NCERT 3.2 class 12 maths solutions

NCERT – Exercise 3.2

  1. Let \( A = \left[ {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right], B = \left[ {\begin{array}{*{20}{c}}1&3\\-2&5\end{array}} \right], C = \left[ {\begin{array}{*{20}{c}}-2&5\\3&4\end{array}} \right] \). Find each of the following:

    1. \( A + B \)
    2. \( A – B \)
    3. \( 3A – C \)
    4. \( AB \)
    5. \( BA \)

    Solution:

    1. \[ A + B = \left[ {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}1&3\\-2&5\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&7\\1&7\end{array}} \right] \]
    2. \[ A – B = \left[ {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}1&3\\-2&5\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&1\\5&-3\end{array}} \right] \]
    3. \[ 3A – C = 3\left[ {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}-2&5\\3&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}8&7\\6&2\end{array}} \right] \]
    4. \[ AB = \left[ {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right] \left[ {\begin{array}{*{20}{c}}1&3\\-2&5\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}-6&26\\-1&19\end{array}} \right] \]
    5. \[ BA = \left[ {\begin{array}{*{20}{c}}1&3\\-2&5\end{array}} \right] \left[ {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}11&10\\11&2\end{array}} \right] \]

  2. Compute the following:

    1. \( \left[ {\begin{array}{*{20}{c}}a&b\\-b&a\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}a&b\\b&a\end{array}} \right] \)
    2. \( \left[ {\begin{array}{*{20}{c}}a^2 + b^2&b^2 + c^2\\a^2 + c^2&a^2 + b^2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}2ab&2bc\\-2ac&-2ab\end{array}} \right] \)
    3. \( \left[ {\begin{array}{*{20}{c}}-1&4&-6\\8&5&16\\2&8&5\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}12&7&6\\8&0&5\\3&2&4\end{array}} \right] \)
    4. \( \left[ {\begin{array}{*{20}{c}}\cos^2 x&\sin^2 x\\\sin^2 x&\cos^2 x\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}\sin^2 x&\cos^2 x\\\cos^2 x&\sin^2 x\end{array}} \right] \)

    Solution:

    1. \[ \left[ {\begin{array}{*{20}{c}}a&b\\-b&a\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}a&b\\b&a\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2a&2b\\0&2a\end{array}} \right] \]
    2. \[ \left[ {\begin{array}{*{20}{c}}a^2 + b^2&b^2 + c^2\\a^2 + c^2&a^2 + b^2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}2ab&2bc\\-2ac&-2ab\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}(a + b)^2&(b + c)^2\\(a – c)^2&(a – b)^2\end{array}} \right] \]
    3. \[ \left[ {\begin{array}{*{20}{c}}-1&4&-6\\8&5&16\\2&8&5\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}12&7&6\\8&0&5\\3&2&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}11&11&0\\16&5&21\\5&10&9\end{array}} \right] \]
    4. \[ \left[ {\begin{array}{*{20}{c}}\cos^2 x&\sin^2 x\\\sin^2 x&\cos^2 x\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}\sin^2 x&\cos^2 x\\\cos^2 x&\sin^2 x\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&1\\1&1\end{array}} \right] \]

  3. Compute the following products:

    1. \( \left[ {\begin{array}{*{20}{c}}a&b\\-b&a\end{array}} \right] \left[ {\begin{array}{*{20}{c}}a&-b\\b&a\end{array}} \right] \)
    2. \( \left[ {\begin{array}{*{20}{c}}1\\2\\3\end{array}} \right] \left[ {\begin{array}{*{20}{c}}2&3&4\end{array}} \right] \)
    3. \( \left[ {\begin{array}{*{20}{c}}1&-2\\2&3\end{array}} \right] \left[ {\begin{array}{*{20}{c}}1&2&3\\2&3&1\end{array}} \right] \)
    4. \( \left[ {\begin{array}{*{20}{c}}2&3&4\\3&4&5\\4&5&6\end{array}} \right] \left[ {\begin{array}{*{20}{c}}1&-3&5\\0&2&4\\3&0&5\end{array}} \right] \)
    5. \( \left[ {\begin{array}{*{20}{c}}2&1\\3&2\\-1&1\end{array}} \right] \left[ {\begin{array}{*{20}{c}}1&0&1\\-1&2&1\end{array}} \right] \)
    6. \( \left[ {\begin{array}{*{20}{c}}3&-1&3\\-1&0&2\end{array}} \right] \left[ {\begin{array}{*{20}{c}}2&-3\\1&0\\3&1\end{array}} \right] \)

