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Mathematics Question on Matrices

Express the following matrices as the sum of a symmetric and a skew-symmetric matrix:
(i)[35\1−1](i)\begin{bmatrix}3&5\\\1&-1\end{bmatrix}
(ii)[6−22 −23−1 2−13](ii)\begin{bmatrix}6&-2&2\\\ -2&3&-1\\\ 2&-1&3\end{bmatrix}
(iii)[33−1 −2−21 −4−52](iii)\begin{bmatrix}3&3&-1\\\ -2&-2&1\\\ -4&-5&2\end{bmatrix}
(iv)[15−12](iv)\begin{bmatrix}1&5\\\\-1&2\end{bmatrix}

Answer

(i)Let A=[35\1−1]A=\begin{bmatrix}3&5\\\1&-1\end{bmatrix},then A′=[31\5−1]A'=\begin{bmatrix}3&1\\\5&-1\end{bmatrix}
NowA+A'=A=\begin{bmatrix}3&5\\\1&-1\end{bmatrix}+\begin{bmatrix}3&1\\\5&-1\end{bmatrix}$$=\begin{bmatrix}6&6\\\6&-2\end{bmatrix}
LetP=12(A+A′)=12[66\6−2]P=\frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix}6&6\\\6&-2\end{bmatrix}
=[33\3−1]=\begin{bmatrix}3&3\\\3&-1\end{bmatrix}
Now P′=[33\3−1]=PP'=\begin{bmatrix}3&3\\\3&-1\end{bmatrix}=P
Thus,P=12(A+A′),P=\frac{1}{2}(A+A')is a symmetric matrix.
Now A−A′=[35\1−1]−[31\5−1]A-A'=\begin{bmatrix}3&5\\\1&-1\end{bmatrix}-\begin{bmatrix}3&1\\\5&-1\end{bmatrix} =[04−40]=\begin{bmatrix}0&4\\\\-4&0\end{bmatrix}
Let Q=12(A−A′)=12[04−40]Q=\frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix}0&4\\\\-4&0\end{bmatrix}
=[02−20]=\begin{bmatrix}0&2\\\\-2&0\end{bmatrix}
Now Q′=[02−20]=−QQ'=\begin{bmatrix}0&2\\\\-2&0\end{bmatrix}=-Q
Thus Q=12(A−A′)Q=\frac{1}{2}(A-A')is a skew-symmetric matrix.
Representing A as the sum of P and Q:
P+Q=[33\3−1]+[02−20]=[35\1−1]=AP+Q=\begin{bmatrix}3&3\\\3&-1\end{bmatrix}+\begin{bmatrix}0&2\\\\-2&0\end{bmatrix}=\begin{bmatrix}3&5\\\1&-1\end{bmatrix}=A

(ii)Let A=[6−22 −23−1 2−13]A=\begin{bmatrix}6&-2&2\\\ -2&3&-1\\\ 2&-1&3\end{bmatrix}then A′=[6−22 −23−1 2−13]A'=\begin{bmatrix}6&-2&2\\\ -2&3&-1\\\ 2&-1&3\end{bmatrix}
Now A+A′=[6−22 −23−1 2−13]+[6−22 −23−1 2−13]A+A'=\begin{bmatrix}6&-2&2\\\ -2&3&-1\\\ 2&-1&3\end{bmatrix}+\begin{bmatrix}6&-2&2\\\ -2&3&-1\\\ 2&-1&3\end{bmatrix}
=[12−44 −46−2 4−26]=\begin{bmatrix}12&-4&4\\\ -4&6&-2\\\ 4&-2&6\end{bmatrix}
let P=12(A+A′)=12[12−44 −46−2 4−26]P=\frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix}12&-4&4\\\ -4&6&-2\\\ 4&-2&6\end{bmatrix}
=[6−22 −23−1 2−13]=\begin{bmatrix}6&-2&2\\\ -2&3&-1\\\ 2&-1&3\end{bmatrix}
Now P′=[6−22 −23−1 2−13]=PP'=\begin{bmatrix}6&-2&2\\\ -2&3&-1\\\ 2&-1&3\end{bmatrix}=P
Thus P=12(A+A′)P=\frac{1}{2}(A+A') is a symmetric matrix.
Now A−A′=[6−22 −23−1 2−13]−[6−22 −23−1 2−13]A-A'=\begin{bmatrix}6&-2&2\\\ -2&3&-1\\\ 2&-1&3\end{bmatrix}-\begin{bmatrix}6&-2&2\\\ -2&3&-1\\\ 2&-1&3\end{bmatrix}
=[000 000 000]=\begin{bmatrix}0&0&0\\\ 0&0&0\\\ 0&0&0\end{bmatrix}
LetQ=12(A−A′)=[000 000 000]Q=\frac{1}{2}(A-A')=\begin{bmatrix}0&0&0\\\ 0&0&0\\\ 0&0&0\end{bmatrix}
Now Q′==[000 000 000]=QQ'==\begin{bmatrix}0&0&0\\\ 0&0&0\\\ 0&0&0\end{bmatrix}=Q
ThusQ=12(A−A′)Q=\frac{1}{2}(A-A') is a skew-symmetric matrix.
Representing A as the sum of P and Q:
P+Q=[6−22 −23−1 2−13]+[000 000 000]P+Q=\begin{bmatrix}6&-2&2\\\ -2&3&-1\\\ 2&-1&3\end{bmatrix}+\begin{bmatrix}0&0&0\\\ 0&0&0\\\ 0&0&0\end{bmatrix}
=[6−22 −23−1 2−13]=A=\begin{bmatrix}6&-2&2\\\ -2&3&-1\\\ 2&-1&3\end{bmatrix}=A

