Answer to Question #119198 in Linear Algebra for xxx

Question

Given the matrices (Each matrices are inside a [ ] )
A= 3 0 , B= 4 -1 , C= 1 4 2 , D= 1 5 2 , E= 0 1 3
-1 2 0 2 3 1 5 -1 0 1 -1 1 2
1 1 3 2 4 4 1 3

1.-4C^2
2.(E-D)^T
3. A(BC)
4. (B^T B)C
5. 3B-AB

Answer 1

A= $\begin{bmatrix} 3 & 0 \\ -1 & 2 \\ 1 & 1 \end{bmatrix}$ , B = $\begin{bmatrix} 4 & -1 \\ 0 & 2 \end{bmatrix}$

C = $\begin{bmatrix} 1 & 4 & 2 \\ 3 & 1 & 5 \\ \end{bmatrix}$ , D = $\begin{bmatrix} 1 & 5 & 2 \\ -1 & 0 & 1 \\ 3 & 2 & 4 \end{bmatrix}$ ,

E= $\begin{bmatrix} 0 & 1 & 3 \\ -1 & 1 & 2 \\ 4 & 1 & 3 \end{bmatrix}$

1. Since C is not a square matrix , C² doesn't exist. Hence -4C² doesn't exist

2. E - D = $\begin{bmatrix} -1 & -4 & 1 \\ 0 & 1 & 1 \\ 1 & -1 & -1 \end{bmatrix}$

So (E-D)^T = $\begin{bmatrix} -1& 0 & 1 \\ -4 & 1 & -1 \\ 1 & 1 & -1 \end{bmatrix}$

3. BC = $\begin{bmatrix} 4 & -1 \\ 0 & 2 \end{bmatrix}$ X $\begin{bmatrix} 1 & 4 & 2 \\ 3 & 1 & 5 \\ \end{bmatrix}$

= $\begin{bmatrix} 1 & 15 & 3 \\ 6 & 2 & 10 \\ \end{bmatrix}$

4. B^T = $\begin{bmatrix} 4 & 0 \\ -1 & 2 \end{bmatrix}$

B^TB = $\begin{bmatrix} 4 & 0 \\ -1 & 2 \end{bmatrix}$ $\begin{bmatrix} 4 & -1 \\ 0 & 2 \end{bmatrix}$

= $\begin{bmatrix} 16 & -4 \\ -4 & 5 \end{bmatrix}$

(B^TB)C = $\begin{bmatrix} 16 & -4 \\ -4 & 5 \end{bmatrix}$ $\begin{bmatrix} 1 & 4 & 2 \\ 3 & 1 & 5 \\ \end{bmatrix}$

= $\begin{bmatrix} 4 & 60 & 12 \\ 11 & -11 & 17 \\ \end{bmatrix}$

5. 3B is of order 2x2 and AB is of order 3x2

So 3B - AB is not conformable for subtraction