78993
A system of linear equations in two variables, \(p\) and \(q\), is given as \((n+1) p+(n+2) q=8, p-(n\) \(+1) q+(n+2)=0, p+q=3\). Which of the following would be one of the values of \(n\) for which the given system of linear equations is consistent?
78994
In a system of linear equations three equation are given as \(5 x+4 y+2 z=13 ; 4 x-y-k z=6\); \(2 x+3 y+3 z=16\). What should be the value of \(k\) so that the equations have no solution?
1 3
2 (c.) -3
3 \(-5\)
Explanation:
(B) : Given, \(5 x+4 y+2 z=13 \tag{i}\) \(4 x-y-k z=6 \tag{ii}\) \(2 x+3 y+3 z=16 \tag{iii}\) Equation (i), (ii) and (iii) be written in matrix form \({\left[\begin{array}{ccc} 5 & 4 & 2 \\ 4 & -1 & -k \\ 2 & 3 & 3 \end{array}\right]\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\left[\begin{array}{c} 13 \\ 6 \\ 16 \end{array}\right]}\) \(\mathrm{A} \mathrm{X}=\mathrm{B}\) If the given system of equation has no solution then. \([\mathrm{A}]=0\) \(\mathrm{A}=\left[\begin{array}{ccc} 5 & 4 & 2 \\ 4 & -1 & -\mathrm{k} \\ 2 & 3 & 3 \end{array}\right]=0\) \(5(-3+3 \mathrm{k})-4(12+2 \mathrm{k})+2(12+2)=0\) \(-15+15 \mathrm{k}-48-8 \mathrm{k}+28=0\) \(7 \mathrm{k}-35=0\) \(7 \mathrm{k}=35\) \(\mathrm{k}=5\)
J&K CET-2017
Matrix and Determinant
78995
The system of equations \(\mathbf{x}+4 \mathrm{y}-3 \mathrm{z}=3\) \(\mathbf{x}-\mathbf{y}+7 \mathrm{z}=11\) \(2 x+8 y-6 z=7\) have
1 unique solution
2 infinitely many solutions
3 no solutions
4 only finite number of solutions
Explanation:
(C)Given, \(x+4 y-3 z=3\) \(x-y+7 z=11\) \(2 x+8 y-6 z=7\) \(\therefore\) Matrix form as, \(\left[\begin{array}{ccc} 1 & 4 & -3 \\ 1 & -1 & 7 \\ 2 & 8 & -6 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 3 \\ 11 \\ 7 \end{array}\right]\) \(\because \mathrm{AX}=\mathrm{B}\) Now, \(|A|=\left|\begin{array}{ccc}1 & 4 & -3 \\ 1 & -1 & 7 \\ 2 & 8 & -6\end{array}\right|\). \(=1(6-56)-4(-6-14)-3(8+2)\) \(=-50+80-30\) \(=-80+80=0\) \(\therefore\) Solution is not unique \(\operatorname{adj}(\mathrm{A})=\left[\begin{array}{ccc} -50 & 0 & 25 \\ 20 & 0 & -10 \\ 10 & 0 & 5 \end{array}\right]\) \(\therefore(\operatorname{adj} A) B=\left[\begin{array}{ccc} -50 & 0 & 25 \\ 20 & 0 & -10 \\ 10 & 0 & 5 \end{array}\right]\left[\begin{array}{l} 3 \\ 11 \\ 7 \end{array}\right] \neq 0\) Hence, the given system of equations have no solution.
