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Mathematics Test - 6
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Mathematics Test - 6
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  • Question 1/15
    3 / -1

    If A=[aij]3×3 is a square matrix so that

    aij=i2−j2, then A is a

    Solutions
    Given \(A=\left[a_{i j}\right]_{(3 \times 3)}\)
    where, \(a_{i j}=i^{2}-j^{2}\)
    \(\therefore a_{i j}=0\) if \(i=j\)
    Now, \(a_{12}=1^{2}-2^{2}=-3\)
    \(a_{13}=1^{2}-3^{2}=-8\)
    \(a_{21}=2^{2}-1^{2}=3\)
    \(a_{23}=2^{2}-3^{2}=-5\)
    \(a_{31}=3^{2}-1^{2}=8\)
    \(a_{32}=3^{2}-2^{2}=5\)
    \(\therefore A=\left[\begin{array}{ccc}0 & -3 & -8 \\ 3 & 0 & -5 \\ 8 & 5 & 0\end{array}\right]\)
    Here, \(A^{T}=-A\)
    \(\therefore A\) is a skew-symmetric matrix.
    Hence option C is correct.
  • Question 2/15
    3 / -1

    Directions: The following question has four choices, out of which ONE or MORE can be correct.

    For \(0<\theta<\frac{\pi}{2},\) the solution \((s)\) of \(\sum_{m=1}^{6} \operatorname{cosec}\left(\theta+\frac{(m-1) \pi}{4}\right) \operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right)=4 \sqrt{2}\) is / are
    Solutions

  • Question 3/15
    3 / -1
    If \(A=\left[\begin{array}{cc}2 & 2 \\ -3 & 2\end{array}\right], B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\) then \(\left(B^{-1} A^{-1}\right)^{-1}=\)
    Solutions
    \(A=\left[\begin{array}{cc}2 & 2 \\ -3 & 2\end{array}\right] \quad B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\)
    \(A^{-1}=\frac{1}{10}\left[\begin{array}{cc}2 & -2 \\ 3 & 2\end{array}\right] \quad B^{-1}=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]\)
    Now. \(B^{-1} A^{-1}=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right] \cdot \frac{1}{10}\left[\begin{array}{cc}2 & -2 \\ 3 & 2\end{array}\right]\)
    \(=\frac{1}{10}\left[\begin{array}{cc}0+3 & 0+2 \\ -2+0 & 2+0\end{array}\right]\)
    \(=\frac{1}{10}\left[\begin{array}{rr}3 & 2 \\ -2 & 2\end{array}\right]=\left[\begin{array}{rr}3 / 10 & 2 / 10 \\ -2 / 10 & 2 / 10\end{array}\right]\)
    Now. \(\left(B^{-1} A^{-1}\right)^{-1}=\left[\begin{array}{rr}2 & -2 \\ 2 & 3\end{array}\right]\)
    Hence option A is correct.
  • Question 4/15
    3 / -1
    If \(0 \leq x \leq 2 \pi\) and \(|\cos x| \leq \sin x,\) then
    Solutions

  • Question 5/15
    3 / -1

    Directions: The following question has four choices, out of which one or more is/are correct.

    For any two events A and B in a sample space

    Solutions

  • Question 6/15
    3 / -1
    \(A=\left|\begin{array}{cc}1 & 1 \\ 2 & 2 \\ 3 & -1\end{array}\right|\) and \(B=\left|\begin{array}{cc}2 & 1 \\ 1 & 2 \\ -1 & 0\end{array}\right|,\) then find matrix \(C\) if \(A+2 B+C=0\)
    Solutions
    \(2 B=2 \begin{array}{cc}2 & 1 \\ 1 & 2 \\ -1 & 0\end{array}|=| \begin{array}{cc}4 & 2 \\ 2 & 4 \\ -2 & 0\end{array} \mid\)
    \(A+2 B=\left|\begin{array}{cc}1 & 1 \\ 2 & 2 \\ 3 & -1\end{array}\right|+\left|\begin{array}{cc}4 & 2 \\ 2 & 4 \\ -2 & 0\end{array}\right|=\left|\begin{array}{cc}1+4 & 2+1 \\ 2+2 & 4+2 \\ 3+(-2) & (-1)+0\end{array}\right|=\left|\begin{array}{cc}5 & 3 \\ 4 & 6 \\ 1 & -1\end{array}\right|\)
    \(C=-(A+2 B)=\left|\begin{array}{cc}5 & 3 \\ 4 & 6 \\ 1 & -1\end{array}\right| \Rightarrow\left|\begin{array}{cc}-5 & -3 \\ -4 & -6 \\ -1 & 1\end{array}\right|\)
    \(\therefore A+2 B+C=0\)
    Hence option A is correct.
  • Question 7/15
    3 / -1
    If \(A=[x, y], B=\left[\begin{array}{ll}a & h \\ h & b\end{array}\right], C=\left[\begin{array}{l}x \\ y\end{array}\right]\)
    then \(A B C=\)
    Solutions

  • Question 8/15
    3 / -1

    Directions: The following question has four choices, out of which ONE or MORE can be correct.

