NCERT Solutions for Class 12 Maths Chapter 3 Matrices Exercise 3.3

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Jasmine Grover

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Class 12 Maths NCERT Solutions Chapter 3 Matrices Exercise 3.3 is provided in this article. Chapter 3 Exercise 3.3 deals with questions on the transpose of a matrix, properties of transpose of a matrix, and symmetric and skew-symmetric matrices.

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Check out the solutions for Class 12 Maths NCERT solutions chapter 3 Matrices Exercise 3.3:

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CBSE CLASS XII Related Questions

  • 1.
    The least value of \[ f(x)=e^{-x} \] in the interval \[ [0,3] \] is:

      • \( e^{-3} \)
      • \( -1 \)
      • \( 1 \)
      • \( -e^3 \)

    • 2.
      In a school, the probability of holding a debate competition is \( \frac{1}{3} \) and that of a quiz competition is \( \frac{2}{3} \). In the two participating teams, A has 4 girls and 6 boys and B has 7 girls and 3 boys. If a debate competition is held, the students are selected from team A and for the quiz competition they are selected from team B. If only two students are to be chosen from the teams, then find the probability that one will be a girl and the other a boy.


        • 3.

          Check whether \[ f:\mathbb{R}-\{3\}\rightarrow\mathbb{R} \] defined as \[ f(x)=\frac{x-2}{x-3} \] is onto or not. 


            • 4.

              Check whether the function \[ f(x)= \begin{cases} \dfrac{|x-3|}{2(x-3)}, & x<3,\\[6pt] \dfrac{x-6}{6}, & x\geq 3 \end{cases} \] is continuous at \(x=3\) or not. 


                • 5.
                  For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]

                    • local maximum value is 2
                    • local minimum value is \( -2 \)
                    • local maximum value is \( -2 \)
                    • local minimum value \( < \) local maximum value

                  • 6.
                    Assertion (A) : In an experiment of throwing an unbiased die, the probability of getting a prime number given that number appearing on the die being odd is \( \frac{2}{3} \).
                    Reason (R) : For any two events \( A \) and \( B \), \( P(A|B) = \frac{P(A \cup B)}{P(B)} \).

                      • Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
                      • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
                      • Assertion (A) is true and Reason (R) is false.
                      • Assertion (A) is false and Reason (R) is true.
                    CBSE CLASS XII Previous Year Papers

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