If the area of a triangle with vertices A(1,3), B(0,0) and C(k,0) is 3 sq. units, then k is: Option 1: ± 1 Option 2: ± 2 Option 3: ± 4 Option 4: ± 3/2

Answer: ± 2. Solution: Area = 1/2, x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2), = 1/2, 1(0-0) + 0(0-3) + k(3-0), = 1/2, 3k, Set equal to 3: (3, k, )/2 = 3, k, = 2 k = ± 2.

Algebra

Matrices & Determinants

Easy

If the area of a triangle with vertices A(1,3), B(0,0) and C(k,0) is 3 sq. units, then k is:

$\pm 1$

$\pm 2$

$\pm 4$

$\pm \frac{3}{2}$

✅ Correct Option: 2

Area $= \frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$

$= \frac{1}{2}|1(0-0) + 0(0-3) + k(3-0)| = \frac{1}{2}|3k|$

Set equal to 3: $\frac{3|k|}{2} = 3 \Rightarrow |k| = 2 \Rightarrow k = \pm 2$ .

2025: 30 May Shift 1

If $A$ and $B$ are invertible matrices of order $3$ then match List-I with List-II

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

Choose the correct answer from the options given below:

2026: 29 May Shift 2

If $\begin{vmatrix}2a & 2a & 3b+c\\3b & 2a+c & 3b\\2a+3b & c & c\end{vmatrix} = 48$ , then abc is equal to

2025: 27 May Shift 1

If $c_{ij}$ denotes the cofactor of element $a_{ij}$ of the matrix $A = \begin{bmatrix} 1 & 2 & -1 \\ 0 & -3 & 2 \\ 4 & 2 & 3 \end{bmatrix}$ then the value of $c_{21} \cdot c_{33}$ is

2025: 3 June Shift 1

The value of k for which the system of equations

$x + y + z = 1$

$x - ky + z = 1$

$x - y + z = 1$

has more than one solutions is

2026: 14 May Shift 2

If $A=\begin{bmatrix}a & 0\\0 & 0\end{bmatrix}$ and $B=\begin{bmatrix}0 & 0\\0 & b\end{bmatrix}$ then the matrix $A^3B^3$ is:

2026: 31 May Shift 2

If $\begin{bmatrix} 0 & x \\ 2 & 1 \end{bmatrix} + 2\begin{bmatrix} 3 & 1 \\ y & 0 \end{bmatrix} = \begin{bmatrix} 6 & 5 \\ 8 & 1 \end{bmatrix}$ , then the value of $2x+3y$ is :

If the area of a triangle with vertices A(1,3), B(0,0) and C(k,0) is 3 sq. units, then k is:

Solution

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