(iii)Let A=[33−1 −2−21 −4−52]A=\begin{bmatrix}3&3&-1\\\ -2&-2&1\\\ -4&-5&2\end{bmatrix},then A′=[3−2−4 3−2−5 −112]A'=\begin{bmatrix}3&-2&-4\\\ 3&-2&-5\\\ -1&1&2\end{bmatrix}
Now A+A′=[33−1 −2−21 −4−52]+[3−2−4 3−2−5 −112]A+A'=\begin{bmatrix}3&3&-1\\\ -2&-2&1\\\ -4&-5&2\end{bmatrix}+\begin{bmatrix}3&-2&-4\\\ 3&-2&-5\\\ -1&1&2\end{bmatrix}
=[61−5 1−4−4 −5−44]=\begin{bmatrix}6&1&-5\\\ 1&-4&-4\\\ -5&-4&4\end{bmatrix}
Let P=12(A+A′)=12=[61−5 1−4−4 −5−44]P=\frac{1}{2}(A+A')=\frac{1}{2}=\begin{bmatrix}6&1&-5\\\ 1&-4&-4\\\ -5&-4&4\end{bmatrix}
=[312−52 12−2−2 −52−22]=\begin{bmatrix}3&\frac{1}{2}&\frac{-5}{2}\\\ \frac{1}{2}& -2&-2\\\ \frac{-5}{2}&-2&2\end{bmatrix}
Now P′==[312−52 12−2−2 −52−22]P'==\begin{bmatrix}3&\frac{1}{2}&\frac{-5}{2}\\\ \frac{1}{2}& -2&-2\\\ \frac{-5}{2}&-2&2\end{bmatrix}
thus P=12(A+A′)P=\frac{1}{2}( A+A') is a symmetric matrix.

Now A−A′=[33−1 −2−21 −4−52]−[3−2−4 3−2−5 −112]A-A'=\begin{bmatrix}3&3&-1\\\ -2&-2&1\\\ -4&-5&2\end{bmatrix}-\begin{bmatrix}3&-2&-4\\\ 3&-2&-5\\\ -1&1&2\end{bmatrix}
=[053 −506 −3−60]=\begin{bmatrix}0&5&3\\\ -5&0&6\\\ -3&-6&0\end{bmatrix}
Let Q=12(A−A′)=12=[053 −506 −3−60]Q=\frac{1}{2}(A-A')=\frac{1}{2}=\begin{bmatrix}0&5&3\\\ -5&0&6\\\ -3&-6&0\end{bmatrix}
=[05232 −5203 −32−30]=\begin{bmatrix}0&\frac{5}{2}&\frac{3}{2}\\\ \frac{-5}{2}&0&3\\\ \frac{-3}{2}&-3&0\end{bmatrix}
Now Q′=[05232 −5203 −32−30]=−QQ'=\begin{bmatrix}0&\frac{5}{2}&\frac{3}{2}\\\ \frac{-5}{2}&0&3\\\ \frac{-3}{2}&-3&0\end{bmatrix}=-Q
Thus,Q=12(A−A′)Q=\frac{1}{2}( A-A')is a skew-symmetric matrix.
Representing A as the sum of P and Q:
P+Q=[312−52 12−2−2 −52−22]+[05232 −5203 −32−30]P+Q=\begin{bmatrix}3&\frac{1}{2}&\frac{-5}{2}\\\ \frac{1}{2}& -2&-2\\\ \frac{-5}{2}&-2&2\end{bmatrix}+\begin{bmatrix}0&\frac{5}{2}&\frac{3}{2}\\\ \frac{-5}{2}&0&3\\\ \frac{-3}{2}&-3&0\end{bmatrix}
=[33−1 −2−21 −4−52]=A=\begin{bmatrix}3&3&-1\\\ -2&-2&1\\\ -4&-5&2\end{bmatrix}=A

(iv)Let A=[15−12]A=\begin{bmatrix}1&5\\\\-1&2\end{bmatrix},Then A′=[1−1\52]A'=\begin{bmatrix}1&-1\\\5&2\end{bmatrix}
Now A+A′=A=[15−12]+[1−1\52]A+A'=A=\begin{bmatrix}1&5\\\\-1&2\end{bmatrix}+\begin{bmatrix}1&-1\\\5&2\end{bmatrix}
=[24\44]=\begin{bmatrix}2&4\\\4&4\end{bmatrix}
Let P=12(A+A′)=[12\22]P=\frac{1}{2}( A+A')=\begin{bmatrix}1&2\\\2&2\end{bmatrix}
Now P′=[12\22]=PP'=\begin{bmatrix}1&2\\\2&2\end{bmatrix}=P
Thus,P=12(A+A′)P=\frac{1}{2}( A+A') is a symmetric matrix.
Now A−A′=[15−12]−[1−1\52]A-A'=\begin{bmatrix}1&5\\\\-1&2\end{bmatrix}-\begin{bmatrix}1&-1\\\5&2\end{bmatrix}
=[06−60]=\begin{bmatrix}0&6\\\\-6&0\end{bmatrix}
Let Q=12(A−A′)=[03−30]Q=\frac{1}{2}( A-A')=\begin{bmatrix}0&3\\\\-3&0\end{bmatrix}
Now Q′=[03−30]=−QQ'=\begin{bmatrix}0&3\\\\-3&0\end{bmatrix}=-Q
Thus,Q=12(A−A′)Q=\frac{1}{2}( A-A') is a skew-symmetric matrix.
Representing A as the sum of P and Q:
P+Q=[12\22]+[03−30]P+Q=\begin{bmatrix}1&2\\\2&2\end{bmatrix}+\begin{bmatrix}0&3\\\\-3&0\end{bmatrix}
=[15−12]=A=\begin{bmatrix}1&5\\\\-1&2\end{bmatrix}=A