79190
If \(S_{r}=\left|\begin{array}{ccc}2 r & x & n(n+1) \\ 6 r^{2}-1 & y & n^{2}(2 n+3) \\ 4 r^{3}-2 n r & z & n^{3}(n+1)\end{array}\right|\), then the value of \(\sum_{r=1}^{n} S_{r}\) is independent of
1 only \(x\)
2 only y
3 only \(\mathrm{n}\)
4 \(x, y, z\) and \(n\)
Explanation:
(D) : We have, \(S_{r} =\left|\begin{array}{ccc} 2 r & x & n(n+1) \\ 6 r^{2}-1 & y & n^{2}(2 n+3) \\ 4 r^{3}-2 n r & z & n^{3}(n+1) \end{array}\right|\) \(\sum_{r=1}^{n} S_{r} =\left|\begin{array}{ccc} 2 \sum_{r=1}^{n} r & x & n(n+1) \\ \sum_{r=1}^{n}\left(6 r^{2}-1\right) & y & n^{2}(2 n+3) \\ \sum_{r=1}^{n}\left(4 r^{3}-2 n r\right) & z & n^{3}(n+1) \end{array}\right|\) \(=\left|\begin{array}{ccc} n(n+1) & x & n(n+1) \\ n^{2}(2 n+3) & y & n^{2}(2 n+3) \\ n^{3}(n+1) & z & n^{3}(n+1) \end{array}\right|\) \(=0\left(\because C_{1} \text { and } C_{3}\right. \text { are identical) }\) Hence, \(\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{r}}\) in independent of \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) and \(\mathrm{n}\).
WB JEE-2018
Matrix and Determinant
79191
If \(P=3 \hat{i}+5 \hat{j}-\hat{k}\) and \(Q=\hat{i}+2 \hat{j}+3 \hat{k}\) are two sides of a triangle, then its area is equal to ..........sq. units.
1 \(\frac{\sqrt{390}}{4}\)
2 \(\sqrt{390}\)
3 \(\frac{\sqrt{390}}{2}\)
4 \(\frac{\sqrt{390}}{8}\)
Explanation:
(C) : Sides of a triangle given as \(P=3 \hat{i}+5 \hat{j}-\hat{k} \text { and } Q=\hat{i}+2 \hat{j}+3 \hat{k}\) \(\therefore\) Area of triangle \(=1 / 2(\mathrm{P} \times \mathrm{Q})\) \(=\frac{1}{2}\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & 5 & -1 \\ 1 & 2 & 3 \end{array}\right|\) \(=\frac{1}{2}|\hat{\mathrm{i}}(15+2)-\hat{\mathrm{j}}(9+1)+\hat{\mathrm{k}}(6-5)|\) \(=\frac{1}{2}|17 \hat{\mathrm{i}}-10 \hat{\mathrm{j}}+\hat{\mathrm{k}}|\) \(=\frac{1}{2} \sqrt{289+100+1}=\frac{1}{2} \sqrt{390} \text { sq. unit }\)
AP EAMCET-2020-21.09.2020
Matrix and Determinant
79192
If \(\left[\begin{array}{rrr}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]\) has no inverse, then the real value of \(x\) is
1 2
2 3
3 0
4 1
Explanation:
(D) : It is given that matrix \(A=\left[\begin{array}{ccc} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{array}\right] \text { have no inverse so }\) \(|A|=0=\left[\begin{array}{ccc} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{array}\right]=0\) \(\left|(x+1)+1(1-x)+x\left(-1-x^{2}\right)\right|=0\) \(x+1+1-x-x-x^{3}=0\) \(x^{3}+x-2=0\) \((x-1)\left(x^{2}+x+2\right)=0\) Either \(\mathrm{x}=1\) or \(\mathrm{x}^{2}+\mathrm{x}+2=0\) But discriminant of quadratic equation \(\mathrm{x}^{2}+\mathrm{x}+2=0\) is negative so no real roots \(\therefore \quad|\mathrm{A}|=0, \mathrm{x}=1\)
AP EAMCET-2020-21.09.2020
Matrix and Determinant
79193
Let \(A=\left(\begin{array}{lll}3 & 0 & 3 \\ 0 & 3 & 0 \\ 3 & 0 & 3\end{array}\right)\). Then, the roots of the equation det \(\left(A-\lambda I_{3}\right)=0\) (where \(I_{3}\) is the identity matrix of order 3 ) are
