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Class 12 RD Sharma Solutions- Chapter 5 Algebra of Matrices – Exercise 5.5
  • Last Updated : 13 Jan, 2021
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Question 1: If\ A=\begin{bmatrix}2&3\\4&5\end{bmatrix},\ prove\ that\ A-A^T\ is\ a\ skew-symmetric\ matrix.

Solution:

Given:

 If\ A=\begin{bmatrix}2&3\\4&5\end{bmatrix},\ that\ A-A^T\ is\ a\ skew-symmetric\ matrix.

Consider,

(A-A^T)=\left(\begin{bmatrix}2&3\\4&5\end{bmatrix}-\begin{bmatrix}2&3\\4&5\end{bmatrix}^T\right)\\ =\left(\begin{bmatrix}2&3\\4&5\end{bmatrix}-\begin{bmatrix}2&3\\4&5\end{bmatrix}\right)\\ =\begin{bmatrix}2-2&3-4\\4-3&5-5\end{bmatrix}\\ (A-A^T)=\begin{bmatrix}0&-1\\1&0\end{bmatrix}---(equation\ 1)\\ -(A-A^T)^T=-\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}^T\\ =-\begin{bmatrix}0 & 1\\-1 & 0\end{bmatrix}\\ -(A-A^T)=\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}---(equation\ 2)



From equation (1) and (2) it can be seen that,

A skew-symmetric matrix is a square matrix whose transpose equal to its negative, that is,

X = −XT

So, A − AT is a skew-symmetric.

Question 2: If \ A=\begin{bmatrix}3 & -4\\1 & -1\end{bmatrix},\ show\ that\ A-A^T\ is\ a\ skew-symmetric\ matrix.

Solution:

Given:

If \ A=\begin{bmatrix}3 & -4\\1 & -1\end{bmatrix},\ that\ A-A^T\ is\ a\ skew-symmetric\ matrix.

Consider,



(A-A^T)=\begin{bmatrix}0 & -5\\5 & 0\end{bmatrix}---(equation\ 1)\\ -(A-A^T)^T=-\begin{bmatrix}0 & -5\\5 & 0\end{bmatrix}^T\\ =-\begin{bmatrix}0 & 5\\-5 & 0\end{bmatrix}\\ -(A-A^T)=\begin{bmatrix}0 & -5\\5 & 0\end{bmatrix}---(equation\ 2)

From equation (1) and (2) it can be seen,

A skew-symmetric matrix is a square matrix whose transpose equals its negative, that is,

X = −XT

Thus, A − AT is a skew-symmetric matrix.

Question 3: If\ the\ matrix\ A=\begin{bmatrix}5 & 2 & x\\y & z & -3\\4 & t & -7\end{bmatrix},\ is\ a\ symmetric\ matrix\ find\ x,\ y,\ z\ and\ t

Solution:

Given:

A=\begin{bmatrix}5 & 2 & x\\y & z & -3\\4 & t & -7\end{bmatrix}is\ a\ symmetric\ matix.

As we know that A = [aij]m×n is a symmetric matrix if aij = aji

Thus,

x = a13 = a31 = 4

y = a21 = a12 = 2

z = a22 = a22 = z

t = a32 = a23 = −3

Hence, x = 4, y = 2, t = −3 and z can have any value.

Questiion 4: Let\ A=\begin{bmatrix}3 & 2 & 7\\1 & 4 & 3\\-2 & 5 & 8\end{bmatrix},\ Find\ matrices\ X\ and\ Y\ such\ that\ X+Y=A,\ where\ X\ is\ a\ symmetric\ and\ y\ is\ a\ skew-symmetric\ matrix.

