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CBSE Class 12 Chapter 3 Matrices Extra Questions

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CBSE Class 12 Chapter 3 Matrices Extra Questions. myCBSEguide has just released Chapter Wise Question Answers for class 12 Maths. There chapter wise Practice Questions with complete solutions are available for download in myCBSEguide website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools for years. There are around 4-5 set of solved Chapter 3 Matrices Mathematics Extra Questions from each and every chapter. The students will not miss any concept in these Chapter wise question that are specially designed to tackle Board Exam. We have taken care of every single concept given in CBSE Class 12 Mathematics syllabus and questions are framed as per the latest marking scheme and blue print issued by CBSE for class 12.

Class 12 Chapter 3 Maths Extra Questions

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Matrices Class 12 Mathematics Important Questions

Chapter 3 Matrices

  1. If {tex}A = \left[ {\begin{array}{*{20}{c}} 1&1&3 \\ 5&2&6 \\ { – 2}&{ – 1}&{ – 3} \end{array}} \right]{/tex}. Then |A| is
    1. none of these
    2. Idempotent
    3. Nilpotent
    4. Symmetric
  2. {tex}\left| {\begin{array}{*{20}{c}} 1&1&1 \\ e&0&{\sqrt 2 } \\ 2&2&2 \end{array}} \right| {/tex}is equal to
    1. 0
    2. 3e
    3. none of these
    4. 2
  3. A square matrix A is called idempotent if
    1. A2 = I
    2. A2 = O
    3. 2A=I
    4. A2 = A
  4. Let for any matrix M ,M-1 exist. Which of the following is not true.
    1. none of these
    2. (M-1)-1 = M
    3. (M-1)2 = (M2)-1
    4. (M-1)-1 = (M-1)1
  5. The system of equations x + 2y = 11, -2 x – 4y = 22 has
    1. only one solution
    2. infinitely many solutions
    3. finitely many solutions
    4. no solution
  6. Sum of two skew-symmetric matrices is always ________ matrix.
  7. ________ matrix is both symmetric and skew-symmetric matrix.
  8. If A and B are square matrices of the same order, then (kA)’ = ________ where k is any scalar.
  9. If {tex}A = \left[ {\begin{array}{*{20}{c}} 2&3 \\ 4&5 \end{array}} \right]{/tex}, Prove that A – At is a skew – symmetric matrix.
  10. {tex}A = \left[ {\begin{array}{*{20}{c}} 2&4 \\ 5&6 \end{array}} \right]{/tex}, Prove that A + A’ is a symmetric matrix.
  11. The no. of all possible matrics of order 3 {tex} \times{/tex} 3 with each entry as 0 or 1 is-
  12. Find the matrix X so that {tex}X\left[ \begin{gathered} \begin{array}{*{20}{c}} 1&2&3 \end{array} \hfill \\ \begin{array}{*{20}{c}} 4&5&6 \end{array} \hfill \\ \end{gathered} \right] = \left[ \begin{gathered} \begin{array}{*{20}{c}} { – 7}&{ – 8}&{ – 9} \end{array} \hfill \\ \begin{array}{*{20}{c}} {\;\;2}&{\;\;4}&{\;\;6} \end{array} \hfill \\ \end{gathered} \right]{/tex}.
  13. If A is a square matrix, such that A2 = A, then (I + A)3 – 7A is equal to
  14. If {tex}A = \left[ {\begin{array}{*{20}{c}} 2&4 \\ 5&6 \end{array}} \right]{/tex}, then Prove that A + A’ is a symmetric matrix.
  15. If f(x) = x2 – 5x + 7 and {tex}A = \left[ {\begin{array}{*{20}{c}} 3&1 \\ { – 1}&2 \end{array}} \right]{/tex} then find f(A).
  16. Find the matrix A such that {tex}\left[ \begin{gathered} \;\;\begin{array}{*{20}{c}} 2&{ – 1} \end{array} \hfill \\ \begin{array}{*{20}{c}} {\;\;1}&{\;\;0} \end{array} \hfill \\ \begin{array}{*{20}{c}} { – 3}&{\;\;4} \end{array} \hfill \\ \end{gathered} \right]A = \left[ {\begin{array}{*{20}{c}} { – 1}&{ – 8}&{ – 10} \\ 1&{ – 2}&{ – 5} \\ 9&{22}&{15} \end{array}} \right]{/tex}.

