Answer with Step-by-step explanation:
We are given that two matrices A and B are square matrices of the same size.
We have to prove that
Tr(C(A+B)=C(Tr(A)+Tr(B))
Where C is constant
We know that tr A=Sum of diagonal elements of A
Therefore,
Tr(A)=Sum of diagonal elements of A
Tr(B)=Sum of diagonal elements of B
C(Tr(A))=
Sum of diagonal elements of A
C(Tr(B))= Sum of diagonal elements of B
Tr(C(A+B)=Sum of diagonal elements of (C(A+B))
Suppose ,A=![\left[\begin{array}{ccc}1&0\\1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D)
B=![\left[\begin{array}{ccc}1&1\\1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D)
Tr(A)=1+1=2
Tr(B)=1+1=2
C(Tr(A)+Tr(B))=C(2+2)=4C
A+B=![\left[\begin{array}{ccc}1&0\\1&1\end{array}\right]+\left[\begin{array}{ccc}1&1\\1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D)
A+B=![\left[\begin{array}{ccc}2&1\\2&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%5C%5C2%262%5Cend%7Barray%7D%5Cright%5D)
C(A+B)=![\left[\begin{array}{ccc}2C&C\\2C&2C\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2C%26C%5C%5C2C%262C%5Cend%7Barray%7D%5Cright%5D)
Tr(C(A+B))=2C+2C=4C
Hence, Tr(C(A+B)=C(Tr(A)+Tr(B))
Hence, proved.
To apply the distributive property to that equation, it would be
12 * 5y + 12 * 4
Answer:
Try 31 degrees/ Hard to explain but try it
Step-by-step explanation: cause between angle a, c, b - and angle a, o , c - their is an intersect. on that intersect onside is 62 degrees and the other side is 118 degrees; now thats a 180 degrees angle cause the base of the intersect is straight. Now this intersect is connected to the obtuse angle of the bottom triangle which is 118 degrees. Knowing that you can figure out the other corners of the traingle. The last two acute angles of to be congruent.
Answer:
I need more information
Step-by-step explanation:
Answer:
8s+8
Step-by-step explanation:
Put it into groups: (4s+4s)+(10-2)=8s+8
Brainliest is appreciated
