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
Let $A = \begin{bmatrix} i & -i \\ -i & i \end{bmatrix}$,where $i = \sqrt{-1}$. Then,the system of linear equations $A^{8} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 64 \end{bmatrix}$ has :
DifficultJEE MAIN 2021
View SolutionIf $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to:
If the system of linear equations
$2x + y - z = 3$
$x - y - z = \alpha$
$3x + 3y + \beta z = 3$
has infinitely many solutions,then $\alpha + \beta - \alpha \beta$ is equal to .... .
$2x + y - z = 3$
$x - y - z = \alpha$
$3x + 3y + \beta z = 3$
has infinitely many solutions,then $\alpha + \beta - \alpha \beta$ is equal to .... .
The number of solutions of the equations $x + y - z = 0$,$3x - y - z = 0$,and $x - 3y + z = 0$ is
Medium
View SolutionIf $\begin{bmatrix} \alpha & \beta & \gamma \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & -5 \\ 1 & 2 & 5 \end{bmatrix} = \begin{bmatrix} 3 & 5 & 2 \end{bmatrix}$, then $\alpha^3 + \beta^3 + \gamma^3 = $
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