Algebra

Matrices & Determinants

Medium

common

Let A be any square matrix of order 3 and $B = \begin{bmatrix} 0 & -4 & 2 \\ 4 & 0 & 3 \\ -2 & -3 & 0 \end{bmatrix}$ . Then the matrix $ABA^T$ is a

Symmetric matrix

Skew symmetric matrix

Null matrix

Diagonal matrix

✅ Correct Option: 2

Given: $B = \begin{bmatrix} 0 & -4 & 2 \\ 4 & 0 & 3 \\ -2 & -3 & 0 \end{bmatrix}$

The transpose of $B$ is:

$B^T = \begin{bmatrix} 0 & 4 & -2 \\ -4 & 0 & -3 \\ 2 & 3 & 0 \end{bmatrix}$

The negative of $B$ is:

$-B = \begin{bmatrix} 0 & 4 & -2 \\ -4 & 0 & -3 \\ 2 & 3 & 0 \end{bmatrix}$

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

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

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

$(ABA^T)^T = (A^T)^T B^T A^T$

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

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

Substituting:

$(ABA^T)^T = A(-B)A^T$

$(ABA^T)^T = -ABA^T$

This satisfies the condition for a skew-symmetric matrix.

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

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Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	$\vert adj\,A \vert$	I.	$\dfrac{1}{\vert A \vert}(adj\,A)$
B.	$(adj\,A)A$	II.	$\vert A \vert^2$
C.	$A^{-1}$	III.	$\dfrac{1}{\vert A \vert}$
D.	$\vert A^{-1} \vert$	IV.	$\vert A \vert I$

Choose the correct answer from the options given below:

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Let A be any square matrix of order 3 and B=[0−42403−2−30]B = \begin{bmatrix} 0 & -4 & 2 \\ 4 & 0 & 3 \\ -2 & -3 & 0 \end{bmatrix}B=​04−2​−40−3​230​​. Then the matrix ABATABA^TABAT is a

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

Let A be any square matrix of order 3 and $B = \begin{bmatrix} 0 & -4 & 2 \\ 4 & 0 & 3 \\ -2 & -3 & 0 \end{bmatrix}$ . Then the matrix $ABA^T$ is a