Class 12Mathematics3 and 4 .Determinants and MatricesSolution of the Linear equations using Matrices
EasyLet $AX=D$ be a system of three linear non-homogeneous equations. If $|A|=0$ and $\operatorname{rank}(A)=\operatorname{rank}([AD])=\alpha$,then
- A$AX=D$ will have infinite number of solutions when $\alpha=3$
- B$AX=D$ will have unique solution when $\alpha < 3$
- C$AX=D$ will have infinite number of solutions when $\alpha < 3$
- D$AX=D$ will have no solution when $\alpha < 3$
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Similar Questions
The number of values of $\alpha$ for which the system of equations: $x+y+z=\alpha$,$\alpha x+2 \alpha y+3 z=-1$,and $x+3 \alpha y+5 z=4$ is inconsistent,is:
Examine the consistency of the system of equations: $2x - y = 5$ and $x + y = 4$.
Medium
View SolutionThe system of linear equations $x + 2y + z = -3$,$3x + 3y - 2z = -1$,and $2x + 7y + 7z = -4$ has:
If ${x^a}{y^b} = {e^m}$,${x^c}{y^d} = {e^n}$,${\Delta _1} = \left| {\begin{array}{*{20}{c}} m & b \\ n & d \end{array}} \right|$,${\Delta _2} = \left| {\begin{array}{*{20}{c}} a & m \\ c & n \end{array}} \right|$,and ${\Delta _3} = \left| {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right|$,then the values of $x$ and $y$ are respectively:
Difficult
View SolutionThe solution of the equation $\begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$ is $(x, y, z) = $
Easy
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