Matrices - Test Papers

 CBSE Test Paper 01

Chapter 3 Matrices

---

1. If A=⎡⎣⎢15−212−136−3⎤⎦⎥$A=\left[\begin{array}{ccc}1& 1& 3\\ 5& 2& 6\\ -2& -1& -3\end{array}\right]$. Then |A| is 
   1. none of these
   2. Idempotent
   3. Nilpotent
   4. Symmetric
2. ∣∣∣∣1e210212–√2∣∣∣∣$\left|\begin{array}{ccc}1& 1& 1\\ e& 0& \sqrt{2}\\ 2& 2& 2\end{array}\right|$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=[2435]$A=\left[\begin{array}{cc}2& 3\\ 4& 5\end{array}\right]$, Prove that A – At is a skew – symmetric matrix. 
10. A=[2546]$A=\left[\begin{array}{cc}2& 4\\ 5& 6\end{array}\right]$, Prove that A + A' is a symmetric matrix. 
11. The no. of all possible matrics of order 3 ×$\times$ 3 with each entry as 0 or 1 is- 
12. Find the matrix X so that X[123456]=[−7−8−9246]$X\left[\begin{array}{c}\begin{array}{ccc}1& 2& 3\end{array}\\ \begin{array}{ccc}4& 5& 6\end{array}\end{array}\right]=\left[\begin{array}{c}\begin{array}{ccc}-7& -8& -9\end{array}\\ \begin{array}{ccc}2& 4& 6\end{array}\end{array}\right]$. 
13. If A is a square matrix, such that A2 = A, then (I + A)3 – 7A is equal to 
14. If A=[2546]$A=\left[\begin{array}{cc}2& 4\\ 5& 6\end{array}\right]$, then Prove that A + A' is a symmetric matrix. 
15. If f(x) = x2 – 5x + 7 and A=[3−112]$A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$ then find f(A). 
16. Find the matrix A such that ⎡⎣⎢2−110−34⎤⎦⎥A=⎡⎣⎢−119−8−222−10−515⎤⎦⎥$\left[\begin{array}{c}\begin{array}{cc}2& -1\end{array}\\ \begin{array}{cc}1& 0\end{array}\\ \begin{array}{cc}-3& 4\end{array}\end{array}\right]A=\left[\begin{array}{ccc}-1& -8& -10\\ 1& -2& -5\\ 9& 22& 15\end{array}\right]$. 
17. Show that: 

    1. [56−17][2314]≠[2314][56−17]$\left[\begin{array}{cc}5& -1\\ 6& 7\end{array}\right]\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]\ne \left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]\left[\begin{array}{cc}5& -1\\ 6& 7\end{array}\right]$
    2. ⎡⎣⎢101211300⎤⎦⎥⎡⎣⎢−1021−13014⎤⎦⎥$\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 0\\ 1& 1& 0\end{array}\right]\left[\begin{array}{ccc}-1& 1& 0\\ 0& -1& 1\\ 2& 3& 4\end{array}\right]$ ≠⎡⎣⎢−1021−13014⎤⎦⎥⎡⎣⎢101211300⎤⎦⎥$\ne \left[\begin{array}{ccc}-1& 1& 0\\ 0& -1& 1\\ 2& 3& 4\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 0\\ 1& 1& 0\end{array}\right]$
18. A=[0tanα2−tanα20]$A=\left[\begin{array}{cc}0& -\tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 0\end{array}\right]$,
    Prove I+A=(I−A)[cosαsinα−sinαcosα]$I+A=(I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$. 

CBSE Test Paper 01
Chapter 3 Matrices

---

Solution

1. 3. Nilpotent
      Explanation: The given matrix A is nilpotent, because |A| = 0,as determinant of a nilpotent matrix is zero.
2. 1. 0
      Explanation: ∣∣∣∣1e210212–√2∣∣∣∣=2∣∣∣∣1e110112–√1∣∣∣∣=0$\left|\begin{array}{ccc}1& 1& 1\\ e& 0& \sqrt{2}\\ 2& 2& 2\end{array}\right|=2\left|\begin{array}{ccc}1& 1& 1\\ e& 0& \sqrt{2}\\ 1& 1& 1\end{array}\right|=0$, because, row 1 and row 3 are identical.
3. 4. A2 = A
      Explanation: If the product of any square matrix with itself is the matrix itself, then the matrix is called Idempotent.
4. 4. (M-1)-1 = (M-1)1
      Explanation: Clearly , (M-1)-1 = (M-1)1 is not true.
5. 4. no solution
      Explanation: For no solution , we have: a1a2=b1b2≠c1c2$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$, for given system of equations we have: 1−2=2−4≠1122$\frac{1}{-2}=\frac{2}{-4}\ne \frac{11}{22}$.
6. skew symmetric

