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Determinants Test - 53

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Determinants Test - 53
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  • Question 1
    1 / -0
    If $$f(x)=\begin{vmatrix} 1 & x & x+1\\ 2x & x(x-1) & (x+1)x\\ 3x(x-1) & x(x-1)(x-2) & (x+1)x(x-1)\end{vmatrix}$$ then $$f(100)$$ is equal to?
    Solution
    $$f\left( x \right) =\begin{vmatrix} 1 & x & x+1 \\ 2x & x\left( n-1 \right)  & \left( n+1 \right) x \\ 3z\left( n-1 \right)  & x\left( n-1 \right) \left( n-2 \right)  & \left( n+1 \right) x\left( n-1 \right)  \end{vmatrix}$$

    $$\left( n-1 \right) \left( n \right) \begin{vmatrix} 1 & x & n+1 \\ 2 & x-1 & n+1 \\ 3 & n-2 & n+1 \end{vmatrix}\begin{matrix} { R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 } \\ \longrightarrow  \\ { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 3 } \end{matrix}$$

    $$\left( n-1 \right) \left( n \right) \begin{vmatrix} -1 & 1 & 0 \\ -1 & 1 & 0 \\ 3 & n-2 & n+1 \end{vmatrix}$$

    $$(n-1)(n) (0)=0$$
    so, $$t(100)=0$$
  • Question 2
    1 / -0
    The value of the determinant $$\left| \begin{matrix} { b }^{ 2 }-ab & b-c & bc-ac \\ ab-{ b }^{ 2 } & a-b & { b }^{ 2 }-ab \\ bc-ac & c-a & ab-{ b }^{ 2 } \end{matrix} \right| $$
    Solution

  • Question 3
    1 / -0
    Solve 
    $$\begin{vmatrix} 1 & bc & a\left( b+c \right)  \\ 1 & ca & b\left( c+a \right)  \\ 1 & ab & c\left( a+b \right)  \end{vmatrix}$$
    Solution
    Given
    $$\begin{vmatrix} 1 & bc & a(b+c) \\ 1 & ca & b(c+a) \\ 1 & ab & c(a+b) \end{vmatrix}$$
    $$=1\times(a\times c)\times(a\times c+b\times c)+b\times c\times(a\times b+b\times c)\times1+$$
    $$+(a\times b+a\times c)\times1\times(a\times b)-1\times(a\times c)\times (a\times b+a\times c)-$$
    $$-a\times b\times (a\times b+b\times c)\times 1-(a\times c+b\times c)\times1\times(b\times c)$$
    $$=0$$
  • Question 4
    1 / -0
    if $$a>0$$ and discriminant of $${ax}^{2}+{2bx}+{c}$$ is $$-ve$$, then $$\left| \begin{matrix} a & b & ax+b \\ b & c & bx+c \\ ax+b & bx+c & 0 \end{matrix} \right|$$ is
    Solution

  • Question 5
    1 / -0
    $$\left| \begin{matrix} \sqrt { 13 } +\sqrt { 3 }  & 2\sqrt { 5 }  & \sqrt { 5 }  \\ \sqrt { 15 } +\sqrt { 26 }  & 5 & \sqrt { 10 }  \\ 3+\sqrt { 65 }  & \sqrt { 15 }  & 5 \end{matrix} \right| =$$ 
    Solution
    $$R_{3}\rightarrow R_{3}-\sqrt{5}R_{1}$$
    $$R_{2}\rightarrow R_{2}-\sqrt{2}R_{1}$$
    $$\begin{vmatrix} \sqrt { 13 } +\sqrt { 3 }  & 2\sqrt { 5 }  & \sqrt { 5 }  \\ \sqrt { 15 } -\sqrt { 6 }  & 5-2\sqrt { 10 }  & 0 \\ 3-\sqrt { 15 }  & \sqrt { 15 } -10 & 0 \end{vmatrix}$$
    $$\Rightarrow \sqrt{5}(\sqrt{15}-\sqrt{6})(\sqrt{15}-10)(5-2\sqrt{10})(3-\sqrt{15})$$
    $$\Rightarrow \sqrt{5}(15-10\sqrt{15}-3\sqrt{10}+10\sqrt{6})-(15-5\sqrt{15}-6\sqrt{10}+2.5.\sqrt{16})$$
    $$\Rightarrow \sqrt{5}(15-10\sqrt{15}-3\sqrt{10}+10\sqrt{6}-15+5\sqrt{15}+6\sqrt{10}-10\sqrt{6})$$
    $$\sqrt{5}(-5\sqrt{15}+3\sqrt{10})$$
    $$\Rightarrow -5.5\sqrt{3}+3.6\sqrt{2}$$
    $$\Rightarrow 15\sqrt{2}-25\sqrt{3}$$
    $$\Rightarrow 5(3\sqrt{2}-5\sqrt{3})$$
  • Question 6
    1 / -0
    The value of $$\left| \begin{matrix} 1+w & { w }^{ 2 } & -w \\ 1+{ w }^{ 2 } & w & -{ w }^{ 2 } \\ { w }^{ 2 }+w & w & -{ w }^{ 2 } \end{matrix} \right| $$ is equal to 
    Solution
    $$\begin{vmatrix} 1+w & { w }^{ 2 } & -w \\ 1+{ w }^{ 2 } & w & { -w }^{ 2 } \\ { w }^{ 2 }+w & w & { -w }^{ 2 } \end{vmatrix}$$
    Multiplying $$C_{2}$$ by $$w$$ and adding to $$C_{3}$$ 
    $$C_{3}\rightarrow wC_{2}+C_{3}$$
    $$\begin{bmatrix} 1+w & { w }^{ 2 } & { w }^{ 2 }-w \\ 1+{ w }^{ 2 } & w & 0 \\ { w }^{ 2 }+w & w & 0 \end{bmatrix}$$
    Now finding determinant
    $$(-w+w^{3})\left[w(1+w^{2})-w(w^{2}+w)\right]$$
    $$(-w+w)(w+w^{3}-w^{3}-w^{2})$$
    $$\because w^{3}=1$$, then 
    $$(-w+1)(w-w^{2})$$
    $$-w^{2}+w^{3}+w-w^{2}$$
    $$-w^{2}+1+w-w^{2}$$
    $$-2w^{2}+1+w+w^{2}-w^{2}$$
    $$(1+w+w^{2})-3w^{2}$$
    As we know , $$1+w+w^{2}=0$$
    $$0-3w^{2}=-3w^{2}$$
    Option $$(d)$$ is correct
  • Question 7
    1 / -0
    If $$\left[ \begin{array}{l}\cos \theta \,\,\,\, - \sin \theta \,\,\,\,\,\,0\\\sin \theta \,\,\,\,\,\,\,\,\cos \theta \,\,\,\,\,\,0\\0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,1\end{array} \right]$$, then $$adjA = $$
    Solution
    $$A=\begin{bmatrix} cos\theta  & -\sin\theta  & 0 \\ \sin\theta  & \cos\theta  & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
    lets write co factor
    $$C_{11}+\cos\theta$$ $$C_{12}=-\sin\theta$$  $$C_{13}=0$$
    $$C_{21}+\sin\theta$$ $$C_{22}=\cos\theta$$ $$C_{23}=0$$
    $$C_{31}=0$$     $$C_{32}=0$$      $$C_{33}=1$$
    $$\therefore$$ co factor matrix 
    $$=\begin{bmatrix} \cos\theta  & -\sin\theta  & 0 \\ \sin\theta  & \cos\theta  & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
    $$\therefore Adj\,A=$$ transpose of cofactor matrix
    $$=\begin{bmatrix} \cos\theta  & -\sin\theta  & 0 \\ \sin\theta  & \cos\theta  & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
    $$=A^T$$
  • Question 8
    1 / -0
    If the points $$A(at^{2}_{1},2at_{1}), B(at^{2}_{2},2at_{2})$$ and $$C(\alpha,0)$$ are collinear, then $$t_{1} t_{2}$$ equals
    Solution