79190
If \(S_{r}=\left|\begin{array}{ccc}2 r & x & n(n+1) \\ 6 r^{2}-1 & y & n^{2}(2 n+3) \\ 4 r^{3}-2 n r & z & n^{3}(n+1)\end{array}\right|\), then the value of \(\sum_{r=1}^{n} S_{r}\) is independent of
1 only \(x\)
2 only y
3 only \(\mathrm{n}\)
4 \(x, y, z\) and \(n\)
Explanation:
(D) : We have, \(S_{r} =\left|\begin{array}{ccc} 2 r & x & n(n+1) \\ 6 r^{2}-1 & y & n^{2}(2 n+3) \\ 4 r^{3}-2 n r & z & n^{3}(n+1) \end{array}\right|\) \(\sum_{r=1}^{n} S_{r} =\left|\begin{array}{ccc} 2 \sum_{r=1}^{n} r & x & n(n+1) \\ \sum_{r=1}^{n}\left(6 r^{2}-1\right) & y & n^{2}(2 n+3) \\ \sum_{r=1}^{n}\left(4 r^{3}-2 n r\right) & z & n^{3}(n+1) \end{array}\right|\) \(=\left|\begin{array}{ccc} n(n+1) & x & n(n+1) \\ n^{2}(2 n+3) & y & n^{2}(2 n+3) \\ n^{3}(n+1) & z & n^{3}(n+1) \end{array}\right|\) \(=0\left(\because C_{1} \text { and } C_{3}\right. \text { are identical) }\) Hence, \(\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{r}}\) in independent of \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) and \(\mathrm{n}\).
WB JEE-2018
Matrix and Determinant
79191
If \(P=3 \hat{i}+5 \hat{j}-\hat{k}\) and \(Q=\hat{i}+2 \hat{j}+3 \hat{k}\) are two sides of a triangle, then its area is equal to ..........sq. units.
1 \(\frac{\sqrt{390}}{4}\)
2 \(\sqrt{390}\)
3 \(\frac{\sqrt{390}}{2}\)
4 \(\frac{\sqrt{390}}{8}\)
Explanation:
(C) : Sides of a triangle given as \(P=3 \hat{i}+5 \hat{j}-\hat{k} \text { and } Q=\hat{i}+2 \hat{j}+3 \hat{k}\) \(\therefore\) Area of triangle \(=1 / 2(\mathrm{P} \times \mathrm{Q})\) \(=\frac{1}{2}\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & 5 & -1 \\ 1 & 2 & 3 \end{array}\right|\) \(=\frac{1}{2}|\hat{\mathrm{i}}(15+2)-\hat{\mathrm{j}}(9+1)+\hat{\mathrm{k}}(6-5)|\) \(=\frac{1}{2}|17 \hat{\mathrm{i}}-10 \hat{\mathrm{j}}+\hat{\mathrm{k}}|\) \(=\frac{1}{2} \sqrt{289+100+1}=\frac{1}{2} \sqrt{390} \text { sq. unit }\)
AP EAMCET-2020-21.09.2020
Matrix and Determinant
79192
If \(\left[\begin{array}{rrr}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]\) has no inverse, then the real value of \(x\) is
1 2
2 3
3 0
4 1
Explanation:
(D) : It is given that matrix \(A=\left[\begin{array}{ccc} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{array}\right] \text { have no inverse so }\) \(|A|=0=\left[\begin{array}{ccc} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{array}\right]=0\) \(\left|(x+1)+1(1-x)+x\left(-1-x^{2}\right)\right|=0\) \(x+1+1-x-x-x^{3}=0\) \(x^{3}+x-2=0\) \((x-1)\left(x^{2}+x+2\right)=0\) Either \(\mathrm{x}=1\) or \(\mathrm{x}^{2}+\mathrm{x}+2=0\) But discriminant of quadratic equation \(\mathrm{x}^{2}+\mathrm{x}+2=0\) is negative so no real roots \(\therefore \quad|\mathrm{A}|=0, \mathrm{x}=1\)
AP EAMCET-2020-21.09.2020
Matrix and Determinant
79193
Let \(A=\left(\begin{array}{lll}3 & 0 & 3 \\ 0 & 3 & 0 \\ 3 & 0 & 3\end{array}\right)\). Then, the roots of the equation det \(\left(A-\lambda I_{3}\right)=0\) (where \(I_{3}\) is the identity matrix of order 3 ) are