7. Null

8. kA'
9. P = A - At
   =[2435]+[−2−3−4−5]$=\left[\begin{array}{cc}2& 3\\ 4& 5\end{array}\right]+\left[\begin{array}{cc}-2& -4\\ -3& -5\end{array}\right]$
   =[01−10]$=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$
   P′=[0−110]$P^{\prime }=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$
   P′=−[01−10]$P^{\prime }=-\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$
   P' = - P
   Proved.
10. P=A+A′=[2546]+[2456]$P=A+A^{\prime }=\left[\begin{array}{cc}2& 4\\ 5& 6\end{array}\right]+\left[\begin{array}{cc}2& 5\\ 4& 6\end{array}\right]$
    P=[49912]$P=\left[\begin{array}{cc}4& 9\\ 9& 12\end{array}\right]$
    P′=[49912]$P^{\prime }=\left[\begin{array}{cc}4& 9\\ 9& 12\end{array}\right]$
    Thus, P' = P.
11. 29=512$2^9=512$
12. Let X=[acbd]$X=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
    ∴[acbd][123456]=[−7−8−9246]$\therefore \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}\begin{array}{ccc}1& 2& 3\end{array}\\ \begin{array}{ccc}4& 5& 6\end{array}\end{array}\right]=\left[\begin{array}{c}\begin{array}{ccc}-7& -8& -9\end{array}\\ \begin{array}{ccc}2& 4& 6\end{array}\end{array}\right]$
    [a+4b2a+5b3a+6bc+4d2c+5d3c+6d]$\left[\begin{array}{c}\begin{array}{ccc}a+4b& 2a+5b& 3a+6b\end{array}\\ \begin{array}{ccc}c+4d& 2c+5d& 3c+6d\end{array}\end{array}\right]$ =[−7−8−9246]$=\left[\begin{array}{c}\begin{array}{ccc}-7& -8& -9\end{array}\\ \begin{array}{ccc}2& 4& 6\end{array}\end{array}\right]$
    Hence, a = 1, b = -2, c = 2, d = 0
    X=[12−20]$X=\left[\begin{array}{cc}1& -2\\ 2& 0\end{array}\right]$
13. (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
14. P = A + A' =[2546]+[2456]$=\left[\begin{array}{cc}2& 4\\ 5& 6\end{array}\right]+\left[\begin{array}{cc}2& 5\\ 4& 6\end{array}\right]$
    P=[49912]$P=\left[\begin{array}{cc}4& 9\\ 9& 12\end{array}\right]$
    P′=[49912]$P^{\prime }=\left[\begin{array}{cc}4& 9\\ 9& 12\end{array}\right]$
    P' = P
    Therefore P = P'
    Hence A+A' is symmetric.

     

15. A=[3−112]$A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$
    A2=AA=[3−112][3−112]=[8−553]$A^2=AA=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]$
    Now, f(A) = A2 - 5A + 7I
    =[8−553]−5[3−112]+7[1001]$=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]-5\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]+7\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
    f(A)=[8−15+7−5+5+05−5+03−10+7]$f(A)=\left[\begin{array}{cc}8-15+7& 5-5+0\\ -5+5+0& 3-10+7\end{array}\right]$=[0000]$=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
    Therefore, f(A)=[0000]$f(A)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

