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Chapter 10: Problem 214

If \(A\) and \(B\) be symmetric matrices of the same order, then show that i. \(A+B\) is a symmetric matrix ii. \(A B-B A\) is a skew-symmetric matrix iii. \(A B+B A\) is a symmetric matrix

Short Answer

Expert verified
The sum \(A+B\) and the sum \(AB+BA\) of two symmetric matrices A and B are symmetric matrices, and the difference \(AB-BA\) is a skew-symmetric matrix.

Step by step solution

01

Proving A+B is symmetric

To prove that \(A+B\) is symmetric, we need to check whether \( (A+B)^T = A+B\). According to the properties of matrix transposition, \( (A+B)^T = A^T + B^T \). Since A and B are symmetric matrices, \( A^T=A\) and \( B^T=B \), thus \( (A+B)^T = A + B\). Therefore, \(A+B\) is a symmetric matrix.
02

Proving AB-BA is skew-symmetric

To prove that \(AB-BA\) is skew-symmetric, we need to check if \((AB - BA)^T = -(AB - BA)\). According to the properties of matrix transposition, \( (AB - BA)^T = B^TA^T - A^TB^T \). But, since A and B are symmetric matrices, \( A^T = A\) and \( B^T = B \), so \( (AB - BA)^T = BA - AB \). Thus, \( (AB - BA)^T = -(AB - BA) \), which makes \( AB - BA\) a skew-symmetric matrix.
03

Proving AB+BA is symmetric

To prove that \(AB + BA\) is symmetric, we need to check if \((AB + BA)^T = AB + BA\). According to the properties of matrix transposition, \( (AB + BA)^T = B^TA^T + A^TB^T \). But, since A and B are symmetric matrices, \( A^T = A\) and \( B^T = B \), so \( (AB + BA)^T = BA + AB \). However, in terms of addition of matrices, \( AB + BA = BA + AB \), Thus, \((AB + BA)^T = AB + BA \), which makes \( AB + BA\) a symmetric matrix.

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Most popular questions from this chapter

EVALUATING DETERMINANTS. $$ \left|\begin{array}{ccc} 1 & \log _{x} y & \log _{x} z \\ \log _{y} x & 1 & \log _{y} z \\ \log _{z} x & \log _{z} y & 1 \end{array}\right| $$

PROVING IDENTITIES BY DETERMINANTS. $$ \left|\begin{array}{ccc} 1 & 1 & 1 \\ b+c & c+a & a+b \\ b^{2}+c^{2} & c^{2}+a^{2} & a^{2}+b^{2} \end{array}\right|=(b-c)(c-a)(a-b) $$

EQUATIONS CONTAINING DETERMINANTS. $$ \left|\begin{array}{ccc} x+a & b & c \\ a & x+b & c \\ a & b & x+c \end{array}\right|=0 $$

PROVING IDENTITIES BY DETERMINANTS. $$ \left|\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right|=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right| $$

PROVING IDENTITIES BY DETERMINANTS. $$ \left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|=\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{array}\right|=(a-b)(b-c)(c-a) $$
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