Transpose of a Matrix is an important topic in the Mathematics section in MHT CET exam. Practising this topic will increase your score overall and make your conceptual grip on MHT CET exam stronger.
This article gives you a full set of MHT CET PYQs for Transpose of a Matrix with explanations for effective preparation. Practice of MHT CET Mathematics PYQs including Transpose of a Matrix questions regularly will improve accuracy, speed, and confidence in the MHT CET 2026 exam.
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MHT CET PYQs for Transpose of a Matrix with Solutions
1.
The negative of \( (p \land (\sim q)) \lor (\sim p) \) is equivalent to:- \( p \land q \)
- \( p \land (\sim q) \)
- \( p \land (q \land (\sim p)) \)
- \( p \lor (q \lor (\sim p)) \)
2.
If \[ B = \begin{bmatrix} 3 & \alpha & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix} \] is the adjoint of a 3x3 matrix \( A \) and \( |A| = 4 \), then \( \alpha \) is equal to:- 1
- 0
- -1
- -2
3.
The inverse of the matrix \[ \begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 3 \\ 5 & 2 & -1 \end{bmatrix} \] is:- \[ \frac{-1}{3} \begin{bmatrix} -3 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 2 & -3 \end{bmatrix} \]
- \[ \frac{-1}{3} \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix} \]
- \[ \frac{-1}{3} \begin{bmatrix} 3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix} \]
- \[ \frac{-1}{3} \begin{bmatrix} -3 & 0 & 0 \\ -3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix} \]
4.
The variance of the following probability distribution is:
\[ \begin{array}{|c|c|} \hline x & P(X) \\ \hline 0 & \frac{9}{16} \\ 1 & \frac{3}{8} \\ 2 & \frac{1}{16} \\ \hline \end{array} \]- \(\frac{1}{8}\)
- \(\frac{5}{8}\)
- \(\frac{1}{4}\)
- \(\frac{3}{8}\)
5.
The converse of \( ((\sim p) \land q) \Rightarrow r \) is:- \( ((\sim p) \lor q) \Rightarrow r \)
- \( (\sim r) \Rightarrow p \land q \)
- \( (p \lor (\sim q)) \Rightarrow (\sim r) \)
- \( (\sim r) \Rightarrow ((\sim p) \land q) \)
6.
If \( A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix} \), then \( A^{-1} \) is:- \[ \frac{1}{2} \begin{bmatrix} -1 & -1 & -1 \\ -8 & 6 & -1 \\ 5 & -3 & 1 \end{bmatrix} \]
- \[ \begin{bmatrix} 1 & -1 & -1 \\ 2 & -3 & 2 \\ -4 & -3 & -1 \end{bmatrix} \]
- \[ \frac{1}{2} \begin{bmatrix} 1 & -1 & 5 \\ -6 & 3 & -1 \\ 1 & 2 & -1 \end{bmatrix} \]
- \[ \begin{bmatrix} -1 & -1 & -2 \\ 1 & -6 & 3 \\ 5 & -3 & 1 \end{bmatrix} \]
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