16. We have, ⎡⎣⎢2−110−34⎤⎦⎥A=⎡⎣⎢−119−8−222−10−515⎤⎦⎥$\left[\begin{array}{c}\begin{array}{cc}2& -1\end{array}\\ \begin{array}{cc}1& 0\end{array}\\ \begin{array}{cc}-3& 4\end{array}\end{array}\right]A=\left[\begin{array}{ccc}-1& -8& -10\\ 1& -2& -5\\ 9& 22& 15\end{array}\right]$
    From the given equation, it is clear that order of A should be 2×3$2\times 3$
    Let A=[abcdef]$A=\left[\begin{array}{c}\begin{array}{ccc}a& b& c\end{array}\\ \begin{array}{ccc}d& e& f\end{array}\end{array}\right]$
    ∴$\therefore$ ⎡⎣⎢21−3−104⎤⎦⎥[adbecf]=⎡⎣⎢−119−8−222−10−515⎤⎦⎥$\left[\begin{array}{cc}2& -1\\ 1& 0\\ -3& 4\end{array}\right]\left[\begin{array}{ccc}a& b& c\\ d& e& f\end{array}\right]=\left[\begin{array}{ccc}-1& -8& -10\\ 1& -2& -5\\ 9& 22& 15\end{array}\right]$
    ⇒⎡⎣⎢2a−da+0d−3a+4d2b−eb+0.e−3b+4e2c−fc+0.f−3c+4f⎤⎦⎥$\Rightarrow \left[\begin{array}{ccc}2a-d& 2b-e& 2c-f\\ a+0d& b+0.e& c+0.f\\ -3a+4d& -3b+4e& -3c+4f\end{array}\right]$ =⎡⎣⎢−119−8−222−10−515⎤⎦⎥$=\left[\begin{array}{ccc}-1& -8& -10\\ 1& -2& -5\\ 9& 22& 15\end{array}\right]$
    ⇒⎡⎣⎢2a−da−3a+4d2b−eb−3b+4e2c−fc−3c+4f⎤⎦⎥$\Rightarrow \left[\begin{array}{ccc}2a-d& 2b-e& 2c-f\\ a& b& c\\ -3a+4d& -3b+4e& -3c+4f\end{array}\right]$ =⎡⎣⎢−119−8−222−10−515⎤⎦⎥$=\left[\begin{array}{ccc}-1& -8& -10\\ 1& -2& -5\\ 9& 22& 15\end{array}\right]$
    By equality of matrices, we get
    a = 1, b = -2, c = -5
    and 2a - d = -1 ⇒$\Rightarrow$ d = 2a + 1 = 3;
    ⇒$\Rightarrow$ 2b - e = -8 ⇒$\Rightarrow$ e = 2(-2) + 8 =4
    2c - f = -10 ⇒$\Rightarrow$ f = 2c + 10 = 0
    ∴A=[1−2−5340]$\therefore A=\left[\begin{array}{c}\begin{array}{ccc}1& -2& -5\end{array}\\ \begin{array}{ccc}3& 4& 0\end{array}\end{array}\right]$
17. 1. L.H.S. =[56−17][2314]$=\left[\begin{array}{cc}5& -1\\ 6& 7\end{array}\right]\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]$ =[5(2)+(−1)36(2)+7(3)5(1)+(−1)46(1)+7(4)]=[733134]$=\left[\begin{array}{cc}5(2)+(-1)3& 5(1)+(-1)4\\ 6(2)+7(3)& 6(1)+7(4)\end{array}\right]=\left[\begin{array}{cc}7& 1\\ 33& 34\end{array}\right]$
       R.H.S. =[2314][56−17]$=\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]\left[\begin{array}{cc}5& -1\\ 6& 7\end{array}\right]$ =[2(5)+1(6)3(5)+4(6)2(−1)+1(7)3(−1)+4(7)]=[1639525]$=\left[\begin{array}{cc}2(5)+1(6)& 2(-1)+1(7)\\ 3(5)+4(6)& 3(-1)+4(7)\end{array}\right]=\left[\begin{array}{cc}16& 5\\ 39& 25\end{array}\right]$
       ∴$\therefore$ L.H.S. ≠$\ne$ R.H.S.
    2. L.H.S. =⎡⎣⎢101211300⎤⎦⎥⎡⎣⎢−1021−13014⎤⎦⎥$=\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 0\\ 1& 1& 0\end{array}\right]\left[\begin{array}{ccc}-1& 1& 0\\ 0& -1& 1\\ 2& 3& 4\end{array}\right]$
       =⎡⎣⎢1(−1)+2(0)+3(2)0(−1)+1(0)+0(2)1(−1)+1(0)+0(2)1(1)+2(−1)+3(3)0(1)+1(−1)+0(3)1(1)+1(−1)+0(3)1(0)+2(1)+3(4)0(0)+1(1)+0(4)1(0)+1(1)+0(4)⎤⎦⎥$=\left[\begin{array}{ccc}1(-1)+2(0)+3(2)& 1(1)+2(-1)+3(3)& 1(0)+2(1)+3(4)\\ 0(-1)+1(0)+0(2)& 0(1)+1(-1)+0(3)& 0(0)+1(1)+0(4)\\ 1(-1)+1(0)+0(2)& 1(1)+1(-1)+0(3)& 1(0)+1(1)+0(4)\end{array}\right]$
       =⎡⎣⎢50−18−101411⎤⎦⎥$=\left[\begin{array}{ccc}5& 8& 14\\ 0& -1& 1\\ -1& 0& 1\end{array}\right]$
