NEET Test Series from KOTA - 10 Papers In MS WORD
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Matrix and Determinant
79008
If \(a \neq b \neq c\), then the value of \(x\) satisfying the equation \(\left|\begin{array}{ccc}0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0\end{array}\right|=0\) is
1 a
2 \(\mathrm{b}\)
3 c
4 0
Explanation:
(D) : It is given that, \(\left|\begin{array}{ccc} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{array}\right|=0\) Suppose, \(\mathrm{x}=0\) Then, the determinant becomes, \(\left|\begin{array}{ccc} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{array}\right|\) \(=0\left(0-c^{2}\right)+a(0+b c)-b(a c+0)\) \(=a b c-a b c=0\) So, L.H.S = R.H.S Hence, the value of \(x=0\)
UPSEE-2008
Matrix and Determinant
79009
If \(a, b, c\) are non-zero distinct real numbers, then \(\left|\begin{array}{lll}\text { bc } & \text { ca } & a b \\ \text { ca } & \text { ab } & b c \\ \text { ab } & \text { bc } & c a\end{array}\right|\) vanishes, when
(A) : It is given that, \(a, b, c\) are non - zero distinct real number then :- \(\left|\begin{array}{lll} \mathrm{bc} & \mathrm{ca} & \mathrm{ab} \\ \mathrm{ca} & \mathrm{ab} & \mathrm{bc} \\ \mathrm{ab} & \mathrm{bc} & \mathrm{ca} \end{array}\right|=0\) Expanding \(\mathrm{R}_{1}\) :- \(b c\left[a^{2} b c-b^{2} c^{2}\right]-c a\left[c^{2} a^{2}-b^{2} a c\right]+a b\left[c^{2} a b-a^{2} b^{2}\right]=0\) \(a^{2} b^{2} c^{2}-b^{3} c^{3}-c^{3} a^{3}+b^{2} a^{2} c^{2}+a^{2} b^{2} c^{2}-a^{3} b^{3}=0\) \(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}-3\left(a^{2} b^{2} c^{2}\right)=0\) \(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}-3(a b)(b c)(c a)=0\) \((a b)^{3}+(b c)^{3}+(c a)^{3}-3(a b)(b c)(c a)=0\) This will be zero only when - \(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}=0 \text { or } \mathrm{ab}=\mathrm{bc}=\mathrm{ca}\) Dividing both side by abc - \(\frac{\mathrm{ab}}{\mathrm{abc}}+\frac{\mathrm{bc}}{\mathrm{abc}}+\frac{\mathrm{ca}}{\mathrm{abc}}=0 \Rightarrow \frac{1}{\mathrm{c}}+\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}=0\) or \(\quad \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}+\frac{1}{\mathrm{c}}=0\)
79008
If \(a \neq b \neq c\), then the value of \(x\) satisfying the equation \(\left|\begin{array}{ccc}0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0\end{array}\right|=0\) is
1 a
2 \(\mathrm{b}\)
3 c
4 0
Explanation:
(D) : It is given that, \(\left|\begin{array}{ccc} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{array}\right|=0\) Suppose, \(\mathrm{x}=0\) Then, the determinant becomes, \(\left|\begin{array}{ccc} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{array}\right|\) \(=0\left(0-c^{2}\right)+a(0+b c)-b(a c+0)\) \(=a b c-a b c=0\) So, L.H.S = R.H.S Hence, the value of \(x=0\)
UPSEE-2008
Matrix and Determinant
79009
If \(a, b, c\) are non-zero distinct real numbers, then \(\left|\begin{array}{lll}\text { bc } & \text { ca } & a b \\ \text { ca } & \text { ab } & b c \\ \text { ab } & \text { bc } & c a\end{array}\right|\) vanishes, when
(A) : It is given that, \(a, b, c\) are non - zero distinct real number then :- \(\left|\begin{array}{lll} \mathrm{bc} & \mathrm{ca} & \mathrm{ab} \\ \mathrm{ca} & \mathrm{ab} & \mathrm{bc} \\ \mathrm{ab} & \mathrm{bc} & \mathrm{ca} \end{array}\right|=0\) Expanding \(\mathrm{R}_{1}\) :- \(b c\left[a^{2} b c-b^{2} c^{2}\right]-c a\left[c^{2} a^{2}-b^{2} a c\right]+a b\left[c^{2} a b-a^{2} b^{2}\right]=0\) \(a^{2} b^{2} c^{2}-b^{3} c^{3}-c^{3} a^{3}+b^{2} a^{2} c^{2}+a^{2} b^{2} c^{2}-a^{3} b^{3}=0\) \(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}-3\left(a^{2} b^{2} c^{2}\right)=0\) \(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}-3(a b)(b c)(c a)=0\) \((a b)^{3}+(b c)^{3}+(c a)^{3}-3(a b)(b c)(c a)=0\) This will be zero only when - \(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}=0 \text { or } \mathrm{ab}=\mathrm{bc}=\mathrm{ca}\) Dividing both side by abc - \(\frac{\mathrm{ab}}{\mathrm{abc}}+\frac{\mathrm{bc}}{\mathrm{abc}}+\frac{\mathrm{ca}}{\mathrm{abc}}=0 \Rightarrow \frac{1}{\mathrm{c}}+\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}=0\) or \(\quad \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}+\frac{1}{\mathrm{c}}=0\)