Solution:

Given:

A=\begin{bmatrix}3 & 2 & 7\\1 & 4 & 3\\-2 & 5 & 8\end{bmatrix}then\\ A=\begin{bmatrix}3 & 1 & -2\\2 & 4 & 5\\7 & 3 & 8\end{bmatrix}\\ X=\frac{1}{2}(A+A^T)\\ =\frac{1}{2}\left(\begin{bmatrix}3 & 2 & 7\\1 & 4 & 3\\-2 & 5 & 8\end{bmatrix}+\begin{bmatrix}3 & 1 & -2\\2 & 4 & 5\\7 & 3 & 8\end{bmatrix}\right)\\ =\frac{1}{2}\begin{bmatrix}3+3 & 2+1 & 7-2\\1+2 & 4+4 & 3+5\\-2+7 & 5+3 & 8+8\end{bmatrix}\\ =\frac{1}{2}\begin{bmatrix}6 & 3 & 5\\3 & 8 & 8\\5 & 8 & 16\end{bmatrix}\\ X=\begin{bmatrix}3 & \frac{3}{2} & \frac{5}{2}\\\frac{3}{2} & 4 & 4\\\frac{5}{2} & 4 & 8\end{bmatrix}\\ Now,\\ Y=\frac{1}{2}(A-A^T)\\ =\frac{1}{2}\left(\begin{bmatrix}3 & 2 & 7\\1 & 4 & 3\\-2 & 5 & 8\end{bmatrix}-\begin{bmatrix}3 & 1 & -2\\2 & 4 & 5\\7 & 3 & 8\end{bmatrix}\right)\\ =\frac{1}{2}\begin{bmatrix}3-3 & 2-1 & 7+2\\1-2 & 4-4 & 3-5\\-2-7 & 5-3 & 8-8\end{bmatrix}\\ =\frac{1}{2}\begin{bmatrix}0 & 1 & 9\\-1 & 0 & -2\\-9 & 2 & 0\end{bmatrix}\\ Y=\begin{bmatrix}0 & \frac{1}{2} & \frac{9}{2}\\-\frac{1}{2} & 0 & -1\\-\frac{9}{2} & 1 & 0\end{bmatrix}\\ Now,\\ X^T=\begin{bmatrix}3 & \frac{3}{2} & \frac{5}{2}\\\frac{3}{2} & 4 & 4\\\frac{5}{2} & 4 & 8\end{bmatrix}^T =\begin{bmatrix}3 & \frac{3}{2} & \frac{5}{2}\\\frac{3}{2} & 4 & 4\\\frac{5}{2} & 4 & 8\end{bmatrix}=X\\ X\ is\ a\ symmetric\ martix\\ Now,\\ -Y^T=-\begin{bmatrix}0 & \frac{1}{2} & \frac{9}{2}\\-\frac{1}{2} & 0 & -1\\-\frac{9}{2} & 1 & 0\end{bmatrix}^T=-\begin{bmatrix}0 & -\frac{1}{2} & -\frac{9}{2}\\\frac{1}{2} & 0 & 1\\\frac{9}{2} & -1 & 0\end{bmatrix}\\ -Y^T=\begin{bmatrix}0 & \frac{1}{2} & \frac{9}{2}\\-\frac{1}{2} & 0 & -1\\-\frac{9}{2} & 1 & 0\end{bmatrix}\\ -Y^T=Y\\ Y\ is\ a\ skew\ symmetric\ matrix.\\ X+Y=\begin{bmatrix}3 & \frac{3}{2} & \frac{5}{2}\\\frac{3}{2} & 4 & 4\\\frac{5}{2} & 4 & 8\end{bmatrix}+\begin{bmatrix}0 & \frac{1}{2} & \frac{9}{2}\\-\frac{1}{2} & 0 & -1\\-\frac{9}{2} & 1 & 0\end{bmatrix}\\ =\begin{bmatrix}3+0 & \frac{3}{2}+\frac{1}{2} & \frac{5}{2}+\frac{9}{2}\\\frac{3}{2}-\frac{1}{2} & 4+0 & 4-1\\\frac{5}{2}-\frac{9}{2} & 4+1 & 8+0\end{bmatrix}\\ =\begin{bmatrix}3 & 2 & 7\\1 & 4 & 3\\-2 & 5 & 8\end{bmatrix}=A\\ Hence, X+Y = A

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