J&K CET-2016
Matrix and Determinant
78996
The number of values of \(k\) for which the following system of equations has at least three solutions is \(8 x+16 y+8 z=25, x+y+z=k\) and \(3 x+y+3 z=k^{2}\)
1 0
2 1
3 2
4 3
Explanation:
(B) : Given, equation has at least three solutions is, \(8 x+16 y+8 z=25\) \(\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{k}\) and \(3 x+y+3 z=k^{2}\) \(|A|=\left|\begin{array}{ccc}8 & 16 & 8 \\ 1 & 1 & 1 \\ 3 & 1 & 3\end{array}\right|\) [Since, \(\mathrm{C}_{1}=\mathrm{C}_{3}\) ] So, \(|\mathrm{A}|=0\) Now, The linear system has infinite solutions if \((\operatorname{adj}(\mathrm{A})(\mathrm{B}))=0\) \(\operatorname{adj} A=\left[\begin{array}{ccc}2 & -40 & 8 \\ 0 & 0 & 0 \\ -2 & 40 & -8\end{array}\right]\) \((\operatorname{adj} A)(B)=\left[\begin{array}{ccc}2 & -40 & 8 \\ 0 & 0 & 0 \\ -2 & 40 & -8\end{array}\right]\left[\begin{array}{l}25 \\ k \\ k^{2}\end{array}\right]\) \(=\left[\begin{array}{c}50-40 \mathrm{k}+8 \mathrm{k}^{2} \\ 0 \\ -50+40 \mathrm{k}-8 \mathrm{k}^{2}\end{array}\right]\) \(8 \mathrm{k}^{2}-40 \mathrm{k}+50=0\) \(4 \mathrm{k}(2 \mathrm{k}-5)-10(2 \mathrm{k}-5)=0\) \((2 \mathrm{k}-5)(4 \mathrm{k}-10)=0\) \(\therefore \mathrm{k}=\frac{5}{2}\) Hence, \(\mathrm{k}\) has only one value.
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Matrix and Determinant
78993
A system of linear equations in two variables, \(p\) and \(q\), is given as \((n+1) p+(n+2) q=8, p-(n\) \(+1) q+(n+2)=0, p+q=3\). Which of the following would be one of the values of \(n\) for which the given system of linear equations is consistent?
78994
In a system of linear equations three equation are given as \(5 x+4 y+2 z=13 ; 4 x-y-k z=6\); \(2 x+3 y+3 z=16\). What should be the value of \(k\) so that the equations have no solution?
1 3
2 (c.) -3
3 \(-5\)
Explanation:
(B) : Given, \(5 x+4 y+2 z=13 \tag{i}\) \(4 x-y-k z=6 \tag{ii}\) \(2 x+3 y+3 z=16 \tag{iii}\) Equation (i), (ii) and (iii) be written in matrix form \({\left[\begin{array}{ccc} 5 & 4 & 2 \\ 4 & -1 & -k \\ 2 & 3 & 3 \end{array}\right]\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\left[\begin{array}{c} 13 \\ 6 \\ 16 \end{array}\right]}\) \(\mathrm{A} \mathrm{X}=\mathrm{B}\) If the given system of equation has no solution then. \([\mathrm{A}]=0\) \(\mathrm{A}=\left[\begin{array}{ccc} 5 & 4 & 2 \\ 4 & -1 & -\mathrm{k} \\ 2 & 3 & 3 \end{array}\right]=0\) \(5(-3+3 \mathrm{k})-4(12+2 \mathrm{k})+2(12+2)=0\) \(-15+15 \mathrm{k}-48-8 \mathrm{k}+28=0\) \(7 \mathrm{k}-35=0\) \(7 \mathrm{k}=35\) \(\mathrm{k}=5\)
J&K CET-2017
Matrix and Determinant
78995
The system of equations \(\mathbf{x}+4 \mathrm{y}-3 \mathrm{z}=3\) \(\mathbf{x}-\mathbf{y}+7 \mathrm{z}=11\) \(2 x+8 y-6 z=7\) have
1 unique solution
2 infinitely many solutions
3 no solutions
4 only finite number of solutions
Explanation:
(C)Given, \(x+4 y-3 z=3\) \(x-y+7 z=11\) \(2 x+8 y-6 z=7\) \(\therefore\) Matrix form as, \(\left[\begin{array}{ccc} 1 & 4 & -3 \\ 1 & -1 & 7 \\ 2 & 8 & -6 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 3 \\ 11 \\ 7 \end{array}\right]\) \(\because \mathrm{AX}=\mathrm{B}\) Now, \(|A|=\left|\begin{array}{ccc}1 & 4 & -3 \\ 1 & -1 & 7 \\ 2 & 8 & -6\end{array}\right|\). \(=1(6-56)-4(-6-14)-3(8+2)\) \(=-50+80-30\) \(=-80+80=0\) \(\therefore\) Solution is not unique \(\operatorname{adj}(\mathrm{A})=\left[\begin{array}{ccc} -50 & 0 & 25 \\ 20 & 0 & -10 \\ 10 & 0 & 5 \end{array}\right]\) \(\therefore(\operatorname{adj} A) B=\left[\begin{array}{ccc} -50 & 0 & 25 \\ 20 & 0 & -10 \\ 10 & 0 & 5 \end{array}\right]\left[\begin{array}{l} 3 \\ 11 \\ 7 \end{array}\right] \neq 0\) Hence, the given system of equations have no solution.
