NCERT 3.3 class 12 maths solutions Matrices

NCERT 3.3 class 12 maths solutions Matrices

1. Find the transpose of each of the following matrices:

   1. \[ \left[ {\begin{array}{*{20}{c}}5\\\frac{1}{2}\\-1\end{array}} \right] \]
   2. \[ \left[ {\begin{array}{*{20}{c}}1&-1\\2&3\end{array}} \right] \]
   3. \[ \left[ {\begin{array}{*{20}{c}}-1&5&6\\\sqrt{3}&5&6\\2&3&-1\end{array}} \right] \]

   Solution:

   1. \[ \left[ {\begin{array}{*{20}{c}}5&\frac{1}{2}&-1\end{array}} \right] \]
   2. \[ \left[ {\begin{array}{*{20}{c}}1&2\\-1&3\end{array}} \right] \]
   3. \[ \left[ {\begin{array}{*{20}{c}}-1&\sqrt{3}&2\\5&5&3\\6&6&-1\end{array}} \right] \]

2. If \( A = \left[ {\begin{array}{*{20}{c}}-1&2&3\\5&7&9\\-2&1&1\end{array}} \right] \) and \( B = \left[ {\begin{array}{*{20}{c}}-4&1&-5\\1&2&0\\1&3&1\end{array}} \right] \), then verify that:

   1. \( (A + B)’ = A’ + B’ \)
   2. \( (A – B)’ = A’ – B’ \)

   Solution:

   1. \[ A + B = \left[ {\begin{array}{*{20}{c}}-5&3&-2\\6&9&9\\-1&4&2\end{array}} \right] \] \[ (A + B)’ = \left[ {\begin{array}{*{20}{c}}-5&6&-1\\3&9&4\\-2&9&2\end{array}} \right] \] \[ A’ + B’ = \left[ {\begin{array}{*{20}{c}}-5&6&-1\\3&9&4\\-2&9&2\end{array}} \right] \] Hence, \( (A + B)’ = A’ + B’ \).
   2. \[ A – B = \left[ {\begin{array}{*{20}{c}}3&1&8\\4&5&9\\-3&-2&0\end{array}} \right] \] \[ (A – B)’ = \left[ {\begin{array}{*{20}{c}}3&4&-3\\1&5&-2\\8&9&0\end{array}} \right] \] \[ A’ – B’ = \left[ {\begin{array}{*{20}{c}}3&4&-3\\1&5&-2\\8&9&0\end{array}} \right] \] Hence, \( (A – B)’ = A’ – B’ \).

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

   1. \( (A + B)’ = A’ + B’ \)
   2. \( (A – B)’ = A’ – B’ \)

   Solution:

   1. \[ A + B = \left[ {\begin{array}{*{20}{c}}2&1&1\\5&4&4\end{array}} \right] \] \[ (A + B)’ = \left[ {\begin{array}{*{20}{c}}2&5\\1&4\\1&4\end{array}} \right] \] \[ A’ + B’ = \left[ {\begin{array}{*{20}{c}}2&5\\1&4\\1&4\end{array}} \right] \] Hence, \( (A + B)’ = A’ + B’ \).
   2. \[ A – B = \left[ {\begin{array}{*{20}{c}}4&-3&-1\\3&0&-2\end{array}} \right] \] \[ (A – B)’ = \left[ {\begin{array}{*{20}{c}}4&3\\-3&0\\-1&-2\end{array}} \right] \] \[ A’ – B’ = \left[ {\begin{array}{*{20}{c}}4&3\\-3&0\\-1&-2\end{array}} \right] \] Hence, \( (A – B)’ = A’ – B’ \).

4. If \( A’ = \left[ {\begin{array}{*{20}{c}}-2&3\\1&2\end{array}} \right] \) and \( B = \left[ {\begin{array}{*{20}{c}}-1&0\\1&2\end{array}} \right] \), then find \( (A + 2B)’ \).

   Solution:

   \[ (A + 2B)’ = \left[ {\begin{array}{*{20}{c}}-4&5\\1&6\end{array}} \right] \]

5. For the matrices \( A \) and \( B \), verify that \( (AB)’ = B’A’ \), where:

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

   Solution:

   1. \[ AB = \left[ {\begin{array}{*{20}{c}}-1&2&1\\4&-8&-4\\-3&6&3\end{array}} \right] \] \[ (AB)’ = \left[ {\begin{array}{*{20}{c}}-1&4&-3\\2&-8&6\\1&-4&3\end{array}} \right] \] \[ B’A’ = \left[ {\begin{array}{*{20}{c}}-1&4&-3\\2&-8&6\\1&-4&3\end{array}} \right] \] Hence, \( (AB)’ = B’A’ \).
   2. \[ AB = \left[ {\begin{array}{*{20}{c}}0&0&0\\1&5&7\\2&10&14\end{array}} \right] \] \[ (AB)’ = \left[ {\begin{array}{*{20}{c}}0&1&2\\0&5&10\\0&7&14\end{array}} \right] \] \[ B’A’ = \left[ {\begin{array}{*{20}{c}}0&1&2\\0&5&10\\0&7&14\end{array}} \right] \] Hence, \( (AB)’ = B’A’ \).

