The value of the determinant vmatrix a & b & 0 \\ -b & a & 0 \\ 0 & 0 & c vmatrix is : Option 1: (a^2 + b^2)c Option 2: (a^2^2 + b^2^2)c Option 3: (a^2^2 - b^2^2)c Option 4: (a^2 + b^2)c^2

Answer: (a^2^2 + b^2^2)c. Solution: Expand along the third column: = c * (a * a - b * (-b)) = c(a^2^2 + b^2^2).

Algebra

Matrices & Determinants

Medium

core

The value of the determinant $\begin{vmatrix} a\cos\theta & b\sin\theta & 0 \\ -b\sin\theta & a\cos\theta & 0 \\ 0 & 0 & c \end{vmatrix}$ is :

$(a^2 + b^2)c$

$(a^2\cos^2\theta + b^2\sin^2\theta)c$

$(a^2\cos^2\theta - b^2\sin^2\theta)c$

$(a^2 + b^2)c^2$

✅ Correct Option: 2

Expand along the third column:

$= c \cdot (a\cos\theta \cdot a\cos\theta - b\sin\theta \cdot (-b\sin\theta))$

$= c(a^2\cos^2\theta + b^2\sin^2\theta)$ .

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The value of the determinant ∣acos⁡θbsin⁡θ0−bsin⁡θacos⁡θ000c∣\begin{vmatrix} a\cos\theta & b\sin\theta & 0 \\ -b\sin\theta & a\cos\theta & 0 \\ 0 & 0 & c \end{vmatrix}​acosθ−bsinθ0​bsinθacosθ0​00c​​ is :

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CUET Mathematics Past Year Questions (Topic-Wise)

The value of the determinant $\begin{vmatrix} a\cos\theta & b\sin\theta & 0 \\ -b\sin\theta & a\cos\theta & 0 \\ 0 & 0 & c \end{vmatrix}$ is :