79177
if \(y \quad y^{2} 1+y^{3}=0, x \neq y \neq z\) then \(1+x y z\) is \(\left.\begin{array}{lll}\mathrm{z} & \mathrm{z}^{2} & 1+\mathrm{z}^{3}\end{array} \right\rvert\,\) equal to
1 0
2 -1
3 1
4 2
Explanation:
(A) : Given, \(\left|\begin{array}{ccc} x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3} \end{array}\right|=0, x \neq y \neq z\) \(\left|\begin{array}{lll} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{array}\right|+\left|\begin{array}{ccc} x & x^{2} & x^{3} \\ y & y^{2} & y^{3} \\ z & z^{2} & z^{3} \end{array}\right|=0\) \(\left|\begin{array}{ccc}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|+\left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right| x y z=0\) \(C_{1} \leftrightarrow C_{2}\) and \(C_{2} \leftrightarrow C_{3}\) \(\left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|+\left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right| x y z=0\) \((1+x y z)\left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|=0\) \(\because \quad \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\) \(\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\) \((1+x y z)\left|\begin{array}{ccc}x & x^{2} & 1 \\ y-x & y^{2}-x^{2} & 0 \\ z-x & z^{2}-x^{2} & 0\end{array}\right|=0\) \(\begin{aligned} & 1+x y z=0 \\ & x y z\end{aligned}=-1\)
AP EAMCET-2010
Matrix and Determinant
79178
If \(\left|\begin{array}{ccc}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right|\) has no inverse, then the real
1 2
2 3
3 0
4 1
Explanation:
(D) : Given, \({\left[\begin{array}{ccc} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{array}\right] \text { has no inverse }}\) \({\left[\begin{array}{ccc} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{array}\right]=0}\) \((x+1)+1(1-x)+x\left(-1-x^{2}\right)=0\) \(2-x-x^{3}=0\) \(x^{3}+x-2=0\) \((x-1)\left(x^{2}+x+2\right)=0\) \(x=1\)
AP EAMCET-2009
Matrix and Determinant
79179
If one of the roots of \(\left|\begin{array}{lll}3 & 5 & x \\ 7 & x & 7 \\ x & 5 & 3\end{array}\right|=0\) is -10 , then
79177
if \(y \quad y^{2} 1+y^{3}=0, x \neq y \neq z\) then \(1+x y z\) is \(\left.\begin{array}{lll}\mathrm{z} & \mathrm{z}^{2} & 1+\mathrm{z}^{3}\end{array} \right\rvert\,\) equal to
1 0
2 -1
3 1
4 2
Explanation:
(A) : Given, \(\left|\begin{array}{ccc} x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3} \end{array}\right|=0, x \neq y \neq z\) \(\left|\begin{array}{lll} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{array}\right|+\left|\begin{array}{ccc} x & x^{2} & x^{3} \\ y & y^{2} & y^{3} \\ z & z^{2} & z^{3} \end{array}\right|=0\) \(\left|\begin{array}{ccc}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|+\left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right| x y z=0\) \(C_{1} \leftrightarrow C_{2}\) and \(C_{2} \leftrightarrow C_{3}\) \(\left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|+\left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right| x y z=0\) \((1+x y z)\left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|=0\) \(\because \quad \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\) \(\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\) \((1+x y z)\left|\begin{array}{ccc}x & x^{2} & 1 \\ y-x & y^{2}-x^{2} & 0 \\ z-x & z^{2}-x^{2} & 0\end{array}\right|=0\) \(\begin{aligned} & 1+x y z=0 \\ & x y z\end{aligned}=-1\)
AP EAMCET-2010
Matrix and Determinant
79178
If \(\left|\begin{array}{ccc}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right|\) has no inverse, then the real
1 2
2 3
3 0
4 1
Explanation:
(D) : Given, \({\left[\begin{array}{ccc} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{array}\right] \text { has no inverse }}\) \({\left[\begin{array}{ccc} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{array}\right]=0}\) \((x+1)+1(1-x)+x\left(-1-x^{2}\right)=0\) \(2-x-x^{3}=0\) \(x^{3}+x-2=0\) \((x-1)\left(x^{2}+x+2\right)=0\) \(x=1\)
AP EAMCET-2009
Matrix and Determinant
79179
If one of the roots of \(\left|\begin{array}{lll}3 & 5 & x \\ 7 & x & 7 \\ x & 5 & 3\end{array}\right|=0\) is -10 , then
79177
if \(y \quad y^{2} 1+y^{3}=0, x \neq y \neq z\) then \(1+x y z\) is \(\left.\begin{array}{lll}\mathrm{z} & \mathrm{z}^{2} & 1+\mathrm{z}^{3}\end{array} \right\rvert\,\) equal to
1 0
2 -1
3 1
4 2
Explanation:
(A) : Given, \(\left|\begin{array}{ccc} x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3} \end{array}\right|=0, x \neq y \neq z\) \(\left|\begin{array}{lll} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{array}\right|+\left|\begin{array}{ccc} x & x^{2} & x^{3} \\ y & y^{2} & y^{3} \\ z & z^{2} & z^{3} \end{array}\right|=0\) \(\left|\begin{array}{ccc}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|+\left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right| x y z=0\) \(C_{1} \leftrightarrow C_{2}\) and \(C_{2} \leftrightarrow C_{3}\) \(\left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|+\left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right| x y z=0\) \((1+x y z)\left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|=0\) \(\because \quad \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\) \(\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\) \((1+x y z)\left|\begin{array}{ccc}x & x^{2} & 1 \\ y-x & y^{2}-x^{2} & 0 \\ z-x & z^{2}-x^{2} & 0\end{array}\right|=0\) \(\begin{aligned} & 1+x y z=0 \\ & x y z\end{aligned}=-1\)