    Solution:

    1. \[ \left[ {\begin{array}{*{20}{c}}a&b\\-b&a\end{array}} \right] \left[ {\begin{array}{*{20}{c}}a&-b\\b&a\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}a^2 + b^2&0\\0&a^2 + b^2\end{array}} \right] \]
    2. \[ \left[ {\begin{array}{*{20}{c}}1\\2\\3\end{array}} \right] \left[ {\begin{array}{*{20}{c}}2&3&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&3&4\\4&6&8\\6&9&12\end{array}} \right] \]
    3. \[ \left[ {\begin{array}{*{20}{c}}1&-2\\2&3\end{array}} \right] \left[ {\begin{array}{*{20}{c}}1&2&3\\2&3&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}-3&-4&1\\8&13&9\end{array}} \right] \]
    4. \[ \left[ {\begin{array}{*{20}{c}}2&3&4\\3&4&5\\4&5&6\end{array}} \right] \left[ {\begin{array}{*{20}{c}}1&-3&5\\0&2&4\\3&0&5\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}14&0&42\\18&-1&56\\22&-2&70\end{array}} \right] \]
    5. \[ \left[ {\begin{array}{*{20}{c}}2&1\\3&2\\-1&1\end{array}} \right] \left[ {\begin{array}{*{20}{c}}1&0&1\\-1&2&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&2&3\\1&4&5\\-2&2&0\end{array}} \right] \]
    6. \[ \left[ {\begin{array}{*{20}{c}}3&-1&3\\-1&0&2\end{array}} \right] \left[ {\begin{array}{*{20}{c}}2&-3\\1&0\\3&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}14&-6\\4&5\end{array}} \right] \]

  4. If \( A = \left[ {\begin{array}{*{20}{c}}1&2&-3\\5&0&2\\1&-1&1\end{array}} \right], B = \left[ {\begin{array}{*{20}{c}}3&-1&2\\4&2&5\\2&0&3\end{array}} \right] \), and \( C = \left[ {\begin{array}{*{20}{c}}4&1&2\\0&3&2\\1&-2&3\end{array}} \right] \), then compute \( (A + B) \) and \( (B – C) \). Also, verify that \( A + (B – C) = (A + B) – C \).

    Solution:

    \[ A + B = \left[ {\begin{array}{*{20}{c}}4&1&-1\\9&2&7\\3&-1&4\end{array}} \right] \]

    \[ B – C = \left[ {\begin{array}{*{20}{c}}-1&-2&0\\4&-1&3\\1&2&0\end{array}} \right] \]

    \[ A + (B – C) = \left[ {\begin{array}{*{20}{c}}0&0&-3\\9&-1&5\\2&1&1\end{array}} \right] \]

    \[ (A + B) – C = \left[ {\begin{array}{*{20}{c}}0&0&-3\\9&-1&5\\2&1&1\end{array}} \right] \]

    Hence, \( A + (B – C) = (A + B) – C \).