79190
If \(S_{r}=\left|\begin{array}{ccc}2 r & x & n(n+1) \\ 6 r^{2}-1 & y & n^{2}(2 n+3) \\ 4 r^{3}-2 n r & z & n^{3}(n+1)\end{array}\right|\), then the value of \(\sum_{r=1}^{n} S_{r}\) is independent of
1 only \(x\)
2 only y
3 only \(\mathrm{n}\)
4 \(x, y, z\) and \(n\)
Explanation:
(D) : We have, \(S_{r} =\left|\begin{array}{ccc} 2 r & x & n(n+1) \\ 6 r^{2}-1 & y & n^{2}(2 n+3) \\ 4 r^{3}-2 n r & z & n^{3}(n+1) \end{array}\right|\) \(\sum_{r=1}^{n} S_{r} =\left|\begin{array}{ccc} 2 \sum_{r=1}^{n} r & x & n(n+1) \\ \sum_{r=1}^{n}\left(6 r^{2}-1\right) & y & n^{2}(2 n+3) \\ \sum_{r=1}^{n}\left(4 r^{3}-2 n r\right) & z & n^{3}(n+1) \end{array}\right|\) \(=\left|\begin{array}{ccc} n(n+1) & x & n(n+1) \\ n^{2}(2 n+3) & y & n^{2}(2 n+3) \\ n^{3}(n+1) & z & n^{3}(n+1) \end{array}\right|\) \(=0\left(\because C_{1} \text { and } C_{3}\right. \text { are identical) }\) Hence, \(\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{r}}\) in independent of \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) and \(\mathrm{n}\).
WB JEE-2018
Matrix and Determinant
79191
If \(P=3 \hat{i}+5 \hat{j}-\hat{k}\) and \(Q=\hat{i}+2 \hat{j}+3 \hat{k}\) are two sides of a triangle, then its area is equal to ..........sq. units.
1 \(\frac{\sqrt{390}}{4}\)
2 \(\sqrt{390}\)
3 \(\frac{\sqrt{390}}{2}\)
4 \(\frac{\sqrt{390}}{8}\)
Explanation:
(C) : Sides of a triangle given as \(P=3 \hat{i}+5 \hat{j}-\hat{k} \text { and } Q=\hat{i}+2 \hat{j}+3 \hat{k}\) \(\therefore\) Area of triangle \(=1 / 2(\mathrm{P} \times \mathrm{Q})\) \(=\frac{1}{2}\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & 5 & -1 \\ 1 & 2 & 3 \end{array}\right|\) \(=\frac{1}{2}|\hat{\mathrm{i}}(15+2)-\hat{\mathrm{j}}(9+1)+\hat{\mathrm{k}}(6-5)|\) \(=\frac{1}{2}|17 \hat{\mathrm{i}}-10 \hat{\mathrm{j}}+\hat{\mathrm{k}}|\) \(=\frac{1}{2} \sqrt{289+100+1}=\frac{1}{2} \sqrt{390} \text { sq. unit }\)
AP EAMCET-2020-21.09.2020
Matrix and Determinant
79192
If \(\left[\begin{array}{rrr}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]\) has no inverse, then the real value of \(x\) is
1 2
2 3
3 0
4 1
Explanation:
(D) : It is given that matrix \(A=\left[\begin{array}{ccc} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{array}\right] \text { have no inverse so }\) \(|A|=0=\left[\begin{array}{ccc} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{array}\right]=0\) \(\left|(x+1)+1(1-x)+x\left(-1-x^{2}\right)\right|=0\) \(x+1+1-x-x-x^{3}=0\) \(x^{3}+x-2=0\) \((x-1)\left(x^{2}+x+2\right)=0\) Either \(\mathrm{x}=1\) or \(\mathrm{x}^{2}+\mathrm{x}+2=0\) But discriminant of quadratic equation \(\mathrm{x}^{2}+\mathrm{x}+2=0\) is negative so no real roots \(\therefore \quad|\mathrm{A}|=0, \mathrm{x}=1\)
AP EAMCET-2020-21.09.2020
Matrix and Determinant
79193
Let \(A=\left(\begin{array}{lll}3 & 0 & 3 \\ 0 & 3 & 0 \\ 3 & 0 & 3\end{array}\right)\). Then, the roots of the equation det \(\left(A-\lambda I_{3}\right)=0\) (where \(I_{3}\) is the identity matrix of order 3 ) are
