79432
If \(a, b, c\) are different and \(\left|\begin{array}{llll}\mathbf{a} & \mathbf{a}^2 & \mathbf{a}^3 & -1 \\ \mathbf{b} & \mathbf{b}^2 & \mathbf{b}^3&-1 \\ \mathbf{c} & \mathbf{c}^2 & \mathbf{c}^3&-1\end{array}\right|=\mathbf{0}\) then abc is equal to
1 0
2 1
3 -1
4 none of these
Explanation:
(B) Given, \(\left|\begin{array}{lll}a & a^{2} & a^{3}-1 \\ b & b^{2} & b^{3}-1 \\ c & c^{2} & c^{3}-1\end{array}\right|=0\) \(\left|\begin{array}{lll}a & a^{2} & a^{3} \\ b & b^{2} & b^{3} \\ c & c^{2} & c^{3}\end{array}\right|-\left|\begin{array}{lll}a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1\end{array}\right|=0\) \(\operatorname{abc}\left|\begin{array}{lll}1 & \mathrm{a} & \mathrm{a}^{2} \\ 1 & \mathrm{~b} & \mathrm{~b}^{2} \\ 1 & \mathrm{c} & \mathrm{c}^{2}\end{array}\right|-1\left|\begin{array}{ccc}1 & \mathrm{a} & \mathrm{a}^{2} \\ 1 & \mathrm{~b} & \mathrm{~b}^{2} \\ 1 & \mathrm{c} & \mathrm{c}^{2}\end{array}\right|=0\) \([\therefore\) Column interchange \(]\) Now, \((a b c-1)\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=0\) Operating \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\), we get:- \((a b c-1)\left|\begin{array}{ccc} 1 & a & a^{2} \\ 0 & b-a & b^{2}-a^{2} \\ 0 & c-a & c^{2}-a^{2} \end{array}\right|=0\) \((a b c-1)(b-a)(c-a)\left|\begin{array}{ccc}1 & a & a \\ 0 & 1 & b+a \\ 0 & 1 & c-a\end{array}\right|=0\) \((a b c-1)(b-a)(c-a)[(c+a)-(b+a)]=0\) \((a b c-1)(b-a)(c-a)[(c-b)=0\) \(\mathrm{abc}=1[\therefore\) since \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are all distinct \(]\)
SRM JEE-2016
Matrix and Determinant
79433
If \(\left|\begin{array}{ccc}6 i & -3 i & 1 \\ 4 & 3 i & -1\end{array}\right|=x+i y\), where \(i=\sqrt{-1}\), then \(\begin{array}{lll}20 & 3 & i\end{array}\) what is \(x\) equal to?
79434
If the three linear equations \(x+4 a y+a z=0 ; x\) \(+3 b y+b z=0, x+2 c y+c z=0\) have a nontrivial solution, where \(\mathbf{a} \neq 0, \mathbf{b} \neq 0, \mathbf{c} \neq 0\), then \(\mathbf{a b}+\mathbf{b c}\) is equal to
1 \(2 \mathrm{ac}\)
2 \(-\mathrm{ac}\)
3 ac
4 \(-2 \mathrm{ac}\)
Explanation:
(A) : Since the given system has non-trivial solution \(\Rightarrow \quad\left|\begin{array}{lll}1 & 4 a & a \\ 1 & 3 b & b \\ 1 & 2 c & c\end{array}\right|=0(\therefore a \neq 0, b \neq 0, c \neq 0)\) On expanding along \(\mathrm{c}_{1}\), we get \((3 b c-2 b c)-(4 a c-2 a c)+4 a b-3 b a=0\) \(\Rightarrow \quad \mathrm{bc}+\mathrm{ab}=2 \mathrm{ac}\)
BITSAT-2008
Matrix and Determinant
79435
The equations \(2 x+3 y+4=0 ; 3 x+4 y+6=0\) and \(4 x+5 y+8=0\) are
1 consistent with unique solution
2 inconsistent
3 consistent with infinitely many solution
4 None of the above
Explanation:
(A) : According to given summation, consider first two equations: \(2 x+3 y=-4 \text { and } 3 x+4 y=-6\) Here, we have \(\Delta=\left|\begin{array}{ll} 2 & 3 \\ 3 & 4 \end{array}\right|=-1 \neq 0\) \(\Delta x=\left|\begin{array}{ll} -4 & 3 \\ -6 & 4 \end{array}\right|=2 \text { and } \Delta y=\left|\begin{array}{ll} 2 & -4 \\ 3 & -6 \end{array}\right|=0\) \(\therefore \mathrm{x}=-2\) and \(\mathrm{y}=0\) Now, this solution satisfies the third, so the equations are consistent with unique solution.