  5. If \( A + \left[ {\begin{array}{*{20}{c}}\frac{2}{3}&1&\frac{5}{3}\\\frac{1}{3}&\frac{2}{3}&\frac{4}{3}\\\frac{7}{3}&2&\frac{2}{3}\end{array}} \right] \) and \( B = \left[ {\begin{array}{*{20}{c}}\frac{2}{5}&\frac{3}{5}&1\\\frac{1}{5}&\frac{2}{5}&\frac{4}{5}\\\frac{7}{5}&\frac{6}{5}&\frac{2}{5}\end{array}} \right] \), then compute \( 3A – 5B \).

    Solution:

    \[ 3A – 5B = \left[ {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right] \]

  6. Simplify \( \cos \theta \left[ {\begin{array}{*{20}{c}}\cos \theta&\sin \theta\\-\sin \theta&\cos \theta\end{array}} \right] + \sin \theta \left[ {\begin{array}{*{20}{c}}\sin \theta&-\cos \theta\\\cos \theta&\sin \theta\end{array}} \right] \).

    Solution:

    \[ \cos \theta \left[ {\begin{array}{*{20}{c}}\cos \theta&\sin \theta\\-\sin \theta&\cos \theta\end{array}} \right] + \sin \theta \left[ {\begin{array}{*{20}{c}}\sin \theta&-\cos \theta\\\cos \theta&\sin \theta\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] \]

  7. Find \( X \) and \( Y \), if:

    1. \( X + Y = \left[ {\begin{array}{*{20}{c}}7&0\\2&5\end{array}} \right] \) and \( X – Y = \left[ {\begin{array}{*{20}{c}}3&0\\0&3\end{array}} \right] \)
    2. \( 2X + 3Y = \left[ {\begin{array}{*{20}{c}}2&3\\4&0\end{array}} \right] \) and \( 3X + 2Y = \left[ {\begin{array}{*{20}{c}}2&-2\\-1&5\end{array}} \right] \)

    Solution:

    1. \[ X = \left[ {\begin{array}{*{20}{c}}5&0\\1&4\end{array}} \right], \quad Y = \left[ {\begin{array}{*{20}{c}}2&0\\1&1\end{array}} \right] \]
    2. \[ X = \left[ {\begin{array}{*{20}{c}}\frac{2}{5}&\frac{-12}{5}\\\frac{-11}{5}&3\end{array}} \right], \quad Y = \left[ {\begin{array}{*{20}{c}}\frac{2}{5}&\frac{13}{5}\\\frac{14}{5}&-2\end{array}} \right] \]

  8. Find \( X \), if \( Y = \left[ {\begin{array}{*{20}{c}}3&2\\1&4\end{array}} \right] \) and \( 2X + Y = \left[ {\begin{array}{*{20}{c}}1&0\\-3&2\end{array}} \right] \).

    Solution:

    \[ X = \left[ {\begin{array}{*{20}{c}}-1&-1\\-2&-1\end{array}} \right] \]

  9. Find \( x \) and \( y \), if \( 2\left[ {\begin{array}{*{20}{c}}1&3\\0&x\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}y&0\\1&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5&6\\1&8\end{array}} \right] \).

    Solution:

    \( x = 3 \) and \( y = 3 \).

  10. Solve the equation for \( x, y, z \), and \( t \), if:

    \[ 2\left[ {\begin{array}{*{20}{c}}x&z\\y&t\end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}}1&-1\\0&2\end{array}} \right] = 3\left[ {\begin{array}{*{20}{c}}3&5\\4&6\end{array}} \right] \]

    Solution:

    \( x = 3, y = 6, z = 9 \), and \( t = 6 \).

  11. If \( x\left[ {\begin{array}{*{20}{c}}2\\3\end{array}} \right] + y\left[ {\begin{array}{*{20}{c}}-1\\1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}10\\5\end{array}} \right] \), find the values of \( x \) and \( y \).

    Solution:

    \( x = 3 \) and \( y = -4 \).

  12. Given \( 3\left[ {\begin{array}{*{20}{c}}x&y\\z&w\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}x&6\\-1&2w\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}4&x+y\\z+w&3\end{array}} \right] \), find the values of \( x, y, z \), and \( w \).

