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Determinants Test - 29

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Determinants Test - 29
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  • Question 1
    1 / -0
    A= $$\begin{bmatrix}
     b^{2}c^{2}& bc & b+c\\
     c^{2}a^{2}& ca &c+a \\
     a^{2}b^{2}& ab & a+b
    \end{bmatrix}$$ then $$\left | A \right |$$ =?
    Solution
    $$|A|=\begin{vmatrix}
    b^2c^2 & bc & b+c\\
    c^2a^2 & ac & c+a\\
    a^2b^2 & ab & a+b
    \end{vmatrix}$$

    $$=b^2c^2[a^2c+abc-abc-a^2b]-bc[a^3c^2+a^2bc^2-a^2b^2c-a^3b^2]+(b+c)[a^3bc^2-a^3b^2c]$$

    $$=a^2b^2c^3 - a^2b^3c^2 -a^3bc^3-a^2b^2c^3+a^2b^3c^2+a^3b^3c+a^3b^2c^2 + a^3bc^3-a^3b^3c-a^3b^3c-a^3b^2c^2$$
    $$=0$$
  • Question 2
    1 / -0

    $$\mathrm{y}=\sin \mathrm{x},\ y_{n}=\displaystyle \frac{d^{n}(\sin x)}{dx^{n}}$$
    then $$\begin{vmatrix}
    y& y_{1} & y_{2}\\
     y_{3}& y_{4} &y_{5} \\
     y_{6}& y_{7} & y_{8}
    \end{vmatrix}$$ =?



    Solution
    Given $$y=sin x$$ and $$Y_n=\dfrac{\delta ^n(sin x)}{dx^{n}}$$
    Let $$A=\begin{vmatrix}y& y_{1} & y_{2}\\  y_{3}& y_{4} &y_{5} \\  y_{6}& y_{7} & y_{8}\end{vmatrix}$$
    =$$\begin{vmatrix}
    sin x & cos x & -sin x\\
    - cos x & sin x & cos x\\
    -sin x & -cos x & sin x
    \end{vmatrix}$$
    $$= sin x(sin^2 x + cos ^2 x)-cos x[0]-sin x[cos^2x + sin ^2 x]$$
    $$=0$$


  • Question 3
    1 / -0
    $${A}=\begin{bmatrix}
    -1 & -2 & -2\\
    2 & 1 & -2\\
    2 & -2 & 1
    \end{bmatrix}$$ then $$Adj(A)=$$ 

    Solution
    $$A_{11}=-3$$
    $$A_{12}=-6$$
    $$A_{13}=-6$$
    clearly B is the ans


  • Question 4
    1 / -0
    If $$\Delta $$ = $$\begin{vmatrix} cos\frac{\theta }{2} &1  &1 \\   1&cos\frac{\theta }{2}  &-cos\frac{\theta }{2} \\ -cos\frac{\theta }{2} &1  & -1
    \end{vmatrix}$$ the minimum of $$\Delta $$ is $$m_{1}$$ and maximum of $$\Delta $$ is $$m_{2}$$ then $$[m_{1} , m_{2}]$$ is
    Solution
    $$\displaystyle \triangle =\begin{vmatrix} \cos { \frac { \theta  }{ 2 }  }  & 1 & 1 \\ 1 & \cos { \frac { \theta  }{ 2 }  }  & -\cos { \frac { \theta  }{ 2 }  }  \\ -\cos { \frac { \theta  }{ 2 }  }  & 1 & -1 \end{vmatrix}$$
    Expanding along $${ R }_{ 1 },$$ we get
    $$\displaystyle =\cos { \frac { \theta  }{ 2 }  } \left( -\cos { \frac { \theta  }{ 2 }  } +\cos { \frac { \theta  }{ 2 }  }  \right) -1\left( -1-\cos ^{ 2 }{ \frac { \theta  }{ 2 }  }  \right) +1\left( 1+\cos ^{ 2 }{ \frac { \theta  }{ 2 }  }  \right) $$
    $$\displaystyle =2\left( 1+\cos ^{ 2 }{ \frac { \theta  }{ 2 }  }  \right) =2+2\cos ^{ 2 }{ \frac { \theta  }{ 2 }  } $$
    As $$\displaystyle -1\le \cos { x } \le 1\Rightarrow 0\le \cos ^{ 2 }{ x } \le 1$$
    $$\therefore$$ Max of $$\triangle $$ i.e., $$\displaystyle { m }_{ 2 }=4$$ and min of $$\triangle$$ i.e., $$\displaystyle { m }_{ 1 }=2$$
  • Question 5
    1 / -0
    $$f(x)=$$ $$\begin{vmatrix}
    \cos x &x &1 \\
    2 \sin x & x^{2} &2x \\
    \tan x& x & 1
    \end{vmatrix}$$ then $$ \displaystyle \lim _{ x\rightarrow 0 }{ f(x) } =$$
    Solution
    $$f(x)=\begin{vmatrix} cos\, \, x & x & 1 \\ 2sin\, \, x & x^{ 2 } & 2x \\ tan\, \, x & x & 1 \end{vmatrix}$$

    $$ { R }_{ 3 }\rightarrow { R }_{ 3 }{ -R }_{ 1 }$$

    $$f(x)=\begin{vmatrix} cos\, \, x & x & 1 \\ 2sin\, \, x & x^{ 2 } & 2x \\ \tan { x } -\cos { x }  & 0 & 0 \end{vmatrix}$$