79190
If \(S_{r}=\left|\begin{array}{ccc}2 r & x & n(n+1) \\ 6 r^{2}-1 & y & n^{2}(2 n+3) \\ 4 r^{3}-2 n r & z & n^{3}(n+1)\end{array}\right|\), then the value of \(\sum_{r=1}^{n} S_{r}\) is independent of
1 only \(x\)
2 only y
3 only \(\mathrm{n}\)
4 \(x, y, z\) and \(n\)
Explanation:
(D) : We have, \(S_{r} =\left|\begin{array}{ccc} 2 r & x & n(n+1) \\ 6 r^{2}-1 & y & n^{2}(2 n+3) \\ 4 r^{3}-2 n r & z & n^{3}(n+1) \end{array}\right|\) \(\sum_{r=1}^{n} S_{r} =\left|\begin{array}{ccc} 2 \sum_{r=1}^{n} r & x & n(n+1) \\ \sum_{r=1}^{n}\left(6 r^{2}-1\right) & y & n^{2}(2 n+3) \\ \sum_{r=1}^{n}\left(4 r^{3}-2 n r\right) & z & n^{3}(n+1) \end{array}\right|\) \(=\left|\begin{array}{ccc} n(n+1) & x & n(n+1) \\ n^{2}(2 n+3) & y & n^{2}(2 n+3) \\ n^{3}(n+1) & z & n^{3}(n+1) \end{array}\right|\) \(=0\left(\because C_{1} \text { and } C_{3}\right. \text { are identical) }\) Hence, \(\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{r}}\) in independent of \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) and \(\mathrm{n}\).
WB JEE-2018
Matrix and Determinant
79191
If \(P=3 \hat{i}+5 \hat{j}-\hat{k}\) and \(Q=\hat{i}+2 \hat{j}+3 \hat{k}\) are two sides of a triangle, then its area is equal to ..........sq. units.
1 \(\frac{\sqrt{390}}{4}\)
2 \(\sqrt{390}\)
3 \(\frac{\sqrt{390}}{2}\)
4 \(\frac{\sqrt{390}}{8}\)
Explanation:
(C) : Sides of a triangle given as \(P=3 \hat{i}+5 \hat{j}-\hat{k} \text { and } Q=\hat{i}+2 \hat{j}+3 \hat{k}\) \(\therefore\) Area of triangle \(=1 / 2(\mathrm{P} \times \mathrm{Q})\) \(=\frac{1}{2}\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & 5 & -1 \\ 1 & 2 & 3 \end{array}\right|\) \(=\frac{1}{2}|\hat{\mathrm{i}}(15+2)-\hat{\mathrm{j}}(9+1)+\hat{\mathrm{k}}(6-5)|\) \(=\frac{1}{2}|17 \hat{\mathrm{i}}-10 \hat{\mathrm{j}}+\hat{\mathrm{k}}|\) \(=\frac{1}{2} \sqrt{289+100+1}=\frac{1}{2} \sqrt{390} \text { sq. unit }\)
AP EAMCET-2020-21.09.2020
Matrix and Determinant
79192
If \(\left[\begin{array}{rrr}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]\) has no inverse, then the real value of \(x\) is
1 2
2 3
3 0
4 1
Explanation:
(D) : It is given that matrix \(A=\left[\begin{array}{ccc} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{array}\right] \text { have no inverse so }\) \(|A|=0=\left[\begin{array}{ccc} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{array}\right]=0\) \(\left|(x+1)+1(1-x)+x\left(-1-x^{2}\right)\right|=0\) \(x+1+1-x-x-x^{3}=0\) \(x^{3}+x-2=0\) \((x-1)\left(x^{2}+x+2\right)=0\) Either \(\mathrm{x}=1\) or \(\mathrm{x}^{2}+\mathrm{x}+2=0\) But discriminant of quadratic equation \(\mathrm{x}^{2}+\mathrm{x}+2=0\) is negative so no real roots \(\therefore \quad|\mathrm{A}|=0, \mathrm{x}=1\)
AP EAMCET-2020-21.09.2020
Matrix and Determinant
79193
Let \(A=\left(\begin{array}{lll}3 & 0 & 3 \\ 0 & 3 & 0 \\ 3 & 0 & 3\end{array}\right)\). Then, the roots of the equation det \(\left(A-\lambda I_{3}\right)=0\) (where \(I_{3}\) is the identity matrix of order 3 ) are