J&K CET-2016
Matrix and Determinant
78996
The number of values of \(k\) for which the following system of equations has at least three solutions is \(8 x+16 y+8 z=25, x+y+z=k\) and \(3 x+y+3 z=k^{2}\)
1 0
2 1
3 2
4 3
Explanation:
(B) : Given, equation has at least three solutions is, \(8 x+16 y+8 z=25\) \(\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{k}\) and \(3 x+y+3 z=k^{2}\) \(|A|=\left|\begin{array}{ccc}8 & 16 & 8 \\ 1 & 1 & 1 \\ 3 & 1 & 3\end{array}\right|\) [Since, \(\mathrm{C}_{1}=\mathrm{C}_{3}\) ] So, \(|\mathrm{A}|=0\) Now, The linear system has infinite solutions if \((\operatorname{adj}(\mathrm{A})(\mathrm{B}))=0\) \(\operatorname{adj} A=\left[\begin{array}{ccc}2 & -40 & 8 \\ 0 & 0 & 0 \\ -2 & 40 & -8\end{array}\right]\) \((\operatorname{adj} A)(B)=\left[\begin{array}{ccc}2 & -40 & 8 \\ 0 & 0 & 0 \\ -2 & 40 & -8\end{array}\right]\left[\begin{array}{l}25 \\ k \\ k^{2}\end{array}\right]\) \(=\left[\begin{array}{c}50-40 \mathrm{k}+8 \mathrm{k}^{2} \\ 0 \\ -50+40 \mathrm{k}-8 \mathrm{k}^{2}\end{array}\right]\) \(8 \mathrm{k}^{2}-40 \mathrm{k}+50=0\) \(4 \mathrm{k}(2 \mathrm{k}-5)-10(2 \mathrm{k}-5)=0\) \((2 \mathrm{k}-5)(4 \mathrm{k}-10)=0\) \(\therefore \mathrm{k}=\frac{5}{2}\) Hence, \(\mathrm{k}\) has only one value.
78993
A system of linear equations in two variables, \(p\) and \(q\), is given as \((n+1) p+(n+2) q=8, p-(n\) \(+1) q+(n+2)=0, p+q=3\). Which of the following would be one of the values of \(n\) for which the given system of linear equations is consistent?
78994
In a system of linear equations three equation are given as \(5 x+4 y+2 z=13 ; 4 x-y-k z=6\); \(2 x+3 y+3 z=16\). What should be the value of \(k\) so that the equations have no solution?
1 3
2 (c.) -3
3 \(-5\)
Explanation:
(B) : Given, \(5 x+4 y+2 z=13 \tag{i}\) \(4 x-y-k z=6 \tag{ii}\) \(2 x+3 y+3 z=16 \tag{iii}\) Equation (i), (ii) and (iii) be written in matrix form \({\left[\begin{array}{ccc} 5 & 4 & 2 \\ 4 & -1 & -k \\ 2 & 3 & 3 \end{array}\right]\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\left[\begin{array}{c} 13 \\ 6 \\ 16 \end{array}\right]}\) \(\mathrm{A} \mathrm{X}=\mathrm{B}\) If the given system of equation has no solution then. \([\mathrm{A}]=0\) \(\mathrm{A}=\left[\begin{array}{ccc} 5 & 4 & 2 \\ 4 & -1 & -\mathrm{k} \\ 2 & 3 & 3 \end{array}\right]=0\) \(5(-3+3 \mathrm{k})-4(12+2 \mathrm{k})+2(12+2)=0\) \(-15+15 \mathrm{k}-48-8 \mathrm{k}+28=0\) \(7 \mathrm{k}-35=0\) \(7 \mathrm{k}=35\) \(\mathrm{k}=5\)
J&K CET-2017
Matrix and Determinant
78995
The system of equations \(\mathbf{x}+4 \mathrm{y}-3 \mathrm{z}=3\) \(\mathbf{x}-\mathbf{y}+7 \mathrm{z}=11\) \(2 x+8 y-6 z=7\) have
1 unique solution
2 infinitely many solutions
3 no solutions
4 only finite number of solutions
Explanation:
(C)Given, \(x+4 y-3 z=3\) \(x-y+7 z=11\) \(2 x+8 y-6 z=7\) \(\therefore\) Matrix form as, \(\left[\begin{array}{ccc} 1 & 4 & -3 \\ 1 & -1 & 7 \\ 2 & 8 & -6 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 3 \\ 11 \\ 7 \end{array}\right]\) \(\because \mathrm{AX}=\mathrm{B}\) Now, \(|A|=\left|\begin{array}{ccc}1 & 4 & -3 \\ 1 & -1 & 7 \\ 2 & 8 & -6\end{array}\right|\). \(=1(6-56)-4(-6-14)-3(8+2)\) \(=-50+80-30\) \(=-80+80=0\) \(\therefore\) Solution is not unique \(\operatorname{adj}(\mathrm{A})=\left[\begin{array}{ccc} -50 & 0 & 25 \\ 20 & 0 & -10 \\ 10 & 0 & 5 \end{array}\right]\) \(\therefore(\operatorname{adj} A) B=\left[\begin{array}{ccc} -50 & 0 & 25 \\ 20 & 0 & -10 \\ 10 & 0 & 5 \end{array}\right]\left[\begin{array}{l} 3 \\ 11 \\ 7 \end{array}\right] \neq 0\) Hence, the given system of equations have no solution.