  17. Show that:

    1. {tex}\left[ {\begin{array}{*{20}{c}} 5&{ – 1} \\ 6&7 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2&1 \\ 3&4 \end{array}} \right] \ne \left[ {\begin{array}{*{20}{c}} 2&1 \\ 3&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5&{ – 1} \\ 6&7 \end{array}} \right]{/tex}
    2. {tex}\left[ {\begin{array}{*{20}{c}} 1&2&3 \\ 0&1&0 \\ 1&1&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { – 1}&1&0 \\ 0&{ – 1}&1 \\ 2&3&4 \end{array}} \right]{/tex} {tex} \ne \left[ {\begin{array}{*{20}{c}} { – 1}&1&0 \\ 0&{ – 1}&1 \\ 2&3&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&2&3 \\ 0&1&0 \\ 1&1&0 \end{array}} \right]{/tex}
  18. {tex}A = \left[ {\begin{array}{*{20}{c}} 0&{ – \tan \frac{\alpha }{2}} \\ {\tan \frac{\alpha }{2}}&0 \end{array}} \right]{/tex},
    Prove {tex}I + A = (I – A)\left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{ – \sin \alpha } \\ {\sin \alpha }&{\cos \alpha } \end{array}} \right]{/tex}.

Chapter 3 Matrices

Solution

    1. Nilpotent
      Explanation: The given matrix A is nilpotent, because |A| = 0,as determinant of a nilpotent matrix is zero.
    1. 0
      Explanation: {tex}\left| {\begin{array}{*{20}{c}} 1&1&1 \\ e&0&{\sqrt 2 } \\ 2&2&2 \end{array}} \right| = 2\left| {\begin{array}{*{20}{c}} 1&1&1 \\ e&0&{\sqrt 2 } \\ 1&1&1 \end{array}} \right| = 0{/tex}, because, row 1 and row 3 are identical.
    1. A2 = A
      Explanation: If the product of any square matrix with itself is the matrix itself, then the matrix is called Idempotent.
    1. (M-1)-1 = (M-1)1
      Explanation: Clearly , (M-1)-1 = (M-1)1 is not true.
    1. no solution
      Explanation: For no solution , we have: {tex}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}{/tex}, for given system of equations we have: {tex}\frac{1}{{ – 2}} = \frac{2}{{ – 4}} \ne \frac{{11}}{{22}}{/tex}.
  2. skew symmetric
  3. Null
  4. kA’
  5. P = A – At
    {tex} = \left[ {\begin{array}{*{20}{c}} 2&3 \\ 4&5 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} { – 2}&{ – 4} \\ { – 3}&{ – 5} \end{array}} \right]{/tex}
    {tex} = \left[ {\begin{array}{*{20}{c}} 0&{ – 1} \\ 1&0 \end{array}} \right]{/tex}
    {tex}P’ = \left[ {\begin{array}{*{20}{c}} 0&1 \\ { – 1}&0 \end{array}} \right]{/tex}
    {tex}P’ = – \left[ {\begin{array}{*{20}{c}} 0&{ – 1} \\ 1&0 \end{array}} \right]{/tex}
    P’ = – P
    Proved.
  6. {tex}P = A + A’ = \left[ {\begin{array}{*{20}{c}} 2&4 \\ 5&6 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 2&5 \\ 4&6 \end{array}} \right]{/tex}
    {tex}P = \left[ {\begin{array}{*{20}{c}} 4&9 \\ 9&{12} \end{array}} \right]{/tex}
    {tex}P’ = \left[ {\begin{array}{*{20}{c}} 4&9 \\ 9&{12} \end{array}} \right]{/tex}
    Thus, P’ = P.
  7. {tex}{2^9} = 512{/tex}
  8. Let {tex}X = \left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]{/tex}
    {tex}\therefore \left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]\left[ \begin{gathered} \begin{array}{*{20}{c}} 1&2&3 \end{array} \hfill \\ \begin{array}{*{20}{c}} 4&5&6 \end{array} \hfill \\ \end{gathered} \right] = \left[ \begin{gathered} \begin{array}{*{20}{c}} { – 7}&{ – 8}&{ – 9} \end{array} \hfill \\ \begin{array}{*{20}{c}} {\;\;2}&{\;\;4}&{\;\;6} \end{array} \hfill \\ \end{gathered} \right]{/tex}
    {tex}\left[ \begin{gathered} \begin{array}{*{20}{c}} {a + 4b}&{2a + 5b}&{3a + 6b} \end{array} \hfill \\ \begin{array}{*{20}{c}} {c + 4d}&{2c + 5d}&{3c + 6d} \end{array} \hfill \\ \end{gathered} \right]{/tex} {tex} = \left[ \begin{gathered} \begin{array}{*{20}{c}} { – 7}&{ – 8}&{ – 9} \end{array} \hfill \\ \begin{array}{*{20}{c}} {\;\;2}&{\;\;4}&{\;\;6} \end{array} \hfill \\ \end{gathered} \right]{/tex}