    Solution:

    \( x = 2, y = 4, z = 1 \), and \( w = 3 \).

  13. If \( F(x) = \left[ {\begin{array}{*{20}{c}}\cos x&-\sin x&0\\\sin x&\cos x&0\\0&0&1\end{array}} \right] \), then show that \( F(x) \cdot F(y) = F(x + y) \).

    Solution:

    Verified to be true.

  14. Show that:

    1. \( \left[ {\begin{array}{*{20}{c}}5&-1\\6&7\end{array}} \right] \left[ {\begin{array}{*{20}{c}}2&1\\3&4\end{array}} \right] \neq \left[ {\begin{array}{*{20}{c}}2&1\\3&4\end{array}} \right] \left[ {\begin{array}{*{20}{c}}5&-1\\6&7\end{array}} \right] \)
    2. \( \left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&0\\1&1&0\end{array}} \right] \left[ {\begin{array}{*{20}{c}}-1&1&0\\0&-1&1\\2&3&4\end{array}} \right] \neq \left[ {\begin{array}{*{20}{c}}-1&1&0\\0&-1&1\\2&3&4\end{array}} \right] \left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&0\\1&1&0\end{array}} \right] \)

    Solution:

    Both inequalities are verified to be true.

  15. Find \( A^2 – 5A + 6I \), if \( A = \left[ {\begin{array}{*{20}{c}}2&0&1\\2&1&3\\1&-1&0\end{array}} \right] \).

    Solution:

    \[ A^2 – 5A + 6I = \left[ {\begin{array}{*{20}{c}}1&-1&-3\\-1&-1&-10\\-5&4&4\end{array}} \right] \]

  16. If \( A = \left[ {\begin{array}{*{20}{c}}1&0&2\\0&2&1\\2&0&3\end{array}} \right] \), prove that \( A^3 – 6A^2 + 7A + 2I = O \).

    Solution:

    Verified to be true.

  17. If \( A = \left[ {\begin{array}{*{20}{c}}3&-2\\4&-2\end{array}} \right] \) and \( I = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] \), find \( k \) so that \( A^2 = kA – 2I \).

    Solution:

    \( k = 1 \).

  18. If \( A = \left[ {\begin{array}{*{20}{c}}0&-\tan \frac{\alpha}{2}\\\tan \frac{\alpha}{2}&0\end{array}} \right] \) and \( I \) is the identity matrix of order 2, then show that \( I + A = (I – A) \left[ {\begin{array}{*{20}{c}}\cos \alpha&-\sin \alpha\\\sin \alpha&\cos \alpha\end{array}} \right] \).

    Solution:

    Verified to be true.

  19. A trust fund has Rs. 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs. 30,000 among the two types of bonds, if the trust fund obtains an annual total interest of:

    1. Rs. 1800
    2. Rs. 2000

    Solution:

    1. Invest Rs. 15,000 at 5% and Rs. 15,000 at 7%.
    2. Invest Rs. 5,000 at 5% and Rs. 25,000 at 7%.

  20. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, and 10 dozen economics books. Their selling prices are Rs. 80, Rs. 60, and Rs. 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

    Solution:

    Total amount received = Rs. 20,160.

  21. Choose the correct answer:

    1. The restriction on \( n, k \), and \( p \) so that \( PY + WY \) will be defined are:
      1. \( k = 3, p = n \)
      2. \( k \) is arbitrary, \( p = 2 \)
      3. \( p \) is arbitrary, \( k = 3 \)
      4. \( k = 2, p = 3 \)

      Solution:

      (a) \( k = 3, p = n \).

    2. If \( n = p \), then the order of the matrix \( 7X – 5Z \) is:
      1. \( p \times 2 \)
      2. \( 2 \times n \)
      3. \( n \times 3 \)
      4. \( p \times n \)

      Solution:

      (b) \( 2 \times n \).