Question

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

Answer 1

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.