AP EAMCET-2010
Matrix and Determinant
79178
If \(\left|\begin{array}{ccc}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right|\) has no inverse, then the real
1 2
2 3
3 0
4 1
Explanation:
(D) : Given, \({\left[\begin{array}{ccc} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{array}\right] \text { has no inverse }}\) \({\left[\begin{array}{ccc} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{array}\right]=0}\) \((x+1)+1(1-x)+x\left(-1-x^{2}\right)=0\) \(2-x-x^{3}=0\) \(x^{3}+x-2=0\) \((x-1)\left(x^{2}+x+2\right)=0\) \(x=1\)
AP EAMCET-2009
Matrix and Determinant
79179
If one of the roots of \(\left|\begin{array}{lll}3 & 5 & x \\ 7 & x & 7 \\ x & 5 & 3\end{array}\right|=0\) is -10 , then
79177
if \(y \quad y^{2} 1+y^{3}=0, x \neq y \neq z\) then \(1+x y z\) is \(\left.\begin{array}{lll}\mathrm{z} & \mathrm{z}^{2} & 1+\mathrm{z}^{3}\end{array} \right\rvert\,\) equal to
1 0
2 -1
3 1
4 2
Explanation:
(A) : Given, \(\left|\begin{array}{ccc} x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3} \end{array}\right|=0, x \neq y \neq z\) \(\left|\begin{array}{lll} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{array}\right|+\left|\begin{array}{ccc} x & x^{2} & x^{3} \\ y & y^{2} & y^{3} \\ z & z^{2} & z^{3} \end{array}\right|=0\) \(\left|\begin{array}{ccc}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|+\left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right| x y z=0\) \(C_{1} \leftrightarrow C_{2}\) and \(C_{2} \leftrightarrow C_{3}\) \(\left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|+\left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right| x y z=0\) \((1+x y z)\left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|=0\) \(\because \quad \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\) \(\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\) \((1+x y z)\left|\begin{array}{ccc}x & x^{2} & 1 \\ y-x & y^{2}-x^{2} & 0 \\ z-x & z^{2}-x^{2} & 0\end{array}\right|=0\) \(\begin{aligned} & 1+x y z=0 \\ & x y z\end{aligned}=-1\)
AP EAMCET-2010
Matrix and Determinant
79178
If \(\left|\begin{array}{ccc}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right|\) has no inverse, then the real
1 2
2 3
3 0
4 1
Explanation:
(D) : Given, \({\left[\begin{array}{ccc} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{array}\right] \text { has no inverse }}\) \({\left[\begin{array}{ccc} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{array}\right]=0}\) \((x+1)+1(1-x)+x\left(-1-x^{2}\right)=0\) \(2-x-x^{3}=0\) \(x^{3}+x-2=0\) \((x-1)\left(x^{2}+x+2\right)=0\) \(x=1\)
AP EAMCET-2009
Matrix and Determinant
79179
If one of the roots of \(\left|\begin{array}{lll}3 & 5 & x \\ 7 & x & 7 \\ x & 5 & 3\end{array}\right|=0\) is -10 , then
79177
if \(y \quad y^{2} 1+y^{3}=0, x \neq y \neq z\) then \(1+x y z\) is \(\left.\begin{array}{lll}\mathrm{z} & \mathrm{z}^{2} & 1+\mathrm{z}^{3}\end{array} \right\rvert\,\) equal to
1 0
2 -1
3 1
4 2
Explanation:
(A) : Given, \(\left|\begin{array}{ccc} x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3} \end{array}\right|=0, x \neq y \neq z\) \(\left|\begin{array}{lll} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{array}\right|+\left|\begin{array}{ccc} x & x^{2} & x^{3} \\ y & y^{2} & y^{3} \\ z & z^{2} & z^{3} \end{array}\right|=0\) \(\left|\begin{array}{ccc}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|+\left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right| x y z=0\) \(C_{1} \leftrightarrow C_{2}\) and \(C_{2} \leftrightarrow C_{3}\) \(\left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|+\left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right| x y z=0\) \((1+x y z)\left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|=0\) \(\because \quad \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\) \(\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\) \((1+x y z)\left|\begin{array}{ccc}x & x^{2} & 1 \\ y-x & y^{2}-x^{2} & 0 \\ z-x & z^{2}-x^{2} & 0\end{array}\right|=0\) \(\begin{aligned} & 1+x y z=0 \\ & x y z\end{aligned}=-1\)
AP EAMCET-2010
Matrix and Determinant
79178
If \(\left|\begin{array}{ccc}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right|\) has no inverse, then the real
1 2
2 3
3 0
4 1
Explanation:
(D) : Given, \({\left[\begin{array}{ccc} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{array}\right] \text { has no inverse }}\) \({\left[\begin{array}{ccc} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{array}\right]=0}\) \((x+1)+1(1-x)+x\left(-1-x^{2}\right)=0\) \(2-x-x^{3}=0\) \(x^{3}+x-2=0\) \((x-1)\left(x^{2}+x+2\right)=0\) \(x=1\)
AP EAMCET-2009
Matrix and Determinant
79179
If one of the roots of \(\left|\begin{array}{lll}3 & 5 & x \\ 7 & x & 7 \\ x & 5 & 3\end{array}\right|=0\) is -10 , then
