79005
If \(\left[\begin{array}{cc}-2 & 5 \\ 3 & -1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{c}3 \\ -1\end{array}\right]\), then \((x, y)\) is
1 \((1,2)\)
2 \((-1,2)\)
3 \((1,-2)\)
4 \((2,1)\)
Explanation:
(D) : We have, \({\left[\begin{array}{cc} -2 & 5 \\ 3 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\left[\begin{array}{c} 3 \\ -1 \end{array}\right]}\) \({\left[\begin{array}{c} -2 x+5 y \\ 3 x-y \end{array}\right]=\left[\begin{array}{l} 1 \\ 5 \end{array}\right]}\) Comparing these value, we get - \(-2 x+5 y=1, \quad 3 x-y=5 \\ y=3 x-5\) \(-2 \mathrm{x}+5(3 \mathrm{x}-5)=1 \\ 13 \mathrm{x}=26 \Rightarrow \mathrm{x}=2\) Substituting \(x=2\), we get- \(y=6-5=1\) Hence, \((\mathrm{x}, \mathrm{y})=(2,1)\)
COMEDK-2014
Matrix and Determinant
79006
If \(f(x)=\left|\begin{array}{ccc}a & -1 & 0 \\ a x & a & 1 \\ a x^{2} & a x & a\end{array}\right|\), then \(f(2 x)-f(x)\) equals
1 \(a(2 a+3 x)\)
2 ax \((2 x+3 a)\)
3 \(a x(2 a+3 x)\)
4 \(x(2 a+3 x)\)
Explanation:
(C) : We have, \(f(x)=\left|\begin{array}{ccc}a & -1 & 0 \\ a x & a & 1 \\ a x^{2} & a x & a\end{array}\right| \Rightarrow f(x)=a\left|\begin{array}{ccc}1 & -1 & 0 \\ x & a & -1 \\ x^{2} & a x & a\end{array}\right|\) \(f(x)=a\left[1\left(a^{2}+a x\right)+1\left(a x+x^{2}\right)+0\right]\) \(=\mathrm{a}\left(\mathrm{a}^{2}+2 \mathrm{ax}+\mathrm{x}^{2}\right)\) \(f(x)=a(a+x)^{2}\) \(\therefore \quad \mathrm{f}(2 \mathrm{x})=\mathrm{a}(\mathrm{a}+2 \mathrm{x})^{2}\) So, \(f(2 x)-f(x)=a\left[(a+2 x)^{2}-(a+x)^{2}\right]\) \(=a\left[a^{2}+4 x^{2}+4 a x-a^{2}-x^{2}-2 a x\right]\) \(=a\left(3 x^{2}+2 a x\right)=a x(2 a+3 x)\)
SRM JEE-2008
Matrix and Determinant
79007
If \(p+q+r=0=a+b+c\), then the value of the determinant \(\left|\begin{array}{lll}p a & q b & \mathbf{r c} \\ \mathbf{q c} & \mathbf{r a} & \mathbf{p b} \\ \mathbf{r b} & \mathbf{p c} & \mathbf{q a}\end{array}\right|\) is
1 0
2 \(\mathrm{pa}+\mathrm{qb}+\mathrm{rc}\)
3 1
4 -1
Explanation:
(A) : Here, Let, \(\quad \Delta=\left|\begin{array}{ccc}\mathrm{pa} & \mathrm{qb} & \mathrm{rc} \\ \mathrm{qc} & \mathrm{ra} & \mathrm{pb} \\ \mathrm{rb} & \mathrm{pc} & \mathrm{qa}\end{array}\right|\) \(\Delta=p a\left(a^{2} r q-p^{2} c b\right)-q b\left(q^{2} c a-b^{2} p r\right)+r c\left(q p c^{2}-r^{2} a b\right)\) \(\Delta=\) prqa \(^{3}-\mathrm{p}^{3} \mathrm{abc}-\mathrm{q}^{3} \mathrm{abc}+\mathrm{pqrb}^{3}+\mathrm{pqrc}^{3}-\mathrm{r}^{3} \mathrm{abc}\) \(=\operatorname{pqr}\left(\mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}\right)-\mathrm{abc}\left(\mathrm{p}^{3}+\mathrm{q}^{3}+\mathrm{r}^{3}\right)\) \(\because \quad \mathrm{a}+\mathrm{b}+\mathrm{c}=0=\mathrm{p}+\mathrm{q}+\mathrm{r}\) (given) \(\therefore \quad a^{3}+b^{3}+c^{3}=3 a b c\) and \(\quad \mathrm{p}^{3}+\mathrm{q}^{3}+\mathrm{r}^{3}=3 \mathrm{pqr}\) So, \(\quad \Delta=\operatorname{pqr}(3 \mathrm{abc})-\mathrm{abc}(3 \mathrm{pqr})\) \(\Delta=0\)
79005
If \(\left[\begin{array}{cc}-2 & 5 \\ 3 & -1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{c}3 \\ -1\end{array}\right]\), then \((x, y)\) is
1 \((1,2)\)
2 \((-1,2)\)
3 \((1,-2)\)
4 \((2,1)\)
Explanation:
(D) : We have, \({\left[\begin{array}{cc} -2 & 5 \\ 3 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\left[\begin{array}{c} 3 \\ -1 \end{array}\right]}\) \({\left[\begin{array}{c} -2 x+5 y \\ 3 x-y \end{array}\right]=\left[\begin{array}{l} 1 \\ 5 \end{array}\right]}\) Comparing these value, we get - \(-2 x+5 y=1, \quad 3 x-y=5 \\ y=3 x-5\) \(-2 \mathrm{x}+5(3 \mathrm{x}-5)=1 \\ 13 \mathrm{x}=26 \Rightarrow \mathrm{x}=2\) Substituting \(x=2\), we get- \(y=6-5=1\) Hence, \((\mathrm{x}, \mathrm{y})=(2,1)\)
COMEDK-2014
Matrix and Determinant
79006
If \(f(x)=\left|\begin{array}{ccc}a & -1 & 0 \\ a x & a & 1 \\ a x^{2} & a x & a\end{array}\right|\), then \(f(2 x)-f(x)\) equals
1 \(a(2 a+3 x)\)
2 ax \((2 x+3 a)\)
3 \(a x(2 a+3 x)\)
4 \(x(2 a+3 x)\)
Explanation:
(C) : We have, \(f(x)=\left|\begin{array}{ccc}a & -1 & 0 \\ a x & a & 1 \\ a x^{2} & a x & a\end{array}\right| \Rightarrow f(x)=a\left|\begin{array}{ccc}1 & -1 & 0 \\ x & a & -1 \\ x^{2} & a x & a\end{array}\right|\) \(f(x)=a\left[1\left(a^{2}+a x\right)+1\left(a x+x^{2}\right)+0\right]\) \(=\mathrm{a}\left(\mathrm{a}^{2}+2 \mathrm{ax}+\mathrm{x}^{2}\right)\) \(f(x)=a(a+x)^{2}\) \(\therefore \quad \mathrm{f}(2 \mathrm{x})=\mathrm{a}(\mathrm{a}+2 \mathrm{x})^{2}\) So, \(f(2 x)-f(x)=a\left[(a+2 x)^{2}-(a+x)^{2}\right]\) \(=a\left[a^{2}+4 x^{2}+4 a x-a^{2}-x^{2}-2 a x\right]\) \(=a\left(3 x^{2}+2 a x\right)=a x(2 a+3 x)\)
SRM JEE-2008
Matrix and Determinant
79007
