The value of a + 1 a + 2 a + 2 1 a + 2 a + 3 a + 3 1 a + 3 a + 4 a + 4 1 is

R 2 → R 2 - R 1 and R 3 → R 3 - R 1 Δ = a + 1 a + 2 a + 2 1 a + 2 a + 3 - a - 1 1 0 a 2 + 7 a + 12 - a 2 - 3 a - 2 2 0 = a 2 + 3 a + 2 a + 2 1 2 a + 2 1 0 4 a + 10 2 0 = 4 a + 2 - 4 a - 10 = 4 a + 8 - 4 a - 10 = - 2

Mathematics - Determinant Question with Solution | TestHub

MathematicsDeterminantProperties of det.Easy2 minPYQ_2021

MathematicsEasysingle choice

The value of $|\begin{array}{l} (a + 1) (a + 2) & a + 2 & 1 \\ (a + 2) (a + 3) & a + 3 & 1 \\ (a + 3) (a + 4) & a + 4 & 1 \end{array}|$ is

Options:

Answer:

Solution:

$R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$

$Δ = |\begin{matrix} (a + 1) (a + 2) & a + 2 & 1 \\ (a + 2) (a + 3 - a - 1) & 1 & 0 \\ a^{2} + 7 a + 12 - a^{2} - 3 a - 2 & 2 & 0 \end{matrix}|$

$= |\begin{matrix} a^{2} + 3 a + 2 & a + 2 & 1 \\ 2 (a + 2) & 1 & 0 \\ 4 a + 10 & 2 & 0 \end{matrix}|$

$= 4 (a + 2) - 4 a - 10$

$= 4 a + 8 - 4 a - 10 = - 2$

Stream:JEESubject:MathematicsTopic:DeterminantSubtopic:Properties of det.

⏱ 2mℹ️ Source: PYQ_2021

Doubts & Discussion

Loading discussions...