BITSAT-2012
Matrix and Determinant
79436
If the lines \(p_{1} x+q_{1} y=1, p_{2} x+q_{2} y=1\) and \(p_{3} x\) \(+q_{3} y=1\) be concurrent, then the points \(\left(p_{1}, q_{1}\right)\), \(\left(p_{2}, q_{2}\right)\) and \(\left(p_{3}, q_{3}\right)\)
1 are collinear
2 form an equilateral triangle
3 form a scalene triangle
4 form a right angled triangle
Explanation:
(A) : Given equations of the lines are. \(p_{1} x+q_{1} y-1=0 \tag{i}\) \(p_{2} x+q_{2} y-1=0 \tag{ii}\) \(p_{3} x+q_{3} y-1=0 \tag{iii}\) As they are concurrent, \(\begin{array}{ll} \Rightarrow & \left|\begin{array}{lll} p_{1} & q_{1} & -1 \\ p_{2} & q_{2} & -1 \\ p_{3} & q_{3} & -1 \end{array}\right|=0 \\ \Rightarrow & \left|\begin{array}{lll} p_{1} & q_{1} & 1 \\ p_{2} & q_{2} & 1 \\ p_{3} & q_{3} & 1 \end{array}\right|=0 \end{array}\) This is also the condition for the points \(\left(p_{1} \cdot q_{1}\right),\left(p_{2} \cdot q_{2}\right)\) and \(\left(p_{3} . q_{3}\right)\) to be collinear.
79432
If \(a, b, c\) are different and \(\left|\begin{array}{llll}\mathbf{a} & \mathbf{a}^2 & \mathbf{a}^3 & -1 \\ \mathbf{b} & \mathbf{b}^2 & \mathbf{b}^3&-1 \\ \mathbf{c} & \mathbf{c}^2 & \mathbf{c}^3&-1\end{array}\right|=\mathbf{0}\) then abc is equal to
1 0
2 1
3 -1
4 none of these
Explanation:
(B) Given, \(\left|\begin{array}{lll}a & a^{2} & a^{3}-1 \\ b & b^{2} & b^{3}-1 \\ c & c^{2} & c^{3}-1\end{array}\right|=0\) \(\left|\begin{array}{lll}a & a^{2} & a^{3} \\ b & b^{2} & b^{3} \\ c & c^{2} & c^{3}\end{array}\right|-\left|\begin{array}{lll}a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1\end{array}\right|=0\) \(\operatorname{abc}\left|\begin{array}{lll}1 & \mathrm{a} & \mathrm{a}^{2} \\ 1 & \mathrm{~b} & \mathrm{~b}^{2} \\ 1 & \mathrm{c} & \mathrm{c}^{2}\end{array}\right|-1\left|\begin{array}{ccc}1 & \mathrm{a} & \mathrm{a}^{2} \\ 1 & \mathrm{~b} & \mathrm{~b}^{2} \\ 1 & \mathrm{c} & \mathrm{c}^{2}\end{array}\right|=0\) \([\therefore\) Column interchange \(]\) Now, \((a b c-1)\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=0\) Operating \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\), we get:- \((a b c-1)\left|\begin{array}{ccc} 1 & a & a^{2} \\ 0 & b-a & b^{2}-a^{2} \\ 0 & c-a & c^{2}-a^{2} \end{array}\right|=0\) \((a b c-1)(b-a)(c-a)\left|\begin{array}{ccc}1 & a & a \\ 0 & 1 & b+a \\ 0 & 1 & c-a\end{array}\right|=0\) \((a b c-1)(b-a)(c-a)[(c+a)-(b+a)]=0\) \((a b c-1)(b-a)(c-a)[(c-b)=0\) \(\mathrm{abc}=1[\therefore\) since \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are all distinct \(]\)
SRM JEE-2016
Matrix and Determinant
79433
If \(\left|\begin{array}{ccc}6 i & -3 i & 1 \\ 4 & 3 i & -1\end{array}\right|=x+i y\), where \(i=\sqrt{-1}\), then \(\begin{array}{lll}20 & 3 & i\end{array}\) what is \(x\) equal to?
