Matrices - Test Papers

 CBSE Test Paper 01

Chapter 3 Matrices


  1. If A=[113526213]. Then |A| is 
    1. none of these
    2. Idempotent
    3. Nilpotent
    4. Symmetric
  2. |111e02222|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 A=[2345], Prove that A – At is a skew – symmetric matrix. 
  10. A=[2456], Prove that A + A' is a symmetric matrix. 
  11. The no. of all possible matrics of order 3 × 3 with each entry as 0 or 1 is- 
  12. Find the matrix X so that X[123456]=[789246]
  13. If A is a square matrix, such that A2 = A, then (I + A)3 – 7A is equal to 
  14. If A=[2456], then Prove that A + A' is a symmetric matrix. 
  15. If f(x) = x2 – 5x + 7 and A=[3112] then find f(A). 
  16. Find the matrix A such that [211034]A=[181012592215]
  17. Show that: 
    1. [5167][2134][2134][5167]
    2. [123010110][110011234] [110011234][123010110]
  18. A=[0tanα2tanα20],
    Prove I+A=(IA)[cosαsinαsinαcosα]

CBSE Test Paper 01
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: |111e02222|=2|111e02111|=0, 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: a1a2=b1b2c1c2, for given system of equations we have: 12=241122.
  1. skew symmetric
  2. Null

  3. kA'
  4. P = A - At
    =[2345]+[2435]
    =[0110]
    P=[0110]
    P=[0110]
    P' = - P
    Proved.
  5. P=A+A=[2456]+[2546]
    P=[49912]
    P=[49912]
    Thus, P' = P.
  6. 29=512
  7. Let X=[abcd]
    [abcd][123456]=[789246]
    [a+4b2a+5b3a+6bc+4d2c+5d3c+6d] =[789246]
    Hence, a = 1, b = -2, c = 2, d = 0
    X=[1220]
  8. (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
  9. P = A + A' =[2456]+[2546]
    P=[49912]
    P=[49912]
    P' = P
    Therefore P = P'
    Hence A+A' is symmetric.

     

  10. A=[3112]
    A2=AA=[3112][3112]=[8553]
    Now, f(A) = A2 - 5A + 7I
    =[8553]5[3112]+7[1001]
    f(A)=[815+755+05+5+0310+7]=[0000]
    Therefore, f(A)=[0000]

  11. We have, [211034]A=[181012592215]
    From the given equation, it is clear that order of A should be 2×3
    Let A=[abcdef]
     [211034][abcdef]=[181012592215]
    [2ad2be2cfa+0db+0.ec+0.f3a+4d3b+4e3c+4f] =[181012592215]
    [2ad2be2cfabc3a+4d3b+4e3c+4f] =[181012592215]
    By equality of matrices, we get
    a = 1, b = -2, c = -5
    and 2a - d = -1  d = 2a + 1 = 3;
     2b - e = -8  e = 2(-2) + 8 =4
    2c - f = -10  f = 2c + 10 = 0
    A=[125340]
    1. L.H.S. =[5167][2134] =[5(2)+(1)35(1)+(1)46(2)+7(3)6(1)+7(4)]=[713334]
      R.H.S. =[2134][5167] =[2(5)+1(6)2(1)+1(7)3(5)+4(6)3(1)+4(7)]=[1653925]
       L.H.S.  R.H.S.
    2. L.H.S. =[123010110][110011234]
      =[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)]
      =[5814011101]
      R.H.S. =[110011234][123010110]
      =[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)]
      =[1131006116]
       L.H.S.  R.H.S.
  12. Put tanα2=t
    A=[0tt0]
    I+A=[1001]+[0tt0]
    =[1tt1]
    IA=[1001][0tt0]
    =[1001]+[0tt0]
    =[1tt1]
    L.H.S. =(IA)[cosαsinαsinαcosα]
    =(IA)[1tan2α21+tan2α22tan2α21+tan2α22tan2α21+tan2α21tan2α21+tan2α2]
    =[1tt1][1t21+t22t1+t22t1+t21t21+t2]
    =[1t21+t2+t.2t1+t22t1+t2+t(1t21+t2)t(1t21+t2)+2t1+t2t(2t1+t2)+(1t21+t2)]
    =[1t2+2t21+t22t+tt31+t2t+t3+2t1+t22t2+1t21+t2]
    =[1+t21+t2t3t1+t2t3+t1+t2t2+11+t2]
    =[1t(1+t2)1+t2t(1+t2)1+t2t2+11+t2]
    =[1tt1]
    L.H.S = R.H.S
    Hence proved