(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

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(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 $D$.

Class 12 Mathematics 3-4. Matrices and Determinants Solution of the Linear equations using Matrices

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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

TS EAMCET 2022 All TS EAMCET

A
$A^{-1}(B+Q)$
B
$(A^{-1})^2 B$
C
$A^{-1} BQ$
D
$(A^{-1})^2 Q$

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3-4. Matrices and Determinants

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Solution of the Linear equations using Matrices

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EasyTS EAMCET 2021

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Let $A$ be a matrix such that $AB$ is a scalar matrix where $B = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$ and $\det(3A) = 27$. Then $3A^{-1} + A^2 =$

MediumTS EAMCET 2023

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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

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