79434
If the three linear equations \(x+4 a y+a z=0 ; x\) \(+3 b y+b z=0, x+2 c y+c z=0\) have a nontrivial solution, where \(\mathbf{a} \neq 0, \mathbf{b} \neq 0, \mathbf{c} \neq 0\), then \(\mathbf{a b}+\mathbf{b c}\) is equal to
1 \(2 \mathrm{ac}\)
2 \(-\mathrm{ac}\)
3 ac
4 \(-2 \mathrm{ac}\)
Explanation:
(A) : Since the given system has non-trivial solution \(\Rightarrow \quad\left|\begin{array}{lll}1 & 4 a & a \\ 1 & 3 b & b \\ 1 & 2 c & c\end{array}\right|=0(\therefore a \neq 0, b \neq 0, c \neq 0)\) On expanding along \(\mathrm{c}_{1}\), we get \((3 b c-2 b c)-(4 a c-2 a c)+4 a b-3 b a=0\) \(\Rightarrow \quad \mathrm{bc}+\mathrm{ab}=2 \mathrm{ac}\)
BITSAT-2008
Matrix and Determinant
79435
The equations \(2 x+3 y+4=0 ; 3 x+4 y+6=0\) and \(4 x+5 y+8=0\) are
1 consistent with unique solution
2 inconsistent
3 consistent with infinitely many solution
4 None of the above
Explanation:
(A) : According to given summation, consider first two equations: \(2 x+3 y=-4 \text { and } 3 x+4 y=-6\) Here, we have \(\Delta=\left|\begin{array}{ll} 2 & 3 \\ 3 & 4 \end{array}\right|=-1 \neq 0\) \(\Delta x=\left|\begin{array}{ll} -4 & 3 \\ -6 & 4 \end{array}\right|=2 \text { and } \Delta y=\left|\begin{array}{ll} 2 & -4 \\ 3 & -6 \end{array}\right|=0\) \(\therefore \mathrm{x}=-2\) and \(\mathrm{y}=0\) Now, this solution satisfies the third, so the equations are consistent with unique solution.
BITSAT-2012
Matrix and Determinant
79436
If the lines \(p_{1} x+q_{1} y=1, p_{2} x+q_{2} y=1\) and \(p_{3} x\) \(+q_{3} y=1\) be concurrent, then the points \(\left(p_{1}, q_{1}\right)\), \(\left(p_{2}, q_{2}\right)\) and \(\left(p_{3}, q_{3}\right)\)
1 are collinear
2 form an equilateral triangle
3 form a scalene triangle
4 form a right angled triangle
Explanation:
(A) : Given equations of the lines are. \(p_{1} x+q_{1} y-1=0 \tag{i}\) \(p_{2} x+q_{2} y-1=0 \tag{ii}\) \(p_{3} x+q_{3} y-1=0 \tag{iii}\) As they are concurrent, \(\begin{array}{ll} \Rightarrow & \left|\begin{array}{lll} p_{1} & q_{1} & -1 \\ p_{2} & q_{2} & -1 \\ p_{3} & q_{3} & -1 \end{array}\right|=0 \\ \Rightarrow & \left|\begin{array}{lll} p_{1} & q_{1} & 1 \\ p_{2} & q_{2} & 1 \\ p_{3} & q_{3} & 1 \end{array}\right|=0 \end{array}\) This is also the condition for the points \(\left(p_{1} \cdot q_{1}\right),\left(p_{2} \cdot q_{2}\right)\) and \(\left(p_{3} . q_{3}\right)\) to be collinear.