J&K CET-2016
Matrix and Determinant
78996
The number of values of \(k\) for which the following system of equations has at least three solutions is \(8 x+16 y+8 z=25, x+y+z=k\) and \(3 x+y+3 z=k^{2}\)
1 0
2 1
3 2
4 3
Explanation:
(B) : Given, equation has at least three solutions is, \(8 x+16 y+8 z=25\) \(\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{k}\) and \(3 x+y+3 z=k^{2}\) \(|A|=\left|\begin{array}{ccc}8 & 16 & 8 \\ 1 & 1 & 1 \\ 3 & 1 & 3\end{array}\right|\) [Since, \(\mathrm{C}_{1}=\mathrm{C}_{3}\) ] So, \(|\mathrm{A}|=0\) Now, The linear system has infinite solutions if \((\operatorname{adj}(\mathrm{A})(\mathrm{B}))=0\) \(\operatorname{adj} A=\left[\begin{array}{ccc}2 & -40 & 8 \\ 0 & 0 & 0 \\ -2 & 40 & -8\end{array}\right]\) \((\operatorname{adj} A)(B)=\left[\begin{array}{ccc}2 & -40 & 8 \\ 0 & 0 & 0 \\ -2 & 40 & -8\end{array}\right]\left[\begin{array}{l}25 \\ k \\ k^{2}\end{array}\right]\) \(=\left[\begin{array}{c}50-40 \mathrm{k}+8 \mathrm{k}^{2} \\ 0 \\ -50+40 \mathrm{k}-8 \mathrm{k}^{2}\end{array}\right]\) \(8 \mathrm{k}^{2}-40 \mathrm{k}+50=0\) \(4 \mathrm{k}(2 \mathrm{k}-5)-10(2 \mathrm{k}-5)=0\) \((2 \mathrm{k}-5)(4 \mathrm{k}-10)=0\) \(\therefore \mathrm{k}=\frac{5}{2}\) Hence, \(\mathrm{k}\) has only one value.
78993
A system of linear equations in two variables, \(p\) and \(q\), is given as \((n+1) p+(n+2) q=8, p-(n\) \(+1) q+(n+2)=0, p+q=3\). Which of the following would be one of the values of \(n\) for which the given system of linear equations is consistent?
78994
In a system of linear equations three equation are given as \(5 x+4 y+2 z=13 ; 4 x-y-k z=6\); \(2 x+3 y+3 z=16\). What should be the value of \(k\) so that the equations have no solution?
