Class 12Mathematics3 and 4 .Determinants and MatricesSolution of the Linear equations using Matrices
EasyIf a system of three linear equations in three unknowns,which is in the matrix equation form of $AX = D$,is inconsistent,then $\frac{\text{rank of } A}{\text{rank of } AD}$ is
- Aless than one
- Bgreater than or equal to one
- Cone
- Dgreater than one
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DifficultJEE MAIN 2019
View SolutionLet $\alpha, \beta$ and $\gamma$ be real numbers. Consider the following system of linear equations:
$x+2y+z=7$
$x+\alpha z=11$
$2x-3y+\beta z=\gamma$
Match each entry in List-$I$ to the correct entries in List-$II$:
List-$I$ List-$II$ $(P)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma=28$,then the system has $(1)$ a unique solution $(Q)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma \neq 28$,then the system has $(2)$ no solution $(R)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$,then the system has $(3)$ infinitely many solutions $(S)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma=28$,then the system has $(4)$ $x=11, y=-2$ and $z=0$ as a solution $(5)$ $x=-15, y=4$ and $z=0$ as a solution
$x+2y+z=7$
$x+\alpha z=11$
$2x-3y+\beta z=\gamma$
Match each entry in List-$I$ to the correct entries in List-$II$:
| List-$I$ | List-$II$ |
| $(P)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma=28$,then the system has | $(1)$ a unique solution |
| $(Q)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma \neq 28$,then the system has | $(2)$ no solution |
| $(R)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$,then the system has | $(3)$ infinitely many solutions |
| $(S)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma=28$,then the system has | $(4)$ $x=11, y=-2$ and $z=0$ as a solution |
| $(5)$ $x=-15, y=4$ and $z=0$ as a solution |
MediumIIT 2023
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