    Let z1 and z2 be complex numbers such that z1 ≠ z2 and | z1 | = | z2 |, If z1 has positive real part and z2 has negative imaginary part, then (z1+z2)/(z1-z2) may be

    Solutions
    Given: \(\left|z_{1}\right|=\left|z_{2}\right|\)
    \(N o w, \frac{z_{1}+z_{2}}{z_{1}-z_{2}} \times \frac{z_{1}}{z_{1}}-z_{2}\)
    \(=\frac{z_{1} z_{1}-z_{1} z_{2}+z_{2} z_{1}-z_{2} z_{2}}{\left|z_{1}-z_{2}\right|^{2}}\)
    \(=\frac{\left|z_{1}\right|^{2}+\left(z_{2} \bar{z}_{1}-z_{1} \bar{z}_{2}\right)-\left|z_{2}\right|^{2}}{\left|z_{1}-z_{2}\right|^{2}}\)
    \(=\frac{z_{2} z_{1}-z_{1} z_{2}}{\left|z_{1}-z_{2}\right|^{2}}\left(\because\left|z_{1}\right|^{2}=\left|z_{2}\right|^{2}\right)\)
    As, we know \(z-\bar{z}=2 i \operatorname{lm}(z)\)
    \(\therefore z_{2} z_{1}-z_{1} z_{2}=2 i \operatorname{lm}\left(z_{2} z_{1}\right)\)
    \(\frac{z_{1}+z_{2}}{z_{1}-z_{2}}=\frac{2 i \operatorname{lm}\left(z_{2} z_{1}\right)}{\left|z_{1}-z_{2}\right|^{2}} ;\) zero.
    Hence option A is correct.
  • Question 9/15
    3 / -1
    Using elementary transformations, find the inverse of matrix, \(A=\left[\begin{array}{ll}1 & 3 \\ 2 & 7\end{array}\right]\)
    Solutions

  • Question 10/15
    3 / -1

    Directions: The following question has four choices out of which ONE or MORE can be correct.

    The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c respectively. Of these subjects, the student has a 75% chance of passing in at least one, 50% chance of passing in at least two, and 40% chance of passing in exactly two. Which of the following relations is/are true?

    Solutions

    Let A, B and C, respectively, denote the events that the student passes in Maths, Physics and Chemistry.

    It is given:

    P (A) = m, P(B) = p and P(C) = c

    Probability of at least one success:

    p + m + c - pm - pc - cm + pmc = 0.75 ------(i)

    Probability of at least two successes:

    mc + mp + pc - 2mcp = 0.5 ------------(ii)

    Probability of exactly two successes:

    mc + mp + pc - 3mcp = 0.4 ------------(iii)

    From eq. (ii) and (iii),

    mcp = 0.5 - 0.4 = 0.1 = 1/10

    mc + mp + pc = 0.5 + 0.2 = 0.7 = 7/10

    From (i),

    p + m + c = 27/20

    Hence option B is correct.

  • Question 11/15
    3 / -1

    Directions: The following question has four choices, out of which ONE or MORE can be correct.

    If \(I_{n}=\int_{-\pi}^{\pi} \frac{\sin n x}{\left(1+\pi^{x}\right) \sin x} d x, n=0,1,2, \ldots,\) then
    Solutions


    Hence option D is correct.

  • Question 12/15
    3 / -1

    If A, B are two idempotent matrices and

    AB = BA = 0, then A + B is

    Solutions

    Given A,B are Idempotent matrices

    A2=A, B2=B

    (A+B)2=A2+B2+AB+BA

    (A+B)2=A+B+AB+BA

    (A+B)2=A+B (since AB=BA=0)

    Hence A+B is Idempotent

    Hence option B is correct.

