MHB Chris 's question at Yahoo Answers (Inverse of cA)

  • Thread starter Thread starter Fernando Revilla
  • Start date Start date
Click For Summary
If A is an invertible matrix and c is a non-zero scalar, then the matrix cA is also invertible. The inverse of cA can be expressed as (cA)^{-1} = (1/c)A^{-1}. This conclusion is derived using properties of scalar multiplication and matrix multiplication. The proof confirms that multiplying cA by its inverse yields the identity matrix. Thus, the relationship between cA and its inverse is established clearly.
Mathematics news on Phys.org
Hello Chris,

Suppose $c\neq 0$. Then, using the well known properties $(\lambda M) N=M(\lambda N)=\lambda (MN)$ and $\lambda(\mu M)=(\lambda\mu)M$ : $$(cA)\left(\frac{1}{c}A^{-1}\right)=\frac{1}{c}\left((cA)A^{-1}\right)=\frac{1}{c}\left(c\;\left(AA^{-1}\right)\right)=\left(\frac{1}{c}\cdot c\right)I=1I=I$$ So, $cA$ is invertivle and $(cA)^{-1}=\dfrac{1}{c}A^{-1}$.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K