79432
If \(a, b, c\) are different and \(\left|\begin{array}{llll}\mathbf{a} & \mathbf{a}^2 & \mathbf{a}^3 & -1 \\ \mathbf{b} & \mathbf{b}^2 & \mathbf{b}^3&-1 \\ \mathbf{c} & \mathbf{c}^2 & \mathbf{c}^3&-1\end{array}\right|=\mathbf{0}\) then abc is equal to
1 0
2 1
3 -1
4 none of these
Explanation:
(B) Given, \(\left|\begin{array}{lll}a & a^{2} & a^{3}-1 \\ b & b^{2} & b^{3}-1 \\ c & c^{2} & c^{3}-1\end{array}\right|=0\) \(\left|\begin{array}{lll}a & a^{2} & a^{3} \\ b & b^{2} & b^{3} \\ c & c^{2} & c^{3}\end{array}\right|-\left|\begin{array}{lll}a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1\end{array}\right|=0\) \(\operatorname{abc}\left|\begin{array}{lll}1 & \mathrm{a} & \mathrm{a}^{2} \\ 1 & \mathrm{~b} & \mathrm{~b}^{2} \\ 1 & \mathrm{c} & \mathrm{c}^{2}\end{array}\right|-1\left|\begin{array}{ccc}1 & \mathrm{a} & \mathrm{a}^{2} \\ 1 & \mathrm{~b} & \mathrm{~b}^{2} \\ 1 & \mathrm{c} & \mathrm{c}^{2}\end{array}\right|=0\) \([\therefore\) Column interchange \(]\) Now, \((a b c-1)\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=0\) Operating \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\), we get:- \((a b c-1)\left|\begin{array}{ccc} 1 & a & a^{2} \\ 0 & b-a & b^{2}-a^{2} \\ 0 & c-a & c^{2}-a^{2} \end{array}\right|=0\) \((a b c-1)(b-a)(c-a)\left|\begin{array}{ccc}1 & a & a \\ 0 & 1 & b+a \\ 0 & 1 & c-a\end{array}\right|=0\) \((a b c-1)(b-a)(c-a)[(c+a)-(b+a)]=0\) \((a b c-1)(b-a)(c-a)[(c-b)=0\) \(\mathrm{abc}=1[\therefore\) since \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are all distinct \(]\)
SRM JEE-2016
Matrix and Determinant
79433
If \(\left|\begin{array}{ccc}6 i & -3 i & 1 \\ 4 & 3 i & -1\end{array}\right|=x+i y\), where \(i=\sqrt{-1}\), then \(\begin{array}{lll}20 & 3 & i\end{array}\) what is \(x\) equal to?
79434
If the three linear equations \(x+4 a y+a z=0 ; x\) \(+3 b y+b z=0, x+2 c y+c z=0\) have a nontrivial solution, where \(\mathbf{a} \neq 0, \mathbf{b} \neq 0, \mathbf{c} \neq 0\), then \(\mathbf{a b}+\mathbf{b c}\) is equal to
1 \(2 \mathrm{ac}\)
2 \(-\mathrm{ac}\)
3 ac
4 \(-2 \mathrm{ac}\)
Explanation:
(A) : Since the given system has non-trivial solution \(\Rightarrow \quad\left|\begin{array}{lll}1 & 4 a & a \\ 1 & 3 b & b \\ 1 & 2 c & c\end{array}\right|=0(\therefore a \neq 0, b \neq 0, c \neq 0)\) On expanding along \(\mathrm{c}_{1}\), we get \((3 b c-2 b c)-(4 a c-2 a c)+4 a b-3 b a=0\) \(\Rightarrow \quad \mathrm{bc}+\mathrm{ab}=2 \mathrm{ac}\)
BITSAT-2008
Matrix and Determinant
79435
The equations \(2 x+3 y+4=0 ; 3 x+4 y+6=0\) and \(4 x+5 y+8=0\) are
1 consistent with unique solution
2 inconsistent
3 consistent with infinitely many solution
4 None of the above
Explanation:
(A) : According to given summation, consider first two equations: \(2 x+3 y=-4 \text { and } 3 x+4 y=-6\) Here, we have \(\Delta=\left|\begin{array}{ll} 2 & 3 \\ 3 & 4 \end{array}\right|=-1 \neq 0\) \(\Delta x=\left|\begin{array}{ll} -4 & 3 \\ -6 & 4 \end{array}\right|=2 \text { and } \Delta y=\left|\begin{array}{ll} 2 & -4 \\ 3 & -6 \end{array}\right|=0\) \(\therefore \mathrm{x}=-2\) and \(\mathrm{y}=0\) Now, this solution satisfies the third, so the equations are consistent with unique solution.
