Answer to Question #133006 in Linear Algebra for Galata

Question

(1) Let;

• A=the first row [-1 -1] second row [3 3]. Find all 2×2 materices, B such that AB=0

(2) Distinguish whether the given matrix is symmetric or not

• A=the first row [2 1 3] second row [1 5 -3] third row [3 -3 7]
• B=the first row[1 1 3] second row[1 2 2] third row[3 2 3]

Answer 1

(1) Given $A=\begin{bmatrix} -1 & -1 \\ 3 & 3 \end{bmatrix}.$ Let $B=\begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}.$

Then

$$AB=\begin{bmatrix} -1 & -1 \\ 3 & 3 \end{bmatrix}\begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}=$$

$$=\begin{bmatrix} -b_{11}-b_{21} & -b_{12}-b_{22} \\ 3b_{11}+3b_{21} & 3b_{12}+3b_{22} \end{bmatrix}$$

If $AB=0,$ then

$$\begin{alignedat}{2} -b_{11}-b_{21}=0 \\ -b_{12}-b_{22}=0 \\ 3b_{11}+3b_{21}=0 \\ 3b_{12}+3b_{22}=0 \end{alignedat}$$

$$\begin{alignedat}{2} b_{21}=-b_{11} \\ b_{12}=-b_{22} \end{alignedat}$$

$$B=\begin{bmatrix} c & -d \\ -c & d \end{bmatrix}, c,d\in \R$$

(2)

$A=\begin{bmatrix} 2 & 1 & 3 \\ 1 & 5 & -3 \\ 3 & -3 & 7 \end{bmatrix}=>A^T=\begin{bmatrix} 2 & 1 & 3 \\ 1 & 5 & -3 \\ 3 & -3 & 7 \end{bmatrix}=A$

Then the matrix $A$ is symmetric.

$B=\begin{bmatrix} 1 & 1 & 3 \\ 1 & 2 & 2 \\ 3 & 2 & 3 \end{bmatrix}=>B^T=\begin{bmatrix} 1 & 1 & 3 \\ 1 & 2 & 2\\ 3 & 2 & 3 \end{bmatrix}=B$

Then the matrix $B$ is symmetric.