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=
B=
Tr(A)=1+1=2
Tr(B)=1+1=2
C(Tr(A)+Tr(B))=C(2+2)=4C
A+B=
A+B=
C(A+B)=
Tr(C(A+B))=2C+2C=4C
Hence, Tr(C(A+B)=C(Tr(A)+Tr(B))
Hence, proved.
105 because 4*18+33 equals 105
Answer:
The number of letters in the one street be 220 .
Step-by-step explanation:
As given
A postman has to deliver 450 letters .
The number of letters delivered in one street is twice the number delivered in other .
Let us assume that the number of letters delivered in second street be x.
Let us assume that the number of letters delivered in first street = 2x
As given
If he is left with 120 letters .
Thus
The number of letter delivered in one and other street = 450 - 120
= 330
Than the equation becomes
x + 2x = 330
3x = 330
x = 110
The number of letters in the one street = 2 × 110
= 220
Therefore the number of letters delivered in first street be 220.