BITSAT-2012
Matrix and Determinant
79436
If the lines \(p_{1} x+q_{1} y=1, p_{2} x+q_{2} y=1\) and \(p_{3} x\) \(+q_{3} y=1\) be concurrent, then the points \(\left(p_{1}, q_{1}\right)\), \(\left(p_{2}, q_{2}\right)\) and \(\left(p_{3}, q_{3}\right)\)
1 are collinear
2 form an equilateral triangle
3 form a scalene triangle
4 form a right angled triangle
Explanation:
(A) : Given equations of the lines are. \(p_{1} x+q_{1} y-1=0 \tag{i}\) \(p_{2} x+q_{2} y-1=0 \tag{ii}\) \(p_{3} x+q_{3} y-1=0 \tag{iii}\) As they are concurrent, \(\begin{array}{ll} \Rightarrow & \left|\begin{array}{lll} p_{1} & q_{1} & -1 \\ p_{2} & q_{2} & -1 \\ p_{3} & q_{3} & -1 \end{array}\right|=0 \\ \Rightarrow & \left|\begin{array}{lll} p_{1} & q_{1} & 1 \\ p_{2} & q_{2} & 1 \\ p_{3} & q_{3} & 1 \end{array}\right|=0 \end{array}\) This is also the condition for the points \(\left(p_{1} \cdot q_{1}\right),\left(p_{2} \cdot q_{2}\right)\) and \(\left(p_{3} . q_{3}\right)\) to be collinear.
79432
If \(a, b, c\) are different and \(\left|\begin{array}{llll}\mathbf{a} & \mathbf{a}^2 & \mathbf{a}^3 & -1 \\ \mathbf{b} & \mathbf{b}^2 & \mathbf{b}^3&-1 \\ \mathbf{c} & \mathbf{c}^2 & \mathbf{c}^3&-1\end{array}\right|=\mathbf{0}\) then abc is equal to
1 0
2 1
3 -1
4 none of these
Explanation:
(B) Given, \(\left|\begin{array}{lll}a & a^{2} & a^{3}-1 \\ b & b^{2} & b^{3}-1 \\ c & c^{2} & c^{3}-1\end{array}\right|=0\) \(\left|\begin{array}{lll}a & a^{2} & a^{3} \\ b & b^{2} & b^{3} \\ c & c^{2} & c^{3}\end{array}\right|-\left|\begin{array}{lll}a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1\end{array}\right|=0\) \(\operatorname{abc}\left|\begin{array}{lll}1 & \mathrm{a} & \mathrm{a}^{2} \\ 1 & \mathrm{~b} & \mathrm{~b}^{2} \\ 1 & \mathrm{c} & \mathrm{c}^{2}\end{array}\right|-1\left|\begin{array}{ccc}1 & \mathrm{a} & \mathrm{a}^{2} \\ 1 & \mathrm{~b} & \mathrm{~b}^{2} \\ 1 & \mathrm{c} & \mathrm{c}^{2}\end{array}\right|=0\) \([\therefore\) Column interchange \(]\) Now, \((a b c-1)\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=0\) Operating \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\), we get:- \((a b c-1)\left|\begin{array}{ccc} 1 & a & a^{2} \\ 0 & b-a & b^{2}-a^{2} \\ 0 & c-a & c^{2}-a^{2} \end{array}\right|=0\) \((a b c-1)(b-a)(c-a)\left|\begin{array}{ccc}1 & a & a \\ 0 & 1 & b+a \\ 0 & 1 & c-a\end{array}\right|=0\) \((a b c-1)(b-a)(c-a)[(c+a)-(b+a)]=0\) \((a b c-1)(b-a)(c-a)[(c-b)=0\) \(\mathrm{abc}=1[\therefore\) since \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are all distinct \(]\)
SRM JEE-2016
Matrix and Determinant
79433
If \(\left|\begin{array}{ccc}6 i & -3 i & 1 \\ 4 & 3 i & -1\end{array}\right|=x+i y\), where \(i=\sqrt{-1}\), then \(\begin{array}{lll}20 & 3 & i\end{array}\) what is \(x\) equal to?