If \(p+q+r=0=a+b+c\), then the value of the determinant \(\left|\begin{array}{lll}p a & q b & \mathbf{r c} \\ \mathbf{q c} & \mathbf{r a} & \mathbf{p b} \\ \mathbf{r b} & \mathbf{p c} & \mathbf{q a}\end{array}\right|\) is
1 0
2 \(\mathrm{pa}+\mathrm{qb}+\mathrm{rc}\)
3 1
4 -1
Explanation:
(A) : Here, Let, \(\quad \Delta=\left|\begin{array}{ccc}\mathrm{pa} & \mathrm{qb} & \mathrm{rc} \\ \mathrm{qc} & \mathrm{ra} & \mathrm{pb} \\ \mathrm{rb} & \mathrm{pc} & \mathrm{qa}\end{array}\right|\) \(\Delta=p a\left(a^{2} r q-p^{2} c b\right)-q b\left(q^{2} c a-b^{2} p r\right)+r c\left(q p c^{2}-r^{2} a b\right)\) \(\Delta=\) prqa \(^{3}-\mathrm{p}^{3} \mathrm{abc}-\mathrm{q}^{3} \mathrm{abc}+\mathrm{pqrb}^{3}+\mathrm{pqrc}^{3}-\mathrm{r}^{3} \mathrm{abc}\) \(=\operatorname{pqr}\left(\mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}\right)-\mathrm{abc}\left(\mathrm{p}^{3}+\mathrm{q}^{3}+\mathrm{r}^{3}\right)\) \(\because \quad \mathrm{a}+\mathrm{b}+\mathrm{c}=0=\mathrm{p}+\mathrm{q}+\mathrm{r}\) (given) \(\therefore \quad a^{3}+b^{3}+c^{3}=3 a b c\) and \(\quad \mathrm{p}^{3}+\mathrm{q}^{3}+\mathrm{r}^{3}=3 \mathrm{pqr}\) So, \(\quad \Delta=\operatorname{pqr}(3 \mathrm{abc})-\mathrm{abc}(3 \mathrm{pqr})\) \(\Delta=0\)
79005
If \(\left[\begin{array}{cc}-2 & 5 \\ 3 & -1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{c}3 \\ -1\end{array}\right]\), then \((x, y)\) is
1 \((1,2)\)
2 \((-1,2)\)
3 \((1,-2)\)
4 \((2,1)\)
Explanation:
(D) : We have, \({\left[\begin{array}{cc} -2 & 5 \\ 3 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\left[\begin{array}{c} 3 \\ -1 \end{array}\right]}\) \({\left[\begin{array}{c} -2 x+5 y \\ 3 x-y \end{array}\right]=\left[\begin{array}{l} 1 \\ 5 \end{array}\right]}\) Comparing these value, we get - \(-2 x+5 y=1, \quad 3 x-y=5 \\ y=3 x-5\) \(-2 \mathrm{x}+5(3 \mathrm{x}-5)=1 \\ 13 \mathrm{x}=26 \Rightarrow \mathrm{x}=2\) Substituting \(x=2\), we get- \(y=6-5=1\) Hence, \((\mathrm{x}, \mathrm{y})=(2,1)\)
COMEDK-2014
Matrix and Determinant
79006
If \(f(x)=\left|\begin{array}{ccc}a & -1 & 0 \\ a x & a & 1 \\ a x^{2} & a x & a\end{array}\right|\), then \(f(2 x)-f(x)\) equals
1 \(a(2 a+3 x)\)
2 ax \((2 x+3 a)\)
3 \(a x(2 a+3 x)\)
4 \(x(2 a+3 x)\)
Explanation:
