The value of vmatrix 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y vmatrix is Option 1: 0 Option 2: -xy Option 3: xy Option 4: x^2y^2

Answer: xy. Solution: The determinant to evaluate is: vmatrix 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y vmatrix Apply row operations to simplify the determinant. Subtract Row 1 from Row 2 and Row 3: R_2 R_2 - R_1 R_3 R_3 - R_1 For Row 2: First element: 1 - 1 = 0 Second element: (x+y) - x = y Third element: y - y = 0 For Row 3: First element: 1 - 1 = 0 Second element: x - x = 0 Third element: (x+y) - y = x The determinant becomes: vmatrix 1 & x & y \\ 0 & y & 0 \\ 0 & 0 & x vmatrix This is an upper triangular matrix with all elements below the main diagonal equal to zero. For a triangular matrix, the determinant equals the product of the diagonal elements. The diagonal elements are: 1, y, x Determinant = 1 x y x x Determinant = xy Therefore, the value of the determinant is xy.

Algebra

Matrices & Determinants

Medium

core

The value of $\begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix}$ is

$-xy$

$xy$

$x^2y^2$

✅ Correct Option: 3

The determinant to evaluate is:

$\begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix}$

Apply row operations to simplify the determinant. Subtract Row 1 from Row 2 and Row 3:

$R_2 \to R_2 - R_1$

$R_3 \to R_3 - R_1$

For Row 2:

First element: $1 - 1 = 0$

Second element: $(x+y) - x = y$

Third element: $y - y = 0$

For Row 3:

First element: $1 - 1 = 0$

Second element: $x - x = 0$

Third element: $(x+y) - y = x$

The determinant becomes:

$\begin{vmatrix} 1 & x & y \\ 0 & y & 0 \\ 0 & 0 & x \end{vmatrix}$

This is an upper triangular matrix with all elements below the main diagonal equal to zero.

For a triangular matrix, the determinant equals the product of the diagonal elements.

The diagonal elements are: $1, y, x$

Determinant $= 1 \times y \times x$

Determinant $= xy$

Therefore, the value of the determinant is $xy$ .

2026: 22 May Shift 2

Match List-I with List-II

List-I	List-II
(A) If $A = [a_{ij}]_{n \times n}$ , where $a_{ij} = \begin{cases}0, & i \ne j \\ k, & i=j, k\ne 0\end{cases}$ then A is ___	(I) Symmetric matrix
(B) If $A = [a_{ij}]_{2\times 2}$ , where $a_{ij} = \begin{cases}1, & i \ne j \\ 0, & i=j\end{cases}$ , then $A^2$ is ___	(II) Skew-Symmetric matrix
(C) If A and B are matrices of the same order, then $(AB^T - BA^T)$ is ___	(III) Identity matrix
(D) If A and B are symmetric matrices of the same order, then $(AB+BA)$ is ___	(IV) Scalar matrix

Choose the correct answer from the options given below:

2025: 22 May Shift 1

If A is a skew-symmetric matrix, then which of the following statements is NOT true?

(A) A is singular if order of A is odd

(B) A is non-singular

(D) $A^{2025}$ is a symmetric matrix

(E) all diagonal elements of A are zeros

Choose the correct answer from the options given below:

2025: 2 June Shift 1

If $A$ is an invertible matrix of order 3, then

Match List-I with List-II

List-I	List-II
(A) $\vert \text{adj} A\vert$	(I) $8\vert A\vert$
(B) $\vert A(\text{adj} A)\vert$	(II) $\vert A\vert ^2$
(C) $\vert 2A\vert$	(III) $\frac{1}{\vert A\vert }$
(D) $\vert A^{-1}\vert$	(IV) $\vert A\vert ^3$

Choose the correct answer from the options given below:

2025: 27 May Shift 1

If $\begin{vmatrix} p-a & 0 & c-r \\ 0 & q-b & c-r \\ a & b & r \end{vmatrix} = 0$ , then the value of $\dfrac{p}{p-a} + \dfrac{q}{q-b} + \dfrac{r}{r-c}$ is

2026: 29 May Shift 1

If $A=\begin{bmatrix}3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2\end{bmatrix}$ and $A(adjA)=xI$ , where $I$ is an identify matrix of order 3, then $x$ is equal to

2026: 29 May Shift 1

Let $A=\begin{bmatrix}1 & 0 \\ -1 & 7\end{bmatrix}$ and $B=\begin{bmatrix}0 & 4 \\ -1 & 7\end{bmatrix}$ then

Match List-I with List-II

List-I	List-II
Determinants	Value
(A) $\left\vert A^{2} \right\vert$	(I) $28$
(B) $\left\vert AB^{T} \right\vert$	(II) $7$
(C) $\left\vert \text{adj } A \right\vert$	(III) $16$
(D) $\left\vert 2B \right\vert$	(IV) $49$

Choose the correct answer from the options given below:

The value of ∣1xy1x+yy1xx+y∣\begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix}​111​xx+yx​yyx+y​​ is

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The value of $\begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix}$ is