6. If:

   1. \( A = \left[ {\begin{array}{*{20}{c}}\cos \alpha&\sin \alpha\\-\sin \alpha&\cos \alpha\end{array}} \right] \), then verify that \( A’A = I \).
   2. \( A = \left[ {\begin{array}{*{20}{c}}\sin \alpha&\cos \alpha\\-\cos \alpha&\sin \alpha\end{array}} \right] \), then verify that \( A’A = I \).

   Solution:

   1. \[ A’A = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = I \] Hence, \( A’A = I \).
   2. \[ A’A = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = I \] Hence, \( A’A = I \).

7. Show that the matrix \( A = \left[ {\begin{array}{*{20}{c}}1&-1&5\\-1&2&1\\5&1&3\end{array}} \right] \) is a symmetric matrix.

   Solution:

   \[ A’ = \left[ {\begin{array}{*{20}{c}}1&-1&5\\-1&2&1\\5&1&3\end{array}} \right] = A \] Hence, \( A \) is a symmetric matrix.

8. Show that the matrix \( A = \left[ {\begin{array}{*{20}{c}}0&1&-1\\-1&0&1\\1&-1&0\end{array}} \right] \) is a skew-symmetric matrix.

   Solution:

   \[ A’ = \left[ {\begin{array}{*{20}{c}}0&-1&1\\1&0&-1\\-1&1&0\end{array}} \right] = -A \] Hence, \( A \) is a skew-symmetric matrix.

9. For the matrix \( A = \left[ {\begin{array}{*{20}{c}}1&5\\6&7\end{array}} \right] \), verify that:

   1. \( (A + A’) \) is a symmetric matrix.
   2. \( (A – A’) \) is a skew-symmetric matrix.

   Solution:

   1. \[ A + A’ = \left[ {\begin{array}{*{20}{c}}2&11\\11&14\end{array}} \right] \] Hence, \( (A + A’) \) is a symmetric matrix.
   2. \[ A – A’ = \left[ {\begin{array}{*{20}{c}}0&-1\\1&0\end{array}} \right] \] Hence, \( (A – A’) \) is a skew-symmetric matrix.

10. Find \( (A + A’) \) and \( \frac{1}{2}(A – A’) \), when \( A = \left[ {\begin{array}{*{20}{c}}0&a&b\\-a&0&c\\-b&-c&0\end{array}} \right] \).

    Solution:

    \[ A + A’ = \left[ {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right] \] \[ \frac{1}{2}(A – A’) = \left[ {\begin{array}{*{20}{c}}0&a&b\\-a&0&c\\-b&-c&0\end{array}} \right] \]

11. Express the following matrices as the sum of a symmetric and a skew-symmetric matrix:

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

    Solution:

    1. \[ P = \left[ {\begin{array}{*{20}{c}}3&3\\3&-1\end{array}} \right], \quad Q = \left[ {\begin{array}{*{20}{c}}0&2\\-2&0\end{array}} \right] \] Hence, \( P + Q = A \).
    2. \[ P = \left[ {\begin{array}{*{20}{c}}6&-2&2\\-2&3&-1\\2&-1&3\end{array}} \right], \quad Q = \left[ {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right] \] Hence, \( P + Q = A \).
    3. \[ P = \left[ {\begin{array}{*{20}{c}}3&\frac{1}{2}&-\frac{5}{2}\\\frac{1}{2}&-2&-2\\-\frac{5}{2}&-2&2\end{array}} \right], \quad Q = \left[ {\begin{array}{*{20}{c}}0&\frac{5}{2}&\frac{3}{2}\\-\frac{5}{2}&0&3\\-\frac{3}{2}&-3&0\end{array}} \right] \] Hence, \( P + Q = A \).
    4. \[ P = \left[ {\begin{array}{*{20}{c}}1&2\\2&2\end{array}} \right], \quad Q = \left[ {\begin{array}{*{20}{c}}0&3\\-3&0\end{array}} \right] \] Hence, \( P + Q = A \).

12. Choose the correct answer:

    1. If \( A, B \) are symmetric matrices of the same order, then \( AB – BA \) is a:
    1. Skew symmetric matrix
    2. Symmetric matrix
    3. Zero matrix
    4. Identity matrix

    Solution:

    (a) Skew symmetric matrix.

    2. If \( A = \left[ {\begin{array}{*{20}{c}}\cos \alpha&-\sin \alpha\\\sin \alpha&\cos \alpha\end{array}} \right] \), then \( A + A’ = I \), if the value of \( \alpha \) is:
    1. \( \frac{\pi}{6} \)
    2. \( \frac{\pi}{3} \)
    3. \( \pi \)
    4. \( \frac{3\pi}{2} \)

    Solution:

    (b) \( \frac{\pi}{3} \).