Algebra

Matrices & Determinants

Easy

common

If $f(x) = \begin{vmatrix} 0 & x+a & x+b \\ x-a & 0 & x+c \\ x-b & x-c & 0 \end{vmatrix}$ then the value of $f(0)$ is equal to:

$abc$

$0$

$1$

$-1$

✅ Correct Option: 2

Putting $x=0$ gives the matrix with rows $(0,a,b)$ , $(-a,0,c)$ , $(-b,-c,0)$ .

Here $a_{ij}=-a_{ji}$ for every entry, so this is a skew-symmetric matrix of order $3$ .

The determinant of a skew-symmetric matrix of odd order is always $0$ , hence $f(0)=0$ . Direct expansion confirms $0\cdot(0+c^2)-a(0+bc)+b(ac-0)=0$ .

2023: 23 May Shift 3

If $x, y$ & $z$ are non zero real numbers, the inverse of matrix $A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$ is :

2025: 13 May Shift 2

If A is an invertible matrix of order 3 and the determinant of A is 9, then the determinant of $A^{-1}$ is:

2025: 16 May Shift 1

If a matrix has 12 elements, then the possible orders it can have, are

(A) 1 × 12

(B) 4 × 3

(D) 6 × 2

Choose the correct answer from the options given below:

2026: 14 May Shift 2

If A is a 3x3 matrix then:

Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	$\vert adj A\vert$	I.	$\vert A\vert^{-1}$
B.	$\vert A.adj A\vert$	II.	$\vert A\vert$
C.	$\vert A^{-1}\vert$	III.	$\vert A\vert^2$
D.	$\vert A^{-1}.adj A\vert$	IV.	$\vert A\vert^3$

Choose the correct answer from the options given below:

2025: 29 May Shift 2

Suppose that A, B and C are matrices of order $m \times n$ , $n \times 5$ and $5 \times q$ respectively. The restriction on $m$ , $n$ and $q$ so that AB-BC is defined are

2025: 19 May Shift 1

If A and B are square matrices of the same order 3, such that det (A) = 3 and AB = 3I, where I is an identity matrix of order 3. Then the value of det (B) is:

If f(x)=∣0x+ax+bx−a0x+cx−bx−c0∣f(x) = \begin{vmatrix} 0 & x+a & x+b \\ x-a & 0 & x+c \\ x-b & x-c & 0 \end{vmatrix}f(x)=​0x−ax−b​x+a0x−c​x+bx+c0​​ then the value of f(0)f(0)f(0) is equal to:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $f(x) = \begin{vmatrix} 0 & x+a & x+b \\ x-a & 0 & x+c \\ x-b & x-c & 0 \end{vmatrix}$ then the value of $f(0)$ is equal to: