79467
The equation whose roots are the values of the equation \(\left|\begin{array}{ccc}1 & -3 & 1 \\ 1 & 6 & 4 \\ 1 & 3 x & x^{2}\end{array}\right|=0\) is
(D) : Let \(r\) be the common ratio of the given GP Now, \(=\left|\begin{array}{lll} \ln \mathrm{a}_{1} & \ln \mathrm{a}_{2} & \ln \mathrm{a}_{3} \\ \ln \mathrm{a}_{4} & \ln \mathrm{a}_{5} & \ln \mathrm{a}_{6} \\ \ln \mathrm{a}_{7} & \ln \mathrm{a}_{8} & \ln \mathrm{a}_{9} \end{array}\right|\) Apply, \(\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}\) and \(\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{2}\) \(=\left|\begin{array}{lll}\ln \mathrm{a}_{1} & \ln \mathrm{a}_{2}-\ln \mathrm{a}_{1} & \ln \mathrm{a}_{3}-\ln \mathrm{a}_{2} \\ \ln \mathrm{a}_{4} & \ln \mathrm{a}_{5}-\ln \mathrm{a}_{4} & \ln \mathrm{a}_{6}-\ln \mathrm{a}_{5} \\ \ln \mathrm{a}_{7} & \ln \mathrm{a}_{8}-\ln \mathrm{a}_{7} & \ln \mathrm{a}_{9}-\ln \mathrm{a}_{8}\end{array}\right|=\left|\begin{array}{ll}\ln \mathrm{a}_{1} & \ln \mathrm{r} \ln \mathrm{r} \\ \ln \mathrm{a}_{4} & \ln \mathrm{r} \ln \mathrm{r} \\ \ln \mathrm{a}_{7} & \ln \mathrm{r} \ln \mathrm{r}\end{array}\right|\) Since \(\frac{a_{p+1}}{a_{p}}=r\) and \(\left.\log a_{p+1}-\log a_{p}=\log \frac{a_{p+1}}{a_{p}}=\log r\right]\) [Since \(\mathrm{C}_{3}=\mathrm{C}_{2}\) ] \(=0\)
AP EAMCET-2021-19.08.2021
Matrix and Determinant
79469
If the following three linear equations have a non-trivial solution, then \(x+4 a y+a x=0\) \(x+3 b y+b z=0\) \(x+2 c y+c z=0\)
1 a,b,c are in AP
2 a,b,c are in GP
3 a,b,c are in HP
4 \(a+b+c=0\)
Explanation:
(C) : For non - trivial solution \(\text { We have, }\left|\begin{array}{ccc} 1 & 4 a & a \\ 1 & 3 b & b \\ 1 & 2 c & c \end{array}\right|=0\) \(=1(3 b c-2 b c)-4 a c+4 a b+2 a c-3 a b=0\) \(=b c-2 a c+a b=0 \Rightarrow=b c+a b=2 a c\) \(\Rightarrow b(a+c)=2 a c\) \(b=\frac{2 a c}{a+c}=a, b, c \text { are in HP. }\)
WB JEE-2018
Matrix and Determinant
79470
\(\left|\begin{array}{ccc}x & 3 x+2 & 2 x-1 \\ 2 x-1 & 4 x & 3 x+1 \\ 7 x-2 & 17 x+6 & 12 x-1\end{array}\right|=0\) is true for
79467
The equation whose roots are the values of the equation \(\left|\begin{array}{ccc}1 & -3 & 1 \\ 1 & 6 & 4 \\ 1 & 3 x & x^{2}\end{array}\right|=0\) is
(D) : Let \(r\) be the common ratio of the given GP Now, \(=\left|\begin{array}{lll} \ln \mathrm{a}_{1} & \ln \mathrm{a}_{2} & \ln \mathrm{a}_{3} \\ \ln \mathrm{a}_{4} & \ln \mathrm{a}_{5} & \ln \mathrm{a}_{6} \\ \ln \mathrm{a}_{7} & \ln \mathrm{a}_{8} & \ln \mathrm{a}_{9} \end{array}\right|\) Apply, \(\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}\) and \(\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{2}\) \(=\left|\begin{array}{lll}\ln \mathrm{a}_{1} & \ln \mathrm{a}_{2}-\ln \mathrm{a}_{1} & \ln \mathrm{a}_{3}-\ln \mathrm{a}_{2} \\ \ln \mathrm{a}_{4} & \ln \mathrm{a}_{5}-\ln \mathrm{a}_{4} & \ln \mathrm{a}_{6}-\ln \mathrm{a}_{5} \\ \ln \mathrm{a}_{7} & \ln \mathrm{a}_{8}-\ln \mathrm{a}_{7} & \ln \mathrm{a}_{9}-\ln \mathrm{a}_{8}\end{array}\right|=\left|\begin{array}{ll}\ln \mathrm{a}_{1} & \ln \mathrm{r} \ln \mathrm{r} \\ \ln \mathrm{a}_{4} & \ln \mathrm{r} \ln \mathrm{r} \\ \ln \mathrm{a}_{7} & \ln \mathrm{r} \ln \mathrm{r}\end{array}\right|\) Since \(\frac{a_{p+1}}{a_{p}}=r\) and \(\left.\log a_{p+1}-\log a_{p}=\log \frac{a_{p+1}}{a_{p}}=\log r\right]\) [Since \(\mathrm{C}_{3}=\mathrm{C}_{2}\) ] \(=0\)
AP EAMCET-2021-19.08.2021
Matrix and Determinant
79469
If the following three linear equations have a non-trivial solution, then \(x+4 a y+a x=0\) \(x+3 b y+b z=0\) \(x+2 c y+c z=0\)
1 a,b,c are in AP
2 a,b,c are in GP
3 a,b,c are in HP
4 \(a+b+c=0\)
Explanation:
(C) : For non - trivial solution \(\text { We have, }\left|\begin{array}{ccc} 1 & 4 a & a \\ 1 & 3 b & b \\ 1 & 2 c & c \end{array}\right|=0\) \(=1(3 b c-2 b c)-4 a c+4 a b+2 a c-3 a b=0\) \(=b c-2 a c+a b=0 \Rightarrow=b c+a b=2 a c\) \(\Rightarrow b(a+c)=2 a c\) \(b=\frac{2 a c}{a+c}=a, b, c \text { are in HP. }\)
WB JEE-2018
Matrix and Determinant
79470
\(\left|\begin{array}{ccc}x & 3 x+2 & 2 x-1 \\ 2 x-1 & 4 x & 3 x+1 \\ 7 x-2 & 17 x+6 & 12 x-1\end{array}\right|=0\) is true for
79467
The equation whose roots are the values of the equation \(\left|\begin{array}{ccc}1 & -3 & 1 \\ 1 & 6 & 4 \\ 1 & 3 x & x^{2}\end{array}\right|=0\) is
(D) : Let \(r\) be the common ratio of the given GP Now, \(=\left|\begin{array}{lll} \ln \mathrm{a}_{1} & \ln \mathrm{a}_{2} & \ln \mathrm{a}_{3} \\ \ln \mathrm{a}_{4} & \ln \mathrm{a}_{5} & \ln \mathrm{a}_{6} \\ \ln \mathrm{a}_{7} & \ln \mathrm{a}_{8} & \ln \mathrm{a}_{9} \end{array}\right|\) Apply, \(\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}\) and \(\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{2}\) \(=\left|\begin{array}{lll}\ln \mathrm{a}_{1} & \ln \mathrm{a}_{2}-\ln \mathrm{a}_{1} & \ln \mathrm{a}_{3}-\ln \mathrm{a}_{2} \\ \ln \mathrm{a}_{4} & \ln \mathrm{a}_{5}-\ln \mathrm{a}_{4} & \ln \mathrm{a}_{6}-\ln \mathrm{a}_{5} \\ \ln \mathrm{a}_{7} & \ln \mathrm{a}_{8}-\ln \mathrm{a}_{7} & \ln \mathrm{a}_{9}-\ln \mathrm{a}_{8}\end{array}\right|=\left|\begin{array}{ll}\ln \mathrm{a}_{1} & \ln \mathrm{r} \ln \mathrm{r} \\ \ln \mathrm{a}_{4} & \ln \mathrm{r} \ln \mathrm{r} \\ \ln \mathrm{a}_{7} & \ln \mathrm{r} \ln \mathrm{r}\end{array}\right|\) Since \(\frac{a_{p+1}}{a_{p}}=r\) and \(\left.\log a_{p+1}-\log a_{p}=\log \frac{a_{p+1}}{a_{p}}=\log r\right]\) [Since \(\mathrm{C}_{3}=\mathrm{C}_{2}\) ] \(=0\)
AP EAMCET-2021-19.08.2021
Matrix and Determinant
79469
If the following three linear equations have a non-trivial solution, then \(x+4 a y+a x=0\) \(x+3 b y+b z=0\) \(x+2 c y+c z=0\)
1 a,b,c are in AP
2 a,b,c are in GP
3 a,b,c are in HP
4 \(a+b+c=0\)
Explanation:
(C) : For non - trivial solution \(\text { We have, }\left|\begin{array}{ccc} 1 & 4 a & a \\ 1 & 3 b & b \\ 1 & 2 c & c \end{array}\right|=0\) \(=1(3 b c-2 b c)-4 a c+4 a b+2 a c-3 a b=0\) \(=b c-2 a c+a b=0 \Rightarrow=b c+a b=2 a c\) \(\Rightarrow b(a+c)=2 a c\) \(b=\frac{2 a c}{a+c}=a, b, c \text { are in HP. }\)
WB JEE-2018
Matrix and Determinant
79470
\(\left|\begin{array}{ccc}x & 3 x+2 & 2 x-1 \\ 2 x-1 & 4 x & 3 x+1 \\ 7 x-2 & 17 x+6 & 12 x-1\end{array}\right|=0\) is true for
79467
The equation whose roots are the values of the equation \(\left|\begin{array}{ccc}1 & -3 & 1 \\ 1 & 6 & 4 \\ 1 & 3 x & x^{2}\end{array}\right|=0\) is
(D) : Let \(r\) be the common ratio of the given GP Now, \(=\left|\begin{array}{lll} \ln \mathrm{a}_{1} & \ln \mathrm{a}_{2} & \ln \mathrm{a}_{3} \\ \ln \mathrm{a}_{4} & \ln \mathrm{a}_{5} & \ln \mathrm{a}_{6} \\ \ln \mathrm{a}_{7} & \ln \mathrm{a}_{8} & \ln \mathrm{a}_{9} \end{array}\right|\) Apply, \(\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}\) and \(\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{2}\) \(=\left|\begin{array}{lll}\ln \mathrm{a}_{1} & \ln \mathrm{a}_{2}-\ln \mathrm{a}_{1} & \ln \mathrm{a}_{3}-\ln \mathrm{a}_{2} \\ \ln \mathrm{a}_{4} & \ln \mathrm{a}_{5}-\ln \mathrm{a}_{4} & \ln \mathrm{a}_{6}-\ln \mathrm{a}_{5} \\ \ln \mathrm{a}_{7} & \ln \mathrm{a}_{8}-\ln \mathrm{a}_{7} & \ln \mathrm{a}_{9}-\ln \mathrm{a}_{8}\end{array}\right|=\left|\begin{array}{ll}\ln \mathrm{a}_{1} & \ln \mathrm{r} \ln \mathrm{r} \\ \ln \mathrm{a}_{4} & \ln \mathrm{r} \ln \mathrm{r} \\ \ln \mathrm{a}_{7} & \ln \mathrm{r} \ln \mathrm{r}\end{array}\right|\) Since \(\frac{a_{p+1}}{a_{p}}=r\) and \(\left.\log a_{p+1}-\log a_{p}=\log \frac{a_{p+1}}{a_{p}}=\log r\right]\) [Since \(\mathrm{C}_{3}=\mathrm{C}_{2}\) ] \(=0\)
AP EAMCET-2021-19.08.2021
Matrix and Determinant
79469
If the following three linear equations have a non-trivial solution, then \(x+4 a y+a x=0\) \(x+3 b y+b z=0\) \(x+2 c y+c z=0\)
1 a,b,c are in AP
2 a,b,c are in GP
3 a,b,c are in HP
