Algebra

Matrices & Determinants

Medium

applied

Given that $[a_{ij}]$ is a square matrix of order n.

Match the LIST-I with LIST-II

LIST-I LIST-II
A. $A=[a_{ij}]; a_{ij}=\begin{cases}1, & \forall i=j\\0, & \forall i\neq j\end{cases}$ I. Identity Matrix
B. $B=[a_{ij}]; a_{ij}=\begin{cases}2i-3j, & \forall i=j\\0, & \forall i\neq j\end{cases}$ II. Diagonal Matrix
C. $C=[a_{ij}]; a_{ij}=\begin{cases}k, & \forall i=j\\0, & \forall i\neq j\end{cases}$ III. Scalar Matrix
D. $D=[a_{ij}]; a_{ij}=\begin{cases}2i-3j, & \forall i\leq j\\0, & \forall i>j\end{cases}$ IV. Upper Triangular Matrix

Choose the correct answer from the options given below:

LIST-I		LIST-II
A.	$A=[a_{ij}]; a_{ij}=\begin{cases}1, & \forall i=j\\0, & \forall i\neq j\end{cases}$	I. Identity Matrix
B.	$B=[a_{ij}]; a_{ij}=\begin{cases}2i-3j, & \forall i=j\\0, & \forall i\neq j\end{cases}$	II. Diagonal Matrix
C.	$C=[a_{ij}]; a_{ij}=\begin{cases}k, & \forall i=j\\0, & \forall i\neq j\end{cases}$	III. Scalar Matrix
D.	$D=[a_{ij}]; a_{ij}=\begin{cases}2i-3j, & \forall i\leq j\\0, & \forall i>j\end{cases}$	IV. Upper Triangular Matrix

A - I, B - II, C - III, D - IV

A - I, B - IV, C - II, D - III

A - I, B - II, C - IV, D - III

A - II, B - III, C - IV, D - I

✅ Correct Option: 1

A has ones on the diagonal and zeros elsewhere, so it is the identity (I). B has diagonal entries $2i-3i=-i$ , which differ, so it is diagonal but not scalar (II). C has a common value $k$ on the diagonal, making it scalar (III). D has zeros below the diagonal, so it is upper triangular (IV).

2025: 2 June Shift 1

If $\begin{bmatrix}3 & 1\\2 & 1\end{bmatrix}A\begin{bmatrix}2 & 1\\1 & 1\end{bmatrix} = \begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$ , then matrix 'A' is

2026: 15 May Shift 1

Let P be a matrix such that $P = \begin{bmatrix} 1 & -1 \\ 0 & 3 \end{bmatrix}$ , then which of the following statements are True?

A. P is a symmetric matrix

B. P is not a skew-symmetric matrix

C. The determinant of P is non-zero

D. P is an invertible matrix

Choose the correct answer from the options given below:

2026: 31 May Shift 2

If A is a square matrix of order $n$ and $k$ is a scalar such that $|kA| = \alpha|A|$ then the value of $\alpha$ is:

2026: 31 May Shift 2

Match the LIST-I with LIST-II

	LIST-I Matrix A		LIST-II $\vert adj\,A \vert$
A.	$\begin{bmatrix}1 & 2 & 3\\ 0 & 2 & 5\\ 0 & 0 & 1\end{bmatrix}$	I.	$9$
B.	$\begin{bmatrix}3 & -1 & 4\\ 0 & 1 & 2\\ 0 & 0 & -1\end{bmatrix}$	II.	$16$
C.	$\begin{bmatrix}4 & 2 & -3\\ 0 & -2 & 5\\ 0 & 0 & 1\end{bmatrix}$	III.	$64$
D.	$\begin{bmatrix}2 & 5 & 3\\ 0 & 2 & 1\\ 0 & 0 & 1\end{bmatrix}$	IV.	$4$

Choose the correct answer from the options given below:

2022: 16 July Shift 2

If P and Q are symmetric matrices of same order, then $(PQ - QP)$ is

2025: 16 May Shift 1

Match List-I with List-II

Let $A$ be any invertible square matrix. Then

List-I	List-II
(A) $A - A^T$	(I) $\vert A\vert A^{-1}$
(B) $A A^T$	(II) Skew-symmetric
(C) $\det (A^{-1})$	(III) Symmetric
(D) $\text{adj} A$	(IV) $[\det(A)]^{-1}$

Choose the correct answer from the options given below:

Solution

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