  • Question 13/15
    3 / -1
    If \(\left[\begin{array}{ll}x+3 & 2 y+x \\ z-1 & 4 a-z\end{array}\right]=\left[\begin{array}{ll}0 & -7 \\ 3 & 2 a\end{array}\right]\), then
    \((x+y+z+a) i s\)
    Solutions
    If \(A=B\), then all the elements of \(A\) and \(B\) should be same. \(\therefore x+3=0\)
    \(2 y+x=-7\)
    \(z-1=3\)
    \(z-4 a=-2 a\)
    \(x=-3\)
    \(y=-2\)
    \(z=4\)
    \(a=2\)
    \(x+y+z+a=-3-2+4+2\)
    \(=1\)
    Hence option C is correct.
  • Question 14/15
    3 / -1
    Find the inverse of \(A=\left|\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right|\) if
    \(A^{-1}=\left[\begin{array}{ccc}1 / 2 & -1 / 2 & 1 / 2 \\ a & 3 & b \\ c & -3 / 2 & 1 / 2\end{array}\right]\)
    \(|a b c| ?\)
    Solutions
    Given \(A=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]\)
    \(\therefore A \mid=0 .(2-3)-1(1-9)+2(1+6)\)
    \(=0+8-10\)
    \(=-2 \neq 0\)
    If \(C\) be the matrix of cofactors of the elements in \(|A| \therefore \quad C=\left[\begin{array}{lll}C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33}\end{array}\right]\)
    \(C_{11}=\left|\begin{array}{ll}2 & 3 \\ 1 & 1\end{array}\right|=-1 ; \quad C_{12}=-\left|\begin{array}{ll}1 & 3 \\ 3 & 1\end{array}\right|=8 ; \quad C_{13}=\left|\begin{array}{ll}1 & 2 \\ 3 & 1\end{array}\right|=-5\)
    \(C_{21}=-\left|\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right|=1 ; \quad C_{22}=\left|\begin{array}{ll}0 & 2 \\ 3 & 1\end{array}\right|=-6 ; \quad C_{23}=-\left|\begin{array}{ll}0 & 1 \\ 3 & 1\end{array}\right|=3\)
    \(C_{31}=\left|\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right|=-1 ; \quad C_{32}=-\left|\begin{array}{ll}0 & 2 \\ 1 & 3\end{array}\right|=2 ; \quad C_{33}=\left|\begin{array}{ll}0 & 1 \\ 1 & 2\end{array}\right|=-1\)
    \(\therefore C=\left[\begin{array}{ccc}-1 & 8 & -5 \\ 1 & -6 & 3 \\ -1 & 2 & -1\end{array}\right]\)
    \(\therefore \quad \operatorname{Adj}(A)=C^{\prime}=\left[\begin{array}{ccc}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right]\)
    Hence, \(A^{-1}=\frac{A d j A}{|A|}=-\frac{1}{2}\left[\begin{array}{ccc}-1 & 1 & -1 \\ 8 & -6 & -2 \\ -5 & 3 & -1\end{array}\right]\)
    \(=\left[\begin{array}{ccc}1 / 2 & -1 / 2 & 1 / 2 \\ -4 & 3 & -1 \\ 5 / 2 & -3 / 2 & 1 / 2\end{array}\right]\)
    \(\Rightarrow|a b c|=10\)
    Hence option C is correct.
  • Question 15/15
    3 / -1
    If \(A=\left[\begin{array}{ll}2 & 5 \\ 4 & 4\end{array}\right], \quad B=\left[\begin{array}{ll}0 & 5 \\ 1 & 6\end{array}\right],\) find \(5 A^{\prime}+3 B^{\prime}\)
    Solutions
    \(A=\left[\begin{array}{ll}2 & 5 \\ 4 & 4\end{array}\right] \quad B=\left[\begin{array}{ll}0 & 5 \\ 1 & 6\end{array}\right]\)
    \(A^{T}=\left[\begin{array}{ll}2 & 4 \\ 5 & 4\end{array}\right]\) (Transposing is inter changing the rows \(1 \&\) columns)
    \(B^{T}=\left[\begin{array}{ll}0 & 1 \\ 5 & 6\end{array}\right]\)
    \(\therefore \quad 5 A^{T}+3 B^{T}\)
    \(=5\left[\begin{array}{ll}2 & 4 \\ 5 & 4\end{array}\right]+3\left[\begin{array}{ll}0 & 1 \\ 5 & 6\end{array}\right]\)
    \(=\left[\begin{array}{ll}10 & 20 \\ 25 & 20\end{array}\right]+\left[\begin{array}{cc}0 & 3 \\ 15 & 18\end{array}\right]\)
    \(=\left[\begin{array}{ll}10 & 23 \\ 40 & 38\end{array}\right]\)
    Hence option A is correct.
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