(C) : We have, \(f(x)=\left|\begin{array}{ccc}a & -1 & 0 \\ a x & a & 1 \\ a x^{2} & a x & a\end{array}\right| \Rightarrow f(x)=a\left|\begin{array}{ccc}1 & -1 & 0 \\ x & a & -1 \\ x^{2} & a x & a\end{array}\right|\) \(f(x)=a\left[1\left(a^{2}+a x\right)+1\left(a x+x^{2}\right)+0\right]\) \(=\mathrm{a}\left(\mathrm{a}^{2}+2 \mathrm{ax}+\mathrm{x}^{2}\right)\) \(f(x)=a(a+x)^{2}\) \(\therefore \quad \mathrm{f}(2 \mathrm{x})=\mathrm{a}(\mathrm{a}+2 \mathrm{x})^{2}\) So, \(f(2 x)-f(x)=a\left[(a+2 x)^{2}-(a+x)^{2}\right]\) \(=a\left[a^{2}+4 x^{2}+4 a x-a^{2}-x^{2}-2 a x\right]\) \(=a\left(3 x^{2}+2 a x\right)=a x(2 a+3 x)\)
SRM JEE-2008
Matrix and Determinant
79007
If \(p+q+r=0=a+b+c\), then the value of the determinant \(\left|\begin{array}{lll}p a & q b & \mathbf{r c} \\ \mathbf{q c} & \mathbf{r a} & \mathbf{p b} \\ \mathbf{r b} & \mathbf{p c} & \mathbf{q a}\end{array}\right|\) is
1 0
2 \(\mathrm{pa}+\mathrm{qb}+\mathrm{rc}\)
3 1
4 -1
Explanation:
(A) : Here, Let, \(\quad \Delta=\left|\begin{array}{ccc}\mathrm{pa} & \mathrm{qb} & \mathrm{rc} \\ \mathrm{qc} & \mathrm{ra} & \mathrm{pb} \\ \mathrm{rb} & \mathrm{pc} & \mathrm{qa}\end{array}\right|\) \(\Delta=p a\left(a^{2} r q-p^{2} c b\right)-q b\left(q^{2} c a-b^{2} p r\right)+r c\left(q p c^{2}-r^{2} a b\right)\) \(\Delta=\) prqa \(^{3}-\mathrm{p}^{3} \mathrm{abc}-\mathrm{q}^{3} \mathrm{abc}+\mathrm{pqrb}^{3}+\mathrm{pqrc}^{3}-\mathrm{r}^{3} \mathrm{abc}\) \(=\operatorname{pqr}\left(\mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}\right)-\mathrm{abc}\left(\mathrm{p}^{3}+\mathrm{q}^{3}+\mathrm{r}^{3}\right)\) \(\because \quad \mathrm{a}+\mathrm{b}+\mathrm{c}=0=\mathrm{p}+\mathrm{q}+\mathrm{r}\) (given) \(\therefore \quad a^{3}+b^{3}+c^{3}=3 a b c\) and \(\quad \mathrm{p}^{3}+\mathrm{q}^{3}+\mathrm{r}^{3}=3 \mathrm{pqr}\) So, \(\quad \Delta=\operatorname{pqr}(3 \mathrm{abc})-\mathrm{abc}(3 \mathrm{pqr})\) \(\Delta=0\)
79005
If \(\left[\begin{array}{cc}-2 & 5 \\ 3 & -1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{c}3 \\ -1\end{array}\right]\), then \((x, y)\) is
1 \((1,2)\)
2 \((-1,2)\)
3 \((1,-2)\)
4 \((2,1)\)
Explanation:
(D) : We have, \({\left[\begin{array}{cc} -2 & 5 \\ 3 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\left[\begin{array}{c} 3 \\ -1 \end{array}\right]}\) \({\left[\begin{array}{c} -2 x+5 y \\ 3 x-y \end{array}\right]=\left[\begin{array}{l} 1 \\ 5 \end{array}\right]}\) Comparing these value, we get - \(-2 x+5 y=1, \quad 3 x-y=5 \\ y=3 x-5\) \(-2 \mathrm{x}+5(3 \mathrm{x}-5)=1 \\ 13 \mathrm{x}=26 \Rightarrow \mathrm{x}=2\) Substituting \(x=2\), we get- \(y=6-5=1\) Hence, \((\mathrm{x}, \mathrm{y})=(2,1)\)