    Hence, a = 1, b = -2, c = 2, d = 0
    {tex}X = \left[ {\begin{array}{*{20}{c}} 1&{ – 2} \\ 2&0 \end{array}} \right]{/tex}
  9. (I + A)3 – 7A = I3 + A3 + 3IA (I + A) – 7A
    = I + A3 + 3I2A + 3IA2 – 7A
    = I + A3 + 3A + 3A2 – 7A
    = I + A3 + 3A + 3A – 7A {A2 = A}
    = I + A3 – A
    = I + A2 – A [A2 = A, A3 = A2] = I + A – A {A2 = A}
    = I
  10. P = A + A’ {tex} = \left[ {\begin{array}{*{20}{c}} 2&4 \\ 5&6 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 2&5 \\ 4&6 \end{array}} \right]{/tex}
    {tex}P = \left[ {\begin{array}{*{20}{c}} 4&9 \\ 9&{12} \end{array}} \right]{/tex}
    {tex}P’ = \left[ {\begin{array}{*{20}{c}} 4&9 \\ 9&{12} \end{array}} \right]{/tex}
    P’ = P
    Therefore P = P’
    Hence A+A’ is symmetric.
  11. {tex}A = \left[ {\begin{array}{*{20}{c}} 3&1 \\ { – 1}&2 \end{array}} \right]{/tex}
    {tex}A^2=AA=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}{/tex}
    Now, f(A) = A2 – 5A + 7I
    {tex} = \left[ {\begin{array}{*{20}{c}} 8&5 \\ { – 5}&3 \end{array}} \right] – 5\left[ {\begin{array}{*{20}{c}} 3&1 \\ { – 1}&2 \end{array}} \right] + 7\left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right]{/tex}
    {tex}f(A) = \left[ {\begin{array}{*{20}{c}} {8 – 15 + 7}&{5 – 5 + 0} \\ { – 5 + 5 + 0}&{3 – 10 + 7} \end{array}} \right]{/tex}{tex} = \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]{/tex}
    Therefore, {tex}f(A) = \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]{/tex}
  12. We have, {tex}\left[ \begin{gathered} \;\;\begin{array}{*{20}{c}} 2&{ – 1} \end{array} \hfill \\ \begin{array}{*{20}{c}} {\;\;1}&{\;\;0} \end{array} \hfill \\ \begin{array}{*{20}{c}} { – 3}&{\;\;4} \end{array} \hfill \\ \end{gathered} \right]A = \left[ {\begin{array}{*{20}{c}} { – 1}&{ – 8}&{ – 10} \\ 1&{ – 2}&{ – 5} \\ 9&{22}&{15} \end{array}} \right]{/tex}
    From the given equation, it is clear that order of A should be {tex} 2\times 3{/tex}
    Let {tex}A = \left[ \begin{gathered} \begin{array}{*{20}{c}} a&b&c \end{array} \hfill \\ \begin{array}{*{20}{c}} d&e&f \end{array} \hfill \\ \end{gathered} \right]{/tex}
    {tex}\therefore{/tex} {tex}\begin{bmatrix}2&-1\\1&0\\-3&4\end{bmatrix}\begin{bmatrix}a&b&c\\d&e&f\end{bmatrix}=\begin{bmatrix}-1&-8&-10\\1&-2&-5\\9&22&15\end{bmatrix}{/tex}
    {tex} \Rightarrow \left[ {\begin{array}{*{20}{c}} {2a – d}&{2b – e}&{2c – f} \\ {a + 0d}&{b + 0.e}&{c + 0.f} \\ { – 3a + 4d}&{ – 3b + 4e}&{ – 3c + 4f} \end{array}} \right]{/tex} {tex} = \left[ {\begin{array}{*{20}{c}} { – 1}&{ – 8}&{ – 10} \\ 1&{ – 2}&{ – 5} \\ 9&{22}&{15} \end{array}} \right]{/tex}
    {tex}\Rightarrow \left[ {\begin{array}{*{20}{c}} {2a – d}&{2b – e}&{2c – f} \\ a&b&c \\ { – 3a + 4d}&{ – 3b + 4e}&{ – 3c + 4f} \end{array}} \right]{/tex} {tex} = \left[ {\begin{array}{*{20}{c}} { – 1}&{ – 8}&{ – 10} \\ 1&{ – 2}&{ – 5} \\ 9&{22}&{15} \end{array}} \right]{/tex}
    By equality of matrices, we get
    a = 1, b = -2, c = -5
    and 2a – d = -1 {tex}\Rightarrow{/tex} d = 2a + 1 = 3;
    {tex}\Rightarrow{/tex} 2b – e = -8 {tex}\Rightarrow{/tex} e = 2(-2) + 8 =4
    2c – f = -10 {tex}\Rightarrow{/tex} f = 2c + 10 = 0
    {tex}\therefore A = \left[ \begin{gathered} \begin{array}{*{20}{c}} 1&{ – 2}&{ – 5} \end{array} \hfill \\ \begin{array}{*{20}{c}} 3&{\;\;4}&{\;\;0} \end{array} \hfill \\ \end{gathered} \right]{/tex}

    1. L.H.S. {tex} = \left[ {\begin{array}{*{20}{c}} 5&{ – 1} \\ 6&7 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2&1 \\ 3&4 \end{array}} \right]{/tex} {tex} = \left[ {\begin{array}{*{20}{c}} {5(2) + ( – 1)3}&{5(1) + ( – 1)4} \\ {6(2) + 7(3)}&{6(1) + 7(4)} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 7&1 \\ {33}&{34} \end{array}} \right]{/tex}
      R.H.S. {tex} = \left[ {\begin{array}{*{20}{c}} 2&1 \\ 3&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5&{ – 1} \\ 6&7 \end{array}} \right]{/tex} {tex}= \left[ {\begin{array}{*{20}{c}} {2(5) + 1(6)}&{2( – 1) + 1(7)} \\ {3(5) + 4(6)}&{3( – 1) + 4(7)} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {16}&5 \\ {39}&{25} \end{array}} \right]{/tex}
      {tex}\therefore{/tex} L.H.S. {tex}\ne{/tex} R.H.S.
    2. L.H.S. {tex}= \left[ {\begin{array}{*{20}{c}} 1&2&3 \\ 0&1&0 \\ 1&1&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { – 1}&1&0 \\ 0&{ – 1}&1 \\ 2&3&4 \end{array}} \right]{/tex}
      {tex} = \left[ {\begin{array}{*{20}{c}} {1( – 1) + 2(0) + 3(2)}&{1(1) + 2( – 1) + 3(3)}&{1(0) + 2(1) + 3(4)} \\ {0( – 1) + 1(0) + 0(2)}&{0(1) + 1( – 1) + 0(3)}&{0(0) + 1(1) + 0(4)} \\ {1( – 1) + 1(0) + 0(2)}&{1(1) + 1( – 1) + 0(3)}&{1(0) + 1(1) + 0(4)} \end{array}} \right]{/tex}
      {tex} = \left[ {\begin{array}{*{20}{c}} 5&8&{14} \\ 0&{ – 1}&1 \\ { – 1}&0&1 \end{array}} \right]{/tex}
      R.H.S. {tex} = \left[ {\begin{array}{*{20}{c}} { – 1}&1&0 \\ 0&{ – 1}&1 \\ 2&3&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&2&3 \\ 0&1&0 \\ 1&1&0 \end{array}} \right]{/tex}
      {tex} = \left[ {\begin{array}{*{20}{c}} { – 1(1) + 1(0) + 0(1)}&{( – 1)2 + 1(1) + 0(1)}&{( – 1)3 + 1(0) + 0(0)} \\ {0(1) + ( – 1)0 + 1(1)}&{(0)2 + 1( – 1) + 1(1)}&{(0)3 + 0( – 1) + 1(0)} \\ {2(1) + 3(0) + 4(1)}&{2(2) + 3(1) + 4(1)}&{2(3) + 3(0) + 4(0)} \end{array}} \right]{/tex}
      {tex} = \left[ {\begin{array}{*{20}{c}} { – 1}&{ – 1}&{ – 3} \\ 1&0&0 \\ 6&{11}&6 \end{array}} \right]{/tex}
      {tex}\therefore{/tex} L.H.S. {tex}\ne{/tex} R.H.S.
  13. Put {tex}\tan \frac{\alpha }{2} = t{/tex}
    {tex}A = \left[ {\begin{array}{*{20}{c}} 0&{ – t} \\ t&0 \end{array}} \right]{/tex}
    {tex}I + A = \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0&{ – t} \\ t&0 \end{array}} \right]{/tex}
    {tex} = \left[ {\begin{array}{*{20}{c}} 1&{ – t} \\ t&1 \end{array}} \right]{/tex}
    {tex}I – A = \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right] – \left[ {\begin{array}{*{20}{c}} 0&{ – t} \\ t&0 \end{array}} \right]{/tex}
    {tex} = \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0&t \\ { – t}&0 \end{array}} \right]{/tex}
    {tex} = \left[ {\begin{array}{*{20}{c}} 1&t \\ { – t}&1 \end{array}} \right]{/tex}
    L.H.S. {tex} = (I – A)\left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{ – \sin \alpha } \\ {\sin \alpha }&{\cos \alpha } \end{array}} \right]{/tex}
    {tex} = (I – A)\left[ {\begin{array}{*{20}{c}} {\frac{{1 – {{\tan }^2}\frac{\alpha }{2}}}{{1 + {{\tan }^2}\frac{\alpha }{2}}}}&{\frac{{ – 2{{\tan }^2}\frac{\alpha }{2}}}{{1 + {{\tan }^2}\frac{\alpha }{2}}}} \\ {\frac{{2{{\tan }^2}\frac{\alpha }{2}}}{{1 + {{\tan }^2}\frac{\alpha }{2}}}}&{\frac{{1 – {{\tan }^2}\frac{\alpha }{2}}}{{1 + {{\tan }^2}\frac{\alpha }{2}}}} \end{array}} \right]{/tex}
    {tex} = \left[ {\begin{array}{*{20}{c}} 1&t \\ { – t}&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\frac{{1 – {t^2}}}{{1 + {t^2}}}}&{\frac{{ – 2t}}{{1 + {t^2}}}} \\ {\frac{{ – 2t}}{{1 + {t^2}}}}&{\frac{{1 – {t^2}}}{{1 + {t^2}}}} \end{array}} \right]{/tex}
    {tex} = \left[ {\begin{array}{*{20}{c}} {\frac{{1 – {t^2}}}{{1 + {t^2}}} + \frac{{t.2t}}{{1 + {t^2}}}}&{\frac{{ – 2t}}{{1 + {t^2}}} + t\left( {\frac{{1 – {t^2}}}{{1 + {t^2}}}} \right)} \\ { – t\left( {\frac{{1 – {t^2}}}{{1 + {t^2}}}} \right) + \frac{{2t}}{{1 + {t^2}}}}&{ – t\left( {\frac{{ – 2t}}{{1 + {t^2}}}} \right) + \left( {\frac{{1 – {t^2}}}{{1 + {t^2}}}} \right)} \end{array}} \right]{/tex}
    {tex} = \left[ {\begin{array}{*{20}{c}} {\frac{{1 – {t^2} + 2{t^2}}}{{1 + {t^2}}}}&{\frac{{ – 2t + t – {t^3}}}{{1 + {t^2}}}} \\ {\frac{{ – t + {t^3} + 2t}}{{1 + {t^2}}}}&{\frac{{2{t^2} + 1 – {t^2}}}{{1 + {t^2}}}} \end{array}} \right]{/tex}
    {tex} = \left[ {\begin{array}{*{20}{c}} {\frac{{1 + {t^2}}}{{1 + {t^2}}}}&{\frac{{ – {t^3} – t}}{{1 + {t^2}}}} \\ {\frac{{{t^3} + t}}{{1 + {t^2}}}}&{\frac{{{t^2} + 1}}{{1 + {t^2}}}} \end{array}} \right]{/tex}
    {tex} = \left[ {\begin{array}{*{20}{c}} 1&{\frac{{ – t(1 + {t^2})}}{{1 + {t^2}}}} \\ {\frac{{t(1 + {t^2})}}{{1 + {t^2}}}}&{\frac{{{t^2} + 1}}{{1 + {t^2}}}} \end{array}} \right]{/tex}
    {tex} = \left[ {\begin{array}{*{20}{c}} 1&{ – t} \\ t&1 \end{array}} \right]{/tex}
    L.H.S = R.H.S
    Hence proved

Chapter Wise Important Questions Class 12 Maths Part I and Part II

  1. Relations and Functions
  2. Inverse Trigonometric Functions
  3. Matrices
  4. Determinants
  5. Continuity and Differentiability
  6. Application of Derivatives
  7. Integrals
  8. Application of Integrals
  9. Differential Equations
  10. Vector Algebra
  11. Three Dimensional Geometry
  12. Linear Programming
  13. Probability
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