    $$=(\tan { x } -\cos { x } )x^{ 2 }$$

    Now, $$\lim _{ x\rightarrow 0 }{ f(x) }$$

    $$ =\lim _{ x\rightarrow 0 }{ (\tan { x } -\cos { x } )x^{ 2 } } $$

    $$=0$$
  • Question 6
    1 / -0
    If $$[\mathrm{x}]$$ stands greatest integer $$\leq \mathrm{x}$$ then the value of $$\begin{vmatrix}
    \left [ e \right ]& \left [ \pi \right ] &\left [ \pi ^{2}-6 \right ]\\
    \left [ \pi \right ] & \pi ^{2}-6 & \left [ e \right ]\\
    \left [ \pi ^{2}-6 \right ]&\left [ e \right ] & \left [ \pi \right ]
    \end{vmatrix}$$ equals 
    Solution
    $$\Delta =\begin{vmatrix} \left[ e \right]  & \left[ \pi  \right]  & \left[ \pi ^{ 2 }-6 \right]  \\ \left[ \pi  \right]  & \pi ^{ 2 }-6 & \left[ e \right]  \\ \left[ \pi ^{ 2 }-6 \right]  & \left[ e \right]  & \left[ \pi  \right]  \end{vmatrix}$$

    $$=\begin{vmatrix} 2 & 3 & 3 \\ 3 & 3 & 2 \\ 3 & 2 & 3 \end{vmatrix}$$
    $${ R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 },{ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 3 }$$
    $$=\begin{vmatrix} -1 & 0 & 1 \\ 0 & 1 & -1 \\ 3 & 2 & 3 \end{vmatrix}$$
    $$\Delta=-8$$
  • Question 7
    1 / -0
    If $$\alpha,\ \beta$$ are the roots of  $$\begin{vmatrix}
    x & 1 & 2\\
    0 & 1& 1\\
    1 & x & 2
    \end{vmatrix}$$ = 0 then $$\alpha^{n}+\beta^{n}=?$$ 


    Solution
    $$\begin{vmatrix}
    x & 1 &2 \\
    0 & 1 & 1\\
    1 & x & 2
    \end{vmatrix}=0$$
    So, $$x(2-x)-1(-1)+2(-1)=0$$
    $$2x-x^2-1=0$$
    $$x^2-2x+1=0$$
    $$x=1$$
    So, Given that $$ \alpha , \beta$$ are roots of $$x^2-2x+1=0$$
    So, $$\alpha=\beta=1$$
    So, $$\alpha^{n} + \beta^{n}=2$$

  • Question 8
    1 / -0
    $$A=\begin{bmatrix}4 &-2&5\end{bmatrix}$$, $$ B=$$ $$\begin{bmatrix}
    2\\
    0\\
    3\end{bmatrix}$$, then $$Adj(BA)=$$ 
    Solution
    $$A=\left [ 4\ -2\ \ 5  \right ]$$,   $$B=\begin{bmatrix}
    2\\
    0\\
    3
    \end{bmatrix}$$

    $$BA=\begin{bmatrix}
    8 & -4 & 0\\
     0& 0 &0 \\
     12& -6 &15
    \end{bmatrix}$$

    $$M_{11}, M_{12}, M_{13}$$
    $$M_{31},M_{32},M_{33} $$ are all clearly zero
    and $$M_{21},M_{22}M_{23}$$ also out  

    So, all the co factors  are also zero
    thus, $$Adj\left ( BA \right )=\begin{bmatrix}
    0 &0  &0 \\
    0 &  0& 0\\
     0& 0 & 0
    \end{bmatrix}$$(Transpose matrix of co factors)
  • Question 9
    1 / -0
    The straight lines $$\mathrm{x}+2\mathrm{y}-9=0,3\mathrm{x}+5\mathrm{y}-5=0$$ and $$\mathrm{a}\mathrm{x}+\mathrm{b}\mathrm{y}-1=0$$ are concurrent if the straight line $$22\mathrm{x}-35\mathrm{y}-1=0$$ passes through the point 

    Solution

    $$x+2y-9=0$$ ---(1)


    $$3x+5y-5=0$$ ---(2)


    $$ax+by-1=0$$ ---(3)

    Solving (1) and (2) simultaneously we get 

    $$y=22, x=-35$$

    Now, equation (1), (2) and (3) will be concurrent, that is they will pass through one point if $$y=22, x=-35$$ satisfy the third equation $$ax+by-1=0$$.

    Substituting the values of 'x' and 'y' in this equation we get $$-35a+22b-1=0$$ ---(4)


    And another equation given is $$22x-35y-1=0$$ ---(5)

    Equation (5) will be of the form of equation (4), if we substitute

    $$x=b$$ & $$y=a$$

    That is $$22x-35y-1=0$$ passes through $$(b,a)$$
  • Question 10
    1 / -0
    If x, y, z are in A.P., then the value of the det A. Where A=$$\begin{vmatrix}
    4 &5 & 6 & x\\
    5& 6 & 7 &y \\
    6& 7& 8 &z \\
    x & y & z & 0
    \end{vmatrix}$$ 
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