79434
If the three linear equations \(x+4 a y+a z=0 ; x\) \(+3 b y+b z=0, x+2 c y+c z=0\) have a nontrivial solution, where \(\mathbf{a} \neq 0, \mathbf{b} \neq 0, \mathbf{c} \neq 0\), then \(\mathbf{a b}+\mathbf{b c}\) is equal to
1 \(2 \mathrm{ac}\)
2 \(-\mathrm{ac}\)
3 ac
4 \(-2 \mathrm{ac}\)
Explanation:
(A) : Since the given system has non-trivial solution \(\Rightarrow \quad\left|\begin{array}{lll}1 & 4 a & a \\ 1 & 3 b & b \\ 1 & 2 c & c\end{array}\right|=0(\therefore a \neq 0, b \neq 0, c \neq 0)\) On expanding along \(\mathrm{c}_{1}\), we get \((3 b c-2 b c)-(4 a c-2 a c)+4 a b-3 b a=0\) \(\Rightarrow \quad \mathrm{bc}+\mathrm{ab}=2 \mathrm{ac}\)
BITSAT-2008
Matrix and Determinant
79435
The equations \(2 x+3 y+4=0 ; 3 x+4 y+6=0\) and \(4 x+5 y+8=0\) are
1 consistent with unique solution
2 inconsistent
3 consistent with infinitely many solution
4 None of the above
Explanation:
(A) : According to given summation, consider first two equations: \(2 x+3 y=-4 \text { and } 3 x+4 y=-6\) Here, we have \(\Delta=\left|\begin{array}{ll} 2 & 3 \\ 3 & 4 \end{array}\right|=-1 \neq 0\) \(\Delta x=\left|\begin{array}{ll} -4 & 3 \\ -6 & 4 \end{array}\right|=2 \text { and } \Delta y=\left|\begin{array}{ll} 2 & -4 \\ 3 & -6 \end{array}\right|=0\) \(\therefore \mathrm{x}=-2\) and \(\mathrm{y}=0\) Now, this solution satisfies the third, so the equations are consistent with unique solution.
BITSAT-2012
Matrix and Determinant
79436
If the lines \(p_{1} x+q_{1} y=1, p_{2} x+q_{2} y=1\) and \(p_{3} x\) \(+q_{3} y=1\) be concurrent, then the points \(\left(p_{1}, q_{1}\right)\), \(\left(p_{2}, q_{2}\right)\) and \(\left(p_{3}, q_{3}\right)\)
1 are collinear
2 form an equilateral triangle
3 form a scalene triangle
4 form a right angled triangle
Explanation:
(A) : Given equations of the lines are. \(p_{1} x+q_{1} y-1=0 \tag{i}\) \(p_{2} x+q_{2} y-1=0 \tag{ii}\) \(p_{3} x+q_{3} y-1=0 \tag{iii}\) As they are concurrent, \(\begin{array}{ll} \Rightarrow & \left|\begin{array}{lll} p_{1} & q_{1} & -1 \\ p_{2} & q_{2} & -1 \\ p_{3} & q_{3} & -1 \end{array}\right|=0 \\ \Rightarrow & \left|\begin{array}{lll} p_{1} & q_{1} & 1 \\ p_{2} & q_{2} & 1 \\ p_{3} & q_{3} & 1 \end{array}\right|=0 \end{array}\) This is also the condition for the points \(\left(p_{1} \cdot q_{1}\right),\left(p_{2} \cdot q_{2}\right)\) and \(\left(p_{3} . q_{3}\right)\) to be collinear.
79432
If \(a, b, c\) are different and \(\left|\begin{array}{llll}\mathbf{a} & \mathbf{a}^2 & \mathbf{a}^3 & -1 \\ \mathbf{b} & \mathbf{b}^2 & \mathbf{b}^3&-1 \\ \mathbf{c} & \mathbf{c}^2 & \mathbf{c}^3&-1\end{array}\right|=\mathbf{0}\) then abc is equal to
1 0
2 1
3 -1
4 none of these
Explanation:
(B) Given, \(\left|\begin{array}{lll}a & a^{2} & a^{3}-1 \\ b & b^{2} & b^{3}-1 \\ c & c^{2} & c^{3}-1\end{array}\right|=0\) \(\left|\begin{array}{lll}a & a^{2} & a^{3} \\ b & b^{2} & b^{3} \\ c & c^{2} & c^{3}\end{array}\right|-\left|\begin{array}{lll}a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1\end{array}\right|=0\) \(\operatorname{abc}\left|\begin{array}{lll}1 & \mathrm{a} & \mathrm{a}^{2} \\ 1 & \mathrm{~b} & \mathrm{~b}^{2} \\ 1 & \mathrm{c} & \mathrm{c}^{2}\end{array}\right|-1\left|\begin{array}{ccc}1 & \mathrm{a} & \mathrm{a}^{2} \\ 1 & \mathrm{~b} & \mathrm{~b}^{2} \\ 1 & \mathrm{c} & \mathrm{c}^{2}\end{array}\right|=0\) \([\therefore\) Column interchange \(]\) Now, \((a b c-1)\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=0\) Operating \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\), we get:- \((a b c-1)\left|\begin{array}{ccc} 1 & a & a^{2} \\ 0 & b-a & b^{2}-a^{2} \\ 0 & c-a & c^{2}-a^{2} \end{array}\right|=0\) \((a b c-1)(b-a)(c-a)\left|\begin{array}{ccc}1 & a & a \\ 0 & 1 & b+a \\ 0 & 1 & c-a\end{array}\right|=0\) \((a b c-1)(b-a)(c-a)[(c+a)-(b+a)]=0\) \((a b c-1)(b-a)(c-a)[(c-b)=0\) \(\mathrm{abc}=1[\therefore\) since \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are all distinct \(]\)
SRM JEE-2016
Matrix and Determinant
79433
If \(\left|\begin{array}{ccc}6 i & -3 i & 1 \\ 4 & 3 i & -1\end{array}\right|=x+i y\), where \(i=\sqrt{-1}\), then \(\begin{array}{lll}20 & 3 & i\end{array}\) what is \(x\) equal to?