1 3
2 (c.) -3
3 \(-5\)
Explanation:
(B) : Given, \(5 x+4 y+2 z=13 \tag{i}\) \(4 x-y-k z=6 \tag{ii}\) \(2 x+3 y+3 z=16 \tag{iii}\) Equation (i), (ii) and (iii) be written in matrix form \({\left[\begin{array}{ccc} 5 & 4 & 2 \\ 4 & -1 & -k \\ 2 & 3 & 3 \end{array}\right]\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\left[\begin{array}{c} 13 \\ 6 \\ 16 \end{array}\right]}\) \(\mathrm{A} \mathrm{X}=\mathrm{B}\) If the given system of equation has no solution then. \([\mathrm{A}]=0\) \(\mathrm{A}=\left[\begin{array}{ccc} 5 & 4 & 2 \\ 4 & -1 & -\mathrm{k} \\ 2 & 3 & 3 \end{array}\right]=0\) \(5(-3+3 \mathrm{k})-4(12+2 \mathrm{k})+2(12+2)=0\) \(-15+15 \mathrm{k}-48-8 \mathrm{k}+28=0\) \(7 \mathrm{k}-35=0\) \(7 \mathrm{k}=35\) \(\mathrm{k}=5\)
J&K CET-2017
Matrix and Determinant
78995
The system of equations \(\mathbf{x}+4 \mathrm{y}-3 \mathrm{z}=3\) \(\mathbf{x}-\mathbf{y}+7 \mathrm{z}=11\) \(2 x+8 y-6 z=7\) have
1 unique solution
2 infinitely many solutions
3 no solutions
4 only finite number of solutions
Explanation:
(C)Given, \(x+4 y-3 z=3\) \(x-y+7 z=11\) \(2 x+8 y-6 z=7\) \(\therefore\) Matrix form as, \(\left[\begin{array}{ccc} 1 & 4 & -3 \\ 1 & -1 & 7 \\ 2 & 8 & -6 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 3 \\ 11 \\ 7 \end{array}\right]\) \(\because \mathrm{AX}=\mathrm{B}\) Now, \(|A|=\left|\begin{array}{ccc}1 & 4 & -3 \\ 1 & -1 & 7 \\ 2 & 8 & -6\end{array}\right|\). \(=1(6-56)-4(-6-14)-3(8+2)\) \(=-50+80-30\) \(=-80+80=0\) \(\therefore\) Solution is not unique \(\operatorname{adj}(\mathrm{A})=\left[\begin{array}{ccc} -50 & 0 & 25 \\ 20 & 0 & -10 \\ 10 & 0 & 5 \end{array}\right]\) \(\therefore(\operatorname{adj} A) B=\left[\begin{array}{ccc} -50 & 0 & 25 \\ 20 & 0 & -10 \\ 10 & 0 & 5 \end{array}\right]\left[\begin{array}{l} 3 \\ 11 \\ 7 \end{array}\right] \neq 0\) Hence, the given system of equations have no solution.
J&K CET-2016
Matrix and Determinant
78996
The number of values of \(k\) for which the following system of equations has at least three solutions is \(8 x+16 y+8 z=25, x+y+z=k\) and \(3 x+y+3 z=k^{2}\)
1 0
2 1
3 2
4 3
Explanation:
(B) : Given, equation has at least three solutions is, \(8 x+16 y+8 z=25\) \(\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{k}\) and \(3 x+y+3 z=k^{2}\) \(|A|=\left|\begin{array}{ccc}8 & 16 & 8 \\ 1 & 1 & 1 \\ 3 & 1 & 3\end{array}\right|\) [Since, \(\mathrm{C}_{1}=\mathrm{C}_{3}\) ] So, \(|\mathrm{A}|=0\) Now, The linear system has infinite solutions if \((\operatorname{adj}(\mathrm{A})(\mathrm{B}))=0\) \(\operatorname{adj} A=\left[\begin{array}{ccc}2 & -40 & 8 \\ 0 & 0 & 0 \\ -2 & 40 & -8\end{array}\right]\) \((\operatorname{adj} A)(B)=\left[\begin{array}{ccc}2 & -40 & 8 \\ 0 & 0 & 0 \\ -2 & 40 & -8\end{array}\right]\left[\begin{array}{l}25 \\ k \\ k^{2}\end{array}\right]\) \(=\left[\begin{array}{c}50-40 \mathrm{k}+8 \mathrm{k}^{2} \\ 0 \\ -50+40 \mathrm{k}-8 \mathrm{k}^{2}\end{array}\right]\) \(8 \mathrm{k}^{2}-40 \mathrm{k}+50=0\) \(4 \mathrm{k}(2 \mathrm{k}-5)-10(2 \mathrm{k}-5)=0\) \((2 \mathrm{k}-5)(4 \mathrm{k}-10)=0\) \(\therefore \mathrm{k}=\frac{5}{2}\) Hence, \(\mathrm{k}\) has only one value.