = [11a + 3c 3a + 6c]
[3 6] [c d] [3c + 6d 3c + 6d]
Setting this equal to A, we can equate the corresponding elements:
11a + 3c = 2 (Equation 1)
3a + 6c = 4 (Equation 2)
3c + 6d = 2 (Equation 3)
3c + 6d = 4 (Equation 4)
We have a system of four equations with four unknowns. Let's solve it:
From Equation 1, we can isolate a:
a = (2 - 3c) / 11
Substituting this value into Equation 2, we get:
3((2 - 3c) / 11) + 6c = 4
(6 - 9c + 6c) / 11 = 4
-9c + 6c = 4 × 11 - 6
-3c = 44 - 6
-3c = 38
c = -38 / 3
Substituting the value of c into Equation 3, we get:
3(-38 / 3) + 6d = 2
-38 + 6d = 6
6d = 6 + 38
6d = 44
d = 44 / 6
d = 22 / 3
Now, we can substitute the values of c and d back into a to find b:
a = (2 - 3c) / 11
a = (2 - 3(-38 / 3)) / 11
a = (2 + 38) / 11
a = 40 / 11
Finally, we have:
a = 40 / 11
b = To be determined
c = -38 / 3
d = 22 / 3
Therefore, matrix C is:
C = [40/11 b ]
[-38/3 22/3]
Note that we were not able to determine the value of b.
To find C, we can solve the equation B*C = A.
Given that B = [11 3] and A = [2 4], we can set up the equation:
[11 3] * [c1 c2] = [2 4]
This gives us two equations:
11c1 + 3c2 = 2
11c1 + 3c2 = 4
Subtracting the first equation from the second equation gives:
11c1 + 3c2 - (11c1 + 3c2) = 4 - 2
0 = 2
Since this equation is not true, it means that there is no solution for C that satisfies B*C = A.