Math, asked by anuishere, 4 hours ago

express the following matrix as the sum of symmetric and skew symmetric
|2 3 - 1 |
|5 - 2 1 |
|-4 - 5 3|​

Answers

Answered by subashadvocate1974
0

Answer:

Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

(i) [

3

1

5

−1

]

(ii)

6

−2

2

−2

3

−1

2

−1

3

(iii)

3

−2

−4

3

−2

−5

−1

1

2

(iv) [

1

−1

5

2

]

Medium

Answered by mathdude500
1

\large\underline{\sf{Solution-}}

Given matrix is

\rm :\longmapsto\:A =  \begin{gathered}\sf \left[\begin{array}{ccc}2&3& - 1\\5& - 2&1\\ - 4&-5&3\end{array}\right]\end{gathered}

We know,

\rm :\longmapsto\:A = \dfrac{1}{2}(2A)

\rm \:  =  \:  \:\dfrac{1}{2}(A + A)

\rm \:  =  \:  \:\dfrac{1}{2}(A + A +  {A}^{T}  -  {A}^{T} )

\rm \:  =  \:  \:\dfrac{1}{2}\bigg((A +  {A}^{T}) + (A -  {A}^{T})\bigg)

\rm \:  =  \:  \:\dfrac{1}{2}(A +  {A}^{T}) + \dfrac{1}{2} (A -  {A}^{T})

\rm \:  =  \:  \:P + Q

\rm :\longmapsto\:where \: P = \dfrac{1}{2}(A +  {A}^{T})

and

\rm :\longmapsto\:where \: Q = \dfrac{1}{2}(A  -   {A}^{T})

We have

\rm :\longmapsto\: \begin{gathered}\sf A=\left[\begin{array}{ccc}2&3& - 1\\5& - 2&1\\ - 4&-5&3\end{array}\right]\end{gathered}

So,

\rm :\longmapsto\: {A}^{T}  =  \begin{gathered}\sf  \left[\begin{array}{ccc}2&5& - 4\\3& - 2& - 5\\ - 1&1&3\end{array}\right]\end{gathered}

Now,

Consider,

\rm :\longmapsto\:A +  {A}^{T}

\rm \:  =  \:  \:\begin{gathered}\sf  \left[\begin{array}{ccc}2&5& - 4\\3& - 2& - 5\\ - 1&1&3\end{array}\right]\end{gathered} + \begin{gathered}\sf \left[\begin{array}{ccc}2&3& - 1\\5& - 2&1\\ - 4&-5&3\end{array}\right]\end{gathered}

\rm \:  =  \:  \:\begin{gathered}\sf \left[\begin{array}{ccc}4&8& - 5\\8& - 4& - 4\\ - 5&-4&6\end{array}\right]\end{gathered}

So,

\rm :\longmapsto \: P = \dfrac{1}{2}(A +  {A}^{T})

\rm \:  =  \:  \:\begin{gathered}\sf  \left[\begin{array}{ccc}2&4& -  \dfrac{5}{2}  \\ \\4& - 2& - 2 \\ \\ -  \dfrac{5}{2} & - 2& - 3\end{array}\right]\end{gathered}

Now,

\rm :\longmapsto\: {P}^{T}   =  \:  \:\begin{gathered}\sf  \left[\begin{array}{ccc}2&4& -  \dfrac{5}{2}  \\ \\4& - 2& - 2 \\ \\ -  \dfrac{5}{2} & - 2& - 3\end{array}\right]\end{gathered} = P

\rm :\implies\:P \: is \: symmetric.

Now,

Consider

\rm :\longmapsto\:A  -   {A}^{T}

\rm \:  =  \:  \:\begin{gathered}\sf \left[\begin{array}{ccc}2&3& - 1\\5& - 2&1\\ - 4&-5&3\end{array}\right]\end{gathered} - \begin{gathered}\sf  \left[\begin{array}{ccc}2&5& - 4\\3& - 2& - 5\\ - 1&1&3\end{array}\right]\end{gathered}

\rm \:  =  \:  \:\begin{gathered}\sf \left[\begin{array}{ccc}0& - 2& 3\\ 2& 0& 6\\  - 3& - 6&0\end{array}\right]\end{gathered}

So,

\rm :\longmapsto \: Q = \dfrac{1}{2}(A  -   {A}^{T})

\rm \:  =  \:  \:\begin{gathered}\sf  \left[\begin{array}{ccc}0& - 1&  \dfrac{3}{2}  \\ \\ 1& 0& 3 \\ \\   - \dfrac{3}{2} &  - 3& 0\end{array}\right]\end{gathered}

Now,

\rm :\longmapsto\: {Q}^{T}  =  \:  \:\begin{gathered}\sf  \left[\begin{array}{ccc}0& 1& - \dfrac{3}{2}  \\ \\ - 1& 0& -  3 \\ \\  \dfrac{3}{2} &  3& 0\end{array}\right]\end{gathered}

\rm \:  =  \:  \: - \begin{gathered}\sf  \left[\begin{array}{ccc}0& - 1& \dfrac{3}{2}  \\ \\ 1& 0& 3 \\ \\   - \dfrac{3}{2} &  - 3& 0\end{array}\right]\end{gathered}

\rm \:  =  \:  \: -  \: Q

\bf\implies \:Q \: is \: skew \: symmetric.

Thus, we have

\rm :\longmapsto\:A = P + Q

\rm :\longmapsto\:\begin{gathered}\sf \left[\begin{array}{ccc}2&3& - 1\\5& - 2&1\\ - 4&-5&3\end{array}\right]\end{gathered}

\rm \: = \begin{gathered}\sf  \left[\begin{array}{ccc}2&4& -  \dfrac{5}{2}  \\ \\4& - 2& - 2 \\ \\ -  \dfrac{5}{2} & - 2& - 3\end{array}\right]\end{gathered} + \begin{gathered}\sf  \left[\begin{array}{ccc}0& - 1&  \dfrac{3}{2}  \\ \\ 1& 0& 3 \\ \\   - \dfrac{3}{2} & - 3& 0\end{array}\right]\end{gathered}

Additional Information :-

1. If A and B are symmetric matrices then

  • AB + BA is symmetric matrix.

  • AB - BA is skew - symmetric.

2. The main diagonal elements of skew - symmetric matrix are always zero.

3. The sum of all the elements of skew symmetric matrix are always zero.

4. The determinant of skew symmetric matrix of odd order is 0.

Similar questions