  • Question 9
    1 / -0
    Let $$F(x)=$$$$\left | \left|  \right| \begin{matrix}1  &1+sin\ x  &1+sin\ x+cos\ x \\  2 &3+2\ sin\ x  &4+3\ sin\ x+2\ cos\ x  \\  3&6+3\ sin\ x&10+6\ sin\ x+3\ cos\ x  \end{matrix} \right |$$  then $$F'\ \left ( \dfrac{\pi}{2} \right )$$ is equal to
    Solution
    $$f(x) = \begin{vmatrix} 1 & 1+\sin x & 1+ \sin x +\cos x \\ 2 & 3+2 \sin x & 4+3 \sin x +2 \cos x \\ 3 & 6+3 \sin x & 10+6 \sin x +3 \cos x \end{vmatrix}$$
    $$f'(x)= \begin{vmatrix} 0 & \cos x & \cos x- \sin x \\ 2 & 3+2 \sin x & 4+3 \sin x + 2 \cos x \\ 3 & 6+3 \sin x & 10+6 \sin x+ \cos x \end{vmatrix}+ \begin{vmatrix} (1)  & (1+ \sin x) & 1+\sin x+ \cos x \\ 0 & 2 \cos x & 3 \cos x-2 \sin x \\ 3  & 6+3 \sin x & 10+6 \sin x+3 \cos \end{vmatrix}+ \begin{vmatrix} 1 & 1+ \sin x & 1+ \sin x + \cos x \\ 2 & 3+2 \sin x & 4+ 3 \sin x + 2 \cos x \\ 0 & 3 \cos x & 6 \cos -3 \sin x \end{vmatrix}$$ (differentiating row $$mse$$ )
    $$f' \left( \dfrac{\pi}{2} \right) = \begin{vmatrix} 0 & 0 & -1 \\ 2 & 5 & 7 \\ 3 & 9 & 16 \end{vmatrix}+ \begin{vmatrix} 1 & 2  & 2 \\ 0 & 0 & -2 \\ 3 & 9 & 16 \end{vmatrix}+ \begin{vmatrix} 1 & 2 & 2 \\ 2 & 5 & 7 \\ 0 & 0 & -3 \end{vmatrix}$$
    $$- (18-15)+ 2 (9-6)-3 (5-4)$$
    $$=-3+6-3=0$$
    $$\therefore $$ Option $$B$$ is correct.
  • Question 10
    1 / -0
    If $$A = \left[ {\begin{array}{*{20}{c}}a&0&0\\0&a&0\\0&0&a\end{array}} \right]$$ then find the value of $$\left| A \right|\left| {adjA} \right|$$
    Solution
    $$A=\begin{bmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{bmatrix}$$
    $$\left( A \right) =\begin{bmatrix} { a }^{ 2 } & 0 & 0 \\ 0 & { a }^{ 2 } & 0 \\ 0 & 0 & { a }^{ 2 } \end{bmatrix}Adj\left( A \right) =\begin{bmatrix} { a }^{ 2 } & 0 & 0 \\ 0 & { a }^{ 2 } & 0 \\ 0 & 0 & { a }^{ 2 } \end{bmatrix}$$
    $$|A|=a^3$$
    $$|Adj (A)|=a^6$$
    $$|A|.|Adj (A)|=a^3a^6=a^9$$
    $$C$$ is correct 

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