If every element of a square non-singular mat | vedclass

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(C) Given that $B = kA$. Since $A$ is a square matrix of order $n$,the determinant of $B$ is given by $|B| = |kA| = k^n|A|$. We know that for any non-

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$|A^{-1}| = k^n|B^{-1}|$

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(C) Given that $B = kA$. Since $A$ is a square matrix of order $n$,the determinant of $B$ is given by $|B| = |kA| = k^n|A|$. We know that for any non-singular matrix $M$,$|M^{-1}| = \frac{1}{|M|}$. Therefore,$|B^{-1}| = \frac{1}{|B|} = \frac{1}{k^n|A|}$. Since $|A^{-1}| = \frac{1}{|A|}$,we can substitute this into the expression for $|B^{-1}|$: $|B^{-1}| = \frac{1}{k^n} \cdot \frac{1}{|A|} = k^{-n} |A^{-1}|$. Rearranging this,we get $|A^{-1}| = k^n |B^{-1}|$. Thus,the correct option is $C$.

If every element of a square non-singular matrix $A$ of order $n$ is multiplied by $k$ and the new matrix is denoted by $B$,then how are $|A^{-1}|$ and $|B^{-1}|$ related?

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