COMEDK-2014
Matrix and Determinant
79006
If \(f(x)=\left|\begin{array}{ccc}a & -1 & 0 \\ a x & a & 1 \\ a x^{2} & a x & a\end{array}\right|\), then \(f(2 x)-f(x)\) equals
1 \(a(2 a+3 x)\)
2 ax \((2 x+3 a)\)
3 \(a x(2 a+3 x)\)
4 \(x(2 a+3 x)\)
Explanation:
(C) : We have, \(f(x)=\left|\begin{array}{ccc}a & -1 & 0 \\ a x & a & 1 \\ a x^{2} & a x & a\end{array}\right| \Rightarrow f(x)=a\left|\begin{array}{ccc}1 & -1 & 0 \\ x & a & -1 \\ x^{2} & a x & a\end{array}\right|\) \(f(x)=a\left[1\left(a^{2}+a x\right)+1\left(a x+x^{2}\right)+0\right]\) \(=\mathrm{a}\left(\mathrm{a}^{2}+2 \mathrm{ax}+\mathrm{x}^{2}\right)\) \(f(x)=a(a+x)^{2}\) \(\therefore \quad \mathrm{f}(2 \mathrm{x})=\mathrm{a}(\mathrm{a}+2 \mathrm{x})^{2}\) So, \(f(2 x)-f(x)=a\left[(a+2 x)^{2}-(a+x)^{2}\right]\) \(=a\left[a^{2}+4 x^{2}+4 a x-a^{2}-x^{2}-2 a x\right]\) \(=a\left(3 x^{2}+2 a x\right)=a x(2 a+3 x)\)
SRM JEE-2008
Matrix and Determinant
79007
If \(p+q+r=0=a+b+c\), then the value of the determinant \(\left|\begin{array}{lll}p a & q b & \mathbf{r c} \\ \mathbf{q c} & \mathbf{r a} & \mathbf{p b} \\ \mathbf{r b} & \mathbf{p c} & \mathbf{q a}\end{array}\right|\) is
1 0
2 \(\mathrm{pa}+\mathrm{qb}+\mathrm{rc}\)
3 1
4 -1
Explanation:
(A) : Here, Let, \(\quad \Delta=\left|\begin{array}{ccc}\mathrm{pa} & \mathrm{qb} & \mathrm{rc} \\ \mathrm{qc} & \mathrm{ra} & \mathrm{pb} \\ \mathrm{rb} & \mathrm{pc} & \mathrm{qa}\end{array}\right|\) \(\Delta=p a\left(a^{2} r q-p^{2} c b\right)-q b\left(q^{2} c a-b^{2} p r\right)+r c\left(q p c^{2}-r^{2} a b\right)\) \(\Delta=\) prqa \(^{3}-\mathrm{p}^{3} \mathrm{abc}-\mathrm{q}^{3} \mathrm{abc}+\mathrm{pqrb}^{3}+\mathrm{pqrc}^{3}-\mathrm{r}^{3} \mathrm{abc}\) \(=\operatorname{pqr}\left(\mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}\right)-\mathrm{abc}\left(\mathrm{p}^{3}+\mathrm{q}^{3}+\mathrm{r}^{3}\right)\) \(\because \quad \mathrm{a}+\mathrm{b}+\mathrm{c}=0=\mathrm{p}+\mathrm{q}+\mathrm{r}\) (given) \(\therefore \quad a^{3}+b^{3}+c^{3}=3 a b c\) and \(\quad \mathrm{p}^{3}+\mathrm{q}^{3}+\mathrm{r}^{3}=3 \mathrm{pqr}\) So, \(\quad \Delta=\operatorname{pqr}(3 \mathrm{abc})-\mathrm{abc}(3 \mathrm{pqr})\) \(\Delta=0\)