79434
If the three linear equations \(x+4 a y+a z=0 ; x\) \(+3 b y+b z=0, x+2 c y+c z=0\) have a nontrivial solution, where \(\mathbf{a} \neq 0, \mathbf{b} \neq 0, \mathbf{c} \neq 0\), then \(\mathbf{a b}+\mathbf{b c}\) is equal to
1 \(2 \mathrm{ac}\)
2 \(-\mathrm{ac}\)
3 ac
4 \(-2 \mathrm{ac}\)
Explanation:
(A) : Since the given system has non-trivial solution \(\Rightarrow \quad\left|\begin{array}{lll}1 & 4 a & a \\ 1 & 3 b & b \\ 1 & 2 c & c\end{array}\right|=0(\therefore a \neq 0, b \neq 0, c \neq 0)\) On expanding along \(\mathrm{c}_{1}\), we get \((3 b c-2 b c)-(4 a c-2 a c)+4 a b-3 b a=0\) \(\Rightarrow \quad \mathrm{bc}+\mathrm{ab}=2 \mathrm{ac}\)
BITSAT-2008
Matrix and Determinant
79435
The equations \(2 x+3 y+4=0 ; 3 x+4 y+6=0\) and \(4 x+5 y+8=0\) are
1 consistent with unique solution
2 inconsistent
3 consistent with infinitely many solution
4 None of the above
Explanation:
(A) : According to given summation, consider first two equations: \(2 x+3 y=-4 \text { and } 3 x+4 y=-6\) Here, we have \(\Delta=\left|\begin{array}{ll} 2 & 3 \\ 3 & 4 \end{array}\right|=-1 \neq 0\) \(\Delta x=\left|\begin{array}{ll} -4 & 3 \\ -6 & 4 \end{array}\right|=2 \text { and } \Delta y=\left|\begin{array}{ll} 2 & -4 \\ 3 & -6 \end{array}\right|=0\) \(\therefore \mathrm{x}=-2\) and \(\mathrm{y}=0\) Now, this solution satisfies the third, so the equations are consistent with unique solution.
BITSAT-2012
Matrix and Determinant
79436
If the lines \(p_{1} x+q_{1} y=1, p_{2} x+q_{2} y=1\) and \(p_{3} x\) \(+q_{3} y=1\) be concurrent, then the points \(\left(p_{1}, q_{1}\right)\), \(\left(p_{2}, q_{2}\right)\) and \(\left(p_{3}, q_{3}\right)\)
1 are collinear
2 form an equilateral triangle
3 form a scalene triangle
4 form a right angled triangle
Explanation:
(A) : Given equations of the lines are. \(p_{1} x+q_{1} y-1=0 \tag{i}\) \(p_{2} x+q_{2} y-1=0 \tag{ii}\) \(p_{3} x+q_{3} y-1=0 \tag{iii}\) As they are concurrent, \(\begin{array}{ll} \Rightarrow & \left|\begin{array}{lll} p_{1} & q_{1} & -1 \\ p_{2} & q_{2} & -1 \\ p_{3} & q_{3} & -1 \end{array}\right|=0 \\ \Rightarrow & \left|\begin{array}{lll} p_{1} & q_{1} & 1 \\ p_{2} & q_{2} & 1 \\ p_{3} & q_{3} & 1 \end{array}\right|=0 \end{array}\) This is also the condition for the points \(\left(p_{1} \cdot q_{1}\right),\left(p_{2} \cdot q_{2}\right)\) and \(\left(p_{3} . q_{3}\right)\) to be collinear.