       R.H.S. =⎡⎣⎢−1021−13014⎤⎦⎥⎡⎣⎢101211300⎤⎦⎥$=\left[\begin{array}{ccc}-1& 1& 0\\ 0& -1& 1\\ 2& 3& 4\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 0\\ 1& 1& 0\end{array}\right]$
       =⎡⎣⎢−1(1)+1(0)+0(1)0(1)+(−1)0+1(1)2(1)+3(0)+4(1)(−1)2+1(1)+0(1)(0)2+1(−1)+1(1)2(2)+3(1)+4(1)(−1)3+1(0)+0(0)(0)3+0(−1)+1(0)2(3)+3(0)+4(0)⎤⎦⎥$=\left[\begin{array}{ccc}-1(1)+1(0)+0(1)& (-1)2+1(1)+0(1)& (-1)3+1(0)+0(0)\\ 0(1)+(-1)0+1(1)& (0)2+1(-1)+1(1)& (0)3+0(-1)+1(0)\\ 2(1)+3(0)+4(1)& 2(2)+3(1)+4(1)& 2(3)+3(0)+4(0)\end{array}\right]$
       =⎡⎣⎢−116−1011−306⎤⎦⎥$=\left[\begin{array}{ccc}-1& -1& -3\\ 1& 0& 0\\ 6& 11& 6\end{array}\right]$
       ∴$\therefore$ L.H.S. ≠$\ne$ R.H.S.
18. Put tanα2=t$\tan \frac{\alpha }{2}=t$
    A=[0t−t0]$A=\left[\begin{array}{cc}0& -t\\ t& 0\end{array}\right]$
    I+A=[1001]+[0t−t0]$I+A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}0& -t\\ t& 0\end{array}\right]$
    =[1t−t1]$=\left[\begin{array}{cc}1& -t\\ t& 1\end{array}\right]$
    I−A=[1001]−[0t−t0]$I-A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}0& -t\\ t& 0\end{array}\right]$
    =[1001]+[0−tt0]$=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}0& t\\ -t& 0\end{array}\right]$
    =[1−tt1]$=\left[\begin{array}{cc}1& t\\ -t& 1\end{array}\right]$
    L.H.S. =(I−A)[cosαsinα−sinαcosα]$=(I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$
    =(I−A)⎡⎣⎢⎢1−tan2α21+tan2α22tan2α21+tan2α2−2tan2α21+tan2α21−tan2α21+tan2α2⎤⎦⎥⎥$=(I-A)\left[\begin{array}{cc}\frac{1-{\tan }^2\frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}& \frac{-2{\tan }^2\frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}\\ \frac{2{\tan }^2\frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}& \frac{1-{\tan }^2\frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}\end{array}\right]$
    =[1−tt1]⎡⎣⎢1−t21+t2−2t1+t2−2t1+t21−t21+t2⎤⎦⎥$=\left[\begin{array}{cc}1& t\\ -t& 1\end{array}\right]\left[\begin{array}{cc}\frac{1-t^2}{1+t^2}& \frac{-2t}{1+t^2}\\ \frac{-2t}{1+t^2}& \frac{1-t^2}{1+t^2}\end{array}\right]$
    =⎡⎣⎢1−t21+t2+t.2t1+t2−t(1−t21+t2)+2t1+t2−2t1+t2+t(1−t21+t2)−t(−2t1+t2)+(1−t21+t2)⎤⎦⎥$=\left[\begin{array}{cc}\frac{1-t^2}{1+t^2}+\frac{t.2t}{1+t^2}& \frac{-2t}{1+t^2}+t\left(\frac{1-t^2}{1+t^2}\right)\\ -t\left(\frac{1-t^2}{1+t^2}\right)+\frac{2t}{1+t^2}& -t\left(\frac{-2t}{1+t^2}\right)+\left(\frac{1-t^2}{1+t^2}\right)\end{array}\right]$
    =⎡⎣⎢1−t2+2t21+t2−t+t3+2t1+t2−2t+t−t31+t22t2+1−t21+t2⎤⎦⎥$=\left[\begin{array}{cc}\frac{1-t^2+2t^2}{1+t^2}& \frac{-2t+t-t^3}{1+t^2}\\ \frac{-t+t^3+2t}{1+t^2}& \frac{2t^2+1-t^2}{1+t^2}\end{array}\right]$
    =⎡⎣⎢1+t21+t2t3+t1+t2−t3−t1+t2t2+11+t2⎤⎦⎥$=\left[\begin{array}{cc}\frac{1+t^2}{1+t^2}& \frac{-t^3-t}{1+t^2}\\ \frac{t^3+t}{1+t^2}& \frac{t^2+1}{1+t^2}\end{array}\right]$
    =⎡⎣⎢1t(1+t2)1+t2−t(1+t2)1+t2t2+11+t2⎤⎦⎥$=\left[\begin{array}{cc}1& \frac{-t(1+t^2)}{1+t^2}\\ \frac{t(1+t^2)}{1+t^2}& \frac{t^2+1}{1+t^2}\end{array}\right]$
    =[1t−t1]$=\left[\begin{array}{cc}1& -t\\ t& 1\end{array}\right]$
    L.H.S = R.H.S
    Hence proved