Properties of Determinants PYQs
Practice Class 12 CBSE Board Previous Year Questions (2014-2026)
Q1
2024 (Set 65/1/1)
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The value of the determinant $\begin{vmatrix} \cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ \end{vmatrix}$ is: 1 Mark
Value $= \cos 15^\circ \cos 75^\circ - \sin 75^\circ \sin 15^\circ$.
Using the identity $\cos A \cos B - \sin A \sin B = \cos(A + B)$:
Value $= \cos(15^\circ + 75^\circ) = \cos 90^\circ = 0$.
0
Q2
2024 (Assertion-Reason)
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Assertion (A): If $A$ is a square matrix of order 3 such that $|A| = 5$, then $|adj A| = 25$.
Reason (R): For a square matrix $A$ of order $n$, $|adj A| = |A|^{n-1}$. 1 Mark
Reason (R) is a standard property: $|adj A| = |A|^{n-1}$. So (R) is true.
Given $n=3$ and $|A|=5$.
Using (R), $|adj A| = 5^{3-1} = 5^2 = 25$. So (A) is true.
Also, (R) correctly explains how (A) was calculated.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
Q3
2019
00:00
If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, find $|A \cdot adj A|$ without calculating $adj A$. 2 Marks
We know $A \cdot adj A = |A| I$.
So $|A \cdot adj A| = ||A| I| = |A|^n |I| = |A|^n$.
Here $n=2$ (order of matrix).
$|A| = (1)(4) - (3)(2) = 4 - 6 = -2$.
Thus, $|A \cdot adj A| = (-2)^2 = 4$.
4
Q4
2016 (All India)
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Using properties of determinants, prove that: 4 Marks
$\begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = abc \left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$
Step 1: Take out $a, b, c$ common from $R_1, R_2, R_3$ respectively.
$\Delta = abc \begin{vmatrix} 1/a+1 & 1/a & 1/a \\ 1/b & 1/b+1 & 1/b \\ 1/c & 1/c & 1/c+1 \end{vmatrix}$.
Step 2: Apply $R_1 \to R_1 + R_2 + R_3$.
$\Delta = abc \begin{vmatrix} 1+1/a+1/b+1/c & 1+1/a+1/b+1/c & 1+1/a+1/b+1/c \\ 1/b & 1/b+1 & 1/b \\ 1/c & 1/c & 1/c+1 \end{vmatrix}$.
Step 3: Take $(1+1/a+1/b+1/c)$ common from $R_1$.
$\Delta = abc(1+1/a+1/b+1/c) \begin{vmatrix} 1 & 1 & 1 \\ 1/b & 1/b+1 & 1/b \\ 1/c & 1/c & 1/c+1 \end{vmatrix}$.
Step 4: Apply $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$.
$\Delta = abc(1+1/a+1/b+1/c) \begin{vmatrix} 1 & 0 & 0 \\ 1/b & 1 & 0 \\ 1/c & 0 & 1 \end{vmatrix}$.
Step 5: Expanding along $R_1$, we get $abc(1+1/a+1/b+1/c)$.
Proved
Q5
2024 (Set 65/3/1)
00:00
If $A$ is a square matrix of order 3 such that $|A| = 2$, then find the value of $|adj(adj A)|$. 3 Marks
Property: $|adj(adj A)| = |A|^{(n-1)^2}$.
Here $n=3$ and $|A|=2$.
$(n-1)^2 = (3-1)^2 = 2^2 = 4$.
So $|adj(adj A)| = 2^4 = 16$.
16
Q8
2025 Board
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If $A$ is a square matrix of order 3 and $|A| = 4$, then $|adj(adj A)|$ is: 1 Mark
Property: $|adj(adj A)| = |A|^{(n-1)^2}$.
Here $n=3, |A|=4$.
So $|adj(adj A)| = 4^{(3-1)^2} = 4^4 = 256$.
256
Q9
2026 Sample Paper
00:00
The value of $\begin{vmatrix} 1+x & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+x \end{vmatrix}$ is: 2 Marks
Apply $C_1 \to C_1 + C_2 + C_3$.
$\Delta = \begin{vmatrix} 3+x & 1 & 1 \\ 3+x & 1+x & 1 \\ 3+x & 1 & 1+x \end{vmatrix}$.
Take $(3+x)$ common from $C_1$.
$\Delta = (3+x) \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+x \end{vmatrix}$.
Apply $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$.
$\Delta = (3+x) \begin{vmatrix} 1 & 1 & 1 \\ 0 & x & 0 \\ 0 & 0 & x \end{vmatrix} = (3+x)x^2$.
$x^2(x+3)$