Consider the simultaneous linear equations AX | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the linear equations $AX=B$ and $AY=Q$. Since $B$ is the unique solution of $AY=Q$,we have $AB=Q$. We need to find the solution for $X$ in $

Vedclass · Accepted Answer

$(A^{-1})^2 Q$

Vedclass · Answer

(D) Given the linear equations $AX=B$ and $AY=Q$. Since $B$ is the unique solution of $AY=Q$,we have $AB=Q$. We need to find the solution for $X$ in $AX=B$. Multiply both sides of $AX=B$ by $A$ on the left: $A(AX) = AB$ $A^2 X = AB$ Since $AB=Q$,we substitute $Q$ into the equation: $A^2 X = Q$ Since $A$ is an invertible matrix,$A^{-1}$ exists. Multiplying both sides by $(A^{-1})^2$: $(A^{-1})^2 (A^2 X) = (A^{-1})^2 Q$ $I X = (A^{-1})^2 Q$ $X = (A^{-1})^2 Q$ Thus,the correct option is $D$.

Consider the simultaneous linear equations $AX=B$ and $AY=Q$. If $A$ is an invertible matrix and $B$ is the unique solution of $AY=Q$,then the solution of $AX=B$ is

Similar Questions

Let $\alpha, \beta (\alpha \neq \beta)$ be the values of $m$ for which the equations $x+y+z=1$,$x+2y+4z=m$,and $x+4y+10z=m^2$ have infinitely many solutions. Then the value of $\sum_{n=1}^{10}(n^\alpha+n^\beta)$ is equal to:

The set of equations $x - y + 3z = 2$,$2x - y + z = 4$,and $x - 2y + \alpha z = 3$ has:

The existence of a unique solution for the system of equations $x+y+z=\beta$,$5x-y+\alpha z=10$,and $2x+3y-z=6$ depends on:

If the system of equations $x + 2y - 3z = 1$,$(k + 3)z = 3$,and $(2k + 1)x + z = 0$ is inconsistent,then the value of $k$ is

If the system of linear equations
$7x + 11y + \alpha z = 13$
$5x + 4y + 7z = \beta$
$175x + 194y + 57z = 361$
has infinitely many solutions,then $\alpha + \beta + 2$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module