4 \(a+b+c=0\)
Explanation:
(C) : For non - trivial solution \(\text { We have, }\left|\begin{array}{ccc} 1 & 4 a & a \\ 1 & 3 b & b \\ 1 & 2 c & c \end{array}\right|=0\) \(=1(3 b c-2 b c)-4 a c+4 a b+2 a c-3 a b=0\) \(=b c-2 a c+a b=0 \Rightarrow=b c+a b=2 a c\) \(\Rightarrow b(a+c)=2 a c\) \(b=\frac{2 a c}{a+c}=a, b, c \text { are in HP. }\)
WB JEE-2018
Matrix and Determinant
79470
\(\left|\begin{array}{ccc}x & 3 x+2 & 2 x-1 \\ 2 x-1 & 4 x & 3 x+1 \\ 7 x-2 & 17 x+6 & 12 x-1\end{array}\right|=0\) is true for
79467
The equation whose roots are the values of the equation \(\left|\begin{array}{ccc}1 & -3 & 1 \\ 1 & 6 & 4 \\ 1 & 3 x & x^{2}\end{array}\right|=0\) is
(D) : Let \(r\) be the common ratio of the given GP Now, \(=\left|\begin{array}{lll} \ln \mathrm{a}_{1} & \ln \mathrm{a}_{2} & \ln \mathrm{a}_{3} \\ \ln \mathrm{a}_{4} & \ln \mathrm{a}_{5} & \ln \mathrm{a}_{6} \\ \ln \mathrm{a}_{7} & \ln \mathrm{a}_{8} & \ln \mathrm{a}_{9} \end{array}\right|\) Apply, \(\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}\) and \(\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{2}\) \(=\left|\begin{array}{lll}\ln \mathrm{a}_{1} & \ln \mathrm{a}_{2}-\ln \mathrm{a}_{1} & \ln \mathrm{a}_{3}-\ln \mathrm{a}_{2} \\ \ln \mathrm{a}_{4} & \ln \mathrm{a}_{5}-\ln \mathrm{a}_{4} & \ln \mathrm{a}_{6}-\ln \mathrm{a}_{5} \\ \ln \mathrm{a}_{7} & \ln \mathrm{a}_{8}-\ln \mathrm{a}_{7} & \ln \mathrm{a}_{9}-\ln \mathrm{a}_{8}\end{array}\right|=\left|\begin{array}{ll}\ln \mathrm{a}_{1} & \ln \mathrm{r} \ln \mathrm{r} \\ \ln \mathrm{a}_{4} & \ln \mathrm{r} \ln \mathrm{r} \\ \ln \mathrm{a}_{7} & \ln \mathrm{r} \ln \mathrm{r}\end{array}\right|\) Since \(\frac{a_{p+1}}{a_{p}}=r\) and \(\left.\log a_{p+1}-\log a_{p}=\log \frac{a_{p+1}}{a_{p}}=\log r\right]\) [Since \(\mathrm{C}_{3}=\mathrm{C}_{2}\) ] \(=0\)
AP EAMCET-2021-19.08.2021
Matrix and Determinant
79469
If the following three linear equations have a non-trivial solution, then \(x+4 a y+a x=0\) \(x+3 b y+b z=0\) \(x+2 c y+c z=0\)
1 a,b,c are in AP
2 a,b,c are in GP
3 a,b,c are in HP
4 \(a+b+c=0\)
Explanation:
(C) : For non - trivial solution \(\text { We have, }\left|\begin{array}{ccc} 1 & 4 a & a \\ 1 & 3 b & b \\ 1 & 2 c & c \end{array}\right|=0\) \(=1(3 b c-2 b c)-4 a c+4 a b+2 a c-3 a b=0\) \(=b c-2 a c+a b=0 \Rightarrow=b c+a b=2 a c\) \(\Rightarrow b(a+c)=2 a c\) \(b=\frac{2 a c}{a+c}=a, b, c \text { are in HP. }\)
WB JEE-2018
Matrix and Determinant
79470
\(\left|\begin{array}{ccc}x & 3 x+2 & 2 x-1 \\ 2 x-1 & 4 x & 3 x+1 \\ 7 x-2 & 17 x+6 & 12 x-1\end{array}\right|=0\) is true for
