A, P, B are 3 times 3 matrices. If |-B|=5, |B | vedclass

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(A) Given that $A, P, B$ are $3 	imes 3$ matrices.$1$. For $|-B|=5$:Since $B$ is a $3 	imes 3$ matrix,$|-B| = (-1)^3 |B| = -|B|$.Thus,$-|B| = 5 \Rig

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$3$

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(A) Given that $A, P, B$ are $3 	imes 3$ matrices.$1$. For $|-B|=5$:Since $B$ is a $3 	imes 3$ matrix,$|-B| = (-1)^3 |B| = -|B|$.Thus,$-|B| = 5 \Rightarrow |B| = -5$.$2$. For $|BA^T|=15$:Using the property $|XY| = |X||Y|$ and $|A^T| = |A|$,we have:$|B||A| = 15$$(-5)|A| = 15 \Rightarrow |A| = -3$.$3$. For $|P^T AP| = -27$:Using the properties $|P^T| = |P|$ and $|XY| = |X||Y|$,we have:$|P^T||A||P| = -27$$|P||A||P| = -27$$|P|^2 (-3) = -27$$|P|^2 = 9$$|P| = \pm 3$.Therefore,one of the values of $|P|$ is $3$.

$A, P, B$ are $3 \times 3$ matrices. If $|-B|=5, |BA^T|=15, |P^T AP|=-27$,then one of the values of $|P|$ is

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