79008
If \(a \neq b \neq c\), then the value of \(x\) satisfying the equation \(\left|\begin{array}{ccc}0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0\end{array}\right|=0\) is
1 a
2 \(\mathrm{b}\)
3 c
4 0
Explanation:
(D) : It is given that, \(\left|\begin{array}{ccc} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{array}\right|=0\) Suppose, \(\mathrm{x}=0\) Then, the determinant becomes, \(\left|\begin{array}{ccc} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{array}\right|\) \(=0\left(0-c^{2}\right)+a(0+b c)-b(a c+0)\) \(=a b c-a b c=0\) So, L.H.S = R.H.S Hence, the value of \(x=0\)
UPSEE-2008
Matrix and Determinant
79009
If \(a, b, c\) are non-zero distinct real numbers, then \(\left|\begin{array}{lll}\text { bc } & \text { ca } & a b \\ \text { ca } & \text { ab } & b c \\ \text { ab } & \text { bc } & c a\end{array}\right|\) vanishes, when
(A) : It is given that, \(a, b, c\) are non - zero distinct real number then :- \(\left|\begin{array}{lll} \mathrm{bc} & \mathrm{ca} & \mathrm{ab} \\ \mathrm{ca} & \mathrm{ab} & \mathrm{bc} \\ \mathrm{ab} & \mathrm{bc} & \mathrm{ca} \end{array}\right|=0\) Expanding \(\mathrm{R}_{1}\) :- \(b c\left[a^{2} b c-b^{2} c^{2}\right]-c a\left[c^{2} a^{2}-b^{2} a c\right]+a b\left[c^{2} a b-a^{2} b^{2}\right]=0\) \(a^{2} b^{2} c^{2}-b^{3} c^{3}-c^{3} a^{3}+b^{2} a^{2} c^{2}+a^{2} b^{2} c^{2}-a^{3} b^{3}=0\) \(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}-3\left(a^{2} b^{2} c^{2}\right)=0\) \(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}-3(a b)(b c)(c a)=0\) \((a b)^{3}+(b c)^{3}+(c a)^{3}-3(a b)(b c)(c a)=0\) This will be zero only when - \(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}=0 \text { or } \mathrm{ab}=\mathrm{bc}=\mathrm{ca}\) Dividing both side by abc - \(\frac{\mathrm{ab}}{\mathrm{abc}}+\frac{\mathrm{bc}}{\mathrm{abc}}+\frac{\mathrm{ca}}{\mathrm{abc}}=0 \Rightarrow \frac{1}{\mathrm{c}}+\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}=0\) or \(\quad \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}+\frac{1}{\mathrm{c}}=0\)
79008
If \(a \neq b \neq c\), then the value of \(x\) satisfying the equation \(\left|\begin{array}{ccc}0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0\end{array}\right|=0\) is
1 a
2 \(\mathrm{b}\)
3 c
4 0
Explanation:
(D) : It is given that, \(\left|\begin{array}{ccc} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{array}\right|=0\) Suppose, \(\mathrm{x}=0\) Then, the determinant becomes, \(\left|\begin{array}{ccc} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{array}\right|\) \(=0\left(0-c^{2}\right)+a(0+b c)-b(a c+0)\) \(=a b c-a b c=0\) So, L.H.S = R.H.S Hence, the value of \(x=0\)
UPSEE-2008
Matrix and Determinant
79009
If \(a, b, c\) are non-zero distinct real numbers, then \(\left|\begin{array}{lll}\text { bc } & \text { ca } & a b \\ \text { ca } & \text { ab } & b c \\ \text { ab } & \text { bc } & c a\end{array}\right|\) vanishes, when
(A) : It is given that, \(a, b, c\) are non - zero distinct real number then :- \(\left|\begin{array}{lll} \mathrm{bc} & \mathrm{ca} & \mathrm{ab} \\ \mathrm{ca} & \mathrm{ab} & \mathrm{bc} \\ \mathrm{ab} & \mathrm{bc} & \mathrm{ca} \end{array}\right|=0\) Expanding \(\mathrm{R}_{1}\) :- \(b c\left[a^{2} b c-b^{2} c^{2}\right]-c a\left[c^{2} a^{2}-b^{2} a c\right]+a b\left[c^{2} a b-a^{2} b^{2}\right]=0\) \(a^{2} b^{2} c^{2}-b^{3} c^{3}-c^{3} a^{3}+b^{2} a^{2} c^{2}+a^{2} b^{2} c^{2}-a^{3} b^{3}=0\) \(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}-3\left(a^{2} b^{2} c^{2}\right)=0\) \(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}-3(a b)(b c)(c a)=0\) \((a b)^{3}+(b c)^{3}+(c a)^{3}-3(a b)(b c)(c a)=0\) This will be zero only when - \(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}=0 \text { or } \mathrm{ab}=\mathrm{bc}=\mathrm{ca}\) Dividing both side by abc - \(\frac{\mathrm{ab}}{\mathrm{abc}}+\frac{\mathrm{bc}}{\mathrm{abc}}+\frac{\mathrm{ca}}{\mathrm{abc}}=0 \Rightarrow \frac{1}{\mathrm{c}}+\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}=0\) or \(\quad \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}+\frac{1}{\mathrm{c}}=0\)
