33) If $$ A $$ and $$ B $$ are two matrices of order $$ 2 imes 2 $$ and $$ |A B|=9,\left|A B^{-1} ight|=4 $$ then $$ |A|+|B| $$ could be equal to $$ \begin{array}{llll} ext { (a) } 5 & ext { (b) } 6 & ext { (c) } \frac{13}{2} & ext { (d) } \frac{15}{2}\end{array} $$

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33) If $$ A $$ and $$ B $$ are two matrices of order $$ 2 	imes 2 $$ and $$ |A B|=9,\left|A B^{-1}ight|=4 $$ then $$ |A|+|B| $$ could be equal to $$ \begin{array}{llll}	ext { (a) } 5 & 	ext { (b) } 6 & 	ext { (c) } \frac{13}{2} & 	ext { (d) } \frac{15}{2}\end{array} $$

UpStudy AI · Accepted Answer

We know the following properties of determinants for matrices $$ A $$ and $$ B $$:
1. $$ |AB| = |A| \cdot |B| $$
2. $$ |A B^{-1}| = |A| \cdot |B^{-1}| = |A| \cdot \frac{1}{|B|} $$

Given:
- $$ |AB| = 9 $$
- $$ |A B^{-1}| = 4 $$

From the first property, we can express $$ |A| \cdot |B| = 9 $$.

From the second property, we can express $$ |A| \cdot \frac{1}{|B|} = 4 $$, which simplifies to $$ |A| = 4 |B| $$.

Now we have two equations:
1. $$ |A| \cdot |B| = 9 $$
2. $$ |A| = 4 |B| $$

We can substitute the second equation into the first to find $$ |B| $$. Let's denote $$ |B| = b $$. Then $$ |A| = 4b $$. Substituting into the first equation gives us:
$$
(4b) \cdot b = 9
$$
This simplifies to:
$$
4b^2 = 9
$$
Now we can solve for $$ b $$.
Solve the quadratic equation by following steps:
- step0: Solve using square roots:
  $$4b^{2}-9=0$$
- step1: Add or subtract both sides:
  $$4b^{2}=9$$
- step2: Divide both sides:
  $$\frac{4b^{2}}{4}=\frac{9}{4}$$
- step3: Divide the numbers:
  $$b^{2}=\frac{9}{4}$$
- step4: Simplify the expression:
  $$b=\pm \sqrt{\frac{9}{4}}$$
- step5: Simplify the expression:
  $$b=\pm \frac{3}{2}$$
- step6: Separate into possible cases:
  $$\begin{align}&b=\frac{3}{2}\&b=-\frac{3}{2}\end{align}$$
- step7: Rewrite:
  $$b_{1}=-\frac{3}{2},b_{2}=\frac{3}{2}$$
From the equation $$ 4b^2 - 9 = 0 $$, we find the solutions for $$ b $$:
$$
b = \frac{3}{2} \quad \text{or} \quad b = -\frac{3}{2}
$$
Since $$ |B| $$ must be non-negative, we take $$ |B| = \frac{3}{2} $$.
Using $$ |A| = 4|B| $$:
$$
|A| = 4 \cdot \frac{3}{2} = 6
$$
Now we can find $$ |A| + |B| $$:
$$
|A| + |B| = 6 + \frac{3}{2} = \frac{12}{2} + \frac{3}{2} = \frac{15}{2}
$$
The value of $$ |A| + |B| $$ could be equal to $$ \frac{15}{2} $$, which corresponds to option (d).

If and are two matrices of order and
then could be equal to

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If and are two matrices of order and then could be equal to