Let A be any square matrix of order 3 and B = …. Then the matrix ABA^T is a

Given: $B = \begin{bmatrix} 0 & -4 & 2 \\ 4 & 0 & 3 \\ -2 & -3 & 0 \end{bmatrix}$B=​04−2​−40−3​230​​

The transpose of $B$B is:

$$B^T = \begin{bmatrix} 0 & 4 & -2 \\ -4 & 0 & -3 \\ 2 & 3 & 0 \end{bmatrix}$$BT=​0−42​403​−2−30​​

The negative of $B$B is:

$$-B = \begin{bmatrix} 0 & 4 & -2 \\ -4 & 0 & -3 \\ 2 & 3 & 0 \end{bmatrix}$$−B=​0−42​403​−2−30​​

Since $B^T = -B$BT=−B, the matrix $B$B is skew-symmetric.

---

To determine the type of matrix $ABA^T$ABAT, we find its transpose.

Using the property $(XYZ)^T = Z^T Y^T X^T$(XYZ)T=ZTYTXT:

$$(ABA^T)^T = (A^T)^T B^T A^T$$(ABAT)T=(AT)TBTAT

$$(ABA^T)^T = AB^T A^T$$(ABAT)T=ABTAT

---

Since $B$B is skew-symmetric, $B^T = -B$BT=−B.

Substituting:

$$(ABA^T)^T = A(-B)A^T$$(ABAT)T=A(−B)AT

$$(ABA^T)^T = -ABA^T$$(ABAT)T=−ABAT

This satisfies the condition for a skew-symmetric matrix.

---

Therefore, $ABA^T$ABAT is a skew-symmetric matrix.