Minors and Cofactors PYQs
Practice Class 12 CBSE Board Previous Year Questions (2014-2026)
Q1
2024 (MCQ)
00:00
If $A$ is a square matrix of order 3 such that $|A| = -4$, then $|adj A|$ is: 1 Mark
$|adj A| = |A|^{n-1} = (-4)^{3-1} = (-4)^2 = 16$.
16
Q2
2024 (Assertion-Reason)
00:00
Assertion (A): The adjoint of a symmetric matrix is also a symmetric matrix.
Reason (R): For any square matrix $A$, $(adj A)^T = adj(A^T)$. 1 Mark
Reason (R) is a standard property of adjoints: $(adj A)^T = adj(A^T)$. So (R) is true.
Assertion (A): If $A$ is symmetric, $A^T = A$.
Then $(adj A)^T = adj(A^T) = adj(A)$.
This means $adj A$ is also symmetric. So (A) is true.
Since (A) was proved using (R), (R) is the correct explanation.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
Q3
2017
00:00
If $A = \begin{pmatrix} 2 & 3 \\ 5 & -2 \end{pmatrix}$, show that $A^{-1} = \frac{1}{19} A$. 2 Marks
Step 1: Calculate $|A| = (2)(-2) - (5)(3) = -4 - 15 = -19$.
Step 2: $adj A = \begin{pmatrix} -2 & -3 \\ -5 & 2 \end{pmatrix}$.
Step 3: $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{-19} \begin{pmatrix} -2 & -3 \\ -5 & 2 \end{pmatrix} = \frac{1}{19} \begin{pmatrix} 2 & 3 \\ 5 & -2 \end{pmatrix} = \frac{1}{19} A$.
Proved