Question

If \[A\] and \[B\] are symmetric matrices, then \[ABA\] is
(a) Symmetric
(b) Skew-symmetric
(c) Diagonal
(d) Triangular

Answer 1

Hint:
Here, we need to find the type of matrix \[ABA\]. We will use the given information to find the transpose of the matrix \[ABA\], and using that, we will determine the type of matrix \[ABA\].

Formula Used:
We will use the formula of the transpose of two matrices \[X\] and \[Y\] can be written as \[{\left( {XY} \right)^T} = {Y^T}{X^T}\].

Complete step by step solution:
A symmetric matrix is a matrix whose transpose is equal to the matrix.
The transpose of a matrix \[A\] is denoted by \[{A^T}\].
Thus, if \[A\] is a symmetric matrix, then \[A = {A^T}\].
It is given that \[A\] and \[B\] are symmetric matrices.
Therefore, we get
\[A = {A^T}\] and \[B = {B^T}\]
Now, we will find the transpose of the matrix \[ABA\].
The transpose of the matrix \[ABA\] is given by \[{\left( {ABA} \right)^T}\].
The transpose of two matrices \[X\] and \[Y\] can be written as \[{\left( {XY} \right)^T} = {Y^T}{X^T}\].
Therefore, we get
\[ \Rightarrow {\left( {ABA} \right)^T} = {A^T}{B^T}{A^T}\]
Substituting \[{A^T} = A\] and \[{B^T} = B\] in the equation, we get
\[ \Rightarrow {\left( {ABA} \right)^T} = ABA\]
We can observe that the transpose of the matrix \[ABA\] is equal to the matrix \[ABA\].
Therefore, the matrix \[ABA\] is a symmetric matrix.

Thus, the correct option is option (a).

Note:
We can verify our solution by taking any two symmetric matrices.
Let \[A = \left[ {\begin{array}{*{20}{l}}2&0&1\\0&3&2\\1&2&4\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{l}}1&2&1\\2&2&3\\1&3&5\end{array}} \right]\].
We will check whether \[{\left( {ABA} \right)^T} = ABA\].
Multiplying the matrices A and B, we get
\[\begin{array}{l} \Rightarrow AB = \left[ {\begin{array}{*{20}{l}}2&0&1\\0&3&2\\1&2&4\end{array}} \right]\left[ {\begin{array}{*{20}{l}}1&2&1\\2&2&3\\1&3&5\end{array}} \right]\\ \Rightarrow AB = \left[ {\begin{array}{*{20}{l}}{2 \times 1 + 0 \times 2 + 1 \times 1}&{2 \times 2 + 0 \times 2 + 1 \times 3}&{2 \times 1 + 0 \times 3 + 1 \times 5}\\{0 \times 1 + 3 \times 2 + 2 \times 1}&{0 \times 2 + 3 \times 2 + 2 \times 3}&{0 \times 1 + 3 \times 3 + 2 \times 5}\\{1 \times 1 + 2 \times 2 + 4 \times 1}&{1 \times 2 + 2 \times 2 + 4 \times 3}&{1 \times 1 + 2 \times 3 + 4 \times 5}\end{array}} \right]\end{array}\]
Multiplying the terms in the matrix, we get
\[ \Rightarrow AB = \left[ {\begin{array}{*{20}{l}}{2 + 0 + 1}&{4 + 0 + 3}&{2 + 0 + 5}\\{0 + 6 + 2}&{0 + 6 + 6}&{0 + 9 + 10}\\{1 + 4 + 4}&{2 + 4 + 12}&{1 + 6 + 20}\end{array}} \right]\]
Adding the terms in the matrix, we get
\[ \Rightarrow AB = \left[ {\begin{array}{*{20}{l}}3&7&7\\8&{12}&{19}\\9&{18}&{27}\end{array}} \right]\]
Now, we will multiply the matrices AB and A.
Therefore, we get
\[\begin{array}{l} \Rightarrow ABA = \left[ {\begin{array}{*{20}{l}}3&7&7\\8&{12}&{19}\\9&{18}&{27}\end{array}} \right]\left[ {\begin{array}{*{20}{l}}2&0&1\\0&3&2\\1&2&4\end{array}} \right]\\ \Rightarrow ABA = \left[ {\begin{array}{*{20}{l}}{3 \times 2 + 7 \times 0 + 7 \times 1}&{3 \times 0 + 7 \times 3 + 7 \times 2}&{3 \times 1 + 7 \times 2 + 7 \times 4}\\{8 \times 2 + 12 \times 0 + 19 \times 1}&{8 \times 0 + 12 \times 3 + 19 \times 2}&{8 \times 1 + 12 \times 2 + 19 \times 4}\\{9 \times 2 + 18 \times 0 + 27 \times 1}&{9 \times 0 + 18 \times 3 + 27 \times 2}&{9 \times 1 + 18 \times 2 + 27 \times 4}\end{array}} \right]\end{array}\]
Multiplying the terms in the matrix, we get
\[ \Rightarrow ABA = \left[ {\begin{array}{*{20}{l}}{6 + 0 + 7}&{0 + 21 + 14}&{3 + 14 + 28}\\{16 + 0 + 19}&{0 + 36 + 38}&{8 + 24 + 76}\\{18 + 0 + 27}&{0 + 54 + 54}&{9 + 36 + 108}\end{array}} \right]\]
Adding the terms in the matrix, we get
\[ \Rightarrow ABA = \left[ {\begin{array}{*{20}{l}}{13}&{35}&{45}\\{35}&{74}&{108}\\{45}&{108}&{153}\end{array}} \right]\]
Now, we will find the transpose of the matrix ABA.
The transpose of a matrix is formed by interchanging the rows and columns of the matrix.
Therefore, we get
\[ \Rightarrow {\left( {ABA} \right)^T} = \left[ {\begin{array}{*{20}{l}}{13}&{35}&{45}\\{35}&{74}&{108}\\{45}&{108}&{153}\end{array}} \right]\]
We can observe that the transpose of the matrix \[ABA\] is equal to the matrix \[ABA\].
Hence, we have verified that the matrix \[ABA\] is a symmetric matrix.