a) How many license plates consisting of 3 different numbers of digits can be made out of given digits 1,2,3,4,5,6 ? [3] b) If A = pmatrix 3 & 2 \\ 5 & 3 pmatrix and B = pmatrix 3 & 2 \\ 5 & 3…

a) Number of License Plates with 3 Different Digits

Given: Digits available: 1, 2, 3, 4, 5, 6 (total of 6 digits). We need to form license plates consisting of 3 different digits.

Solution: This is a permutation problem because the order of digits matters (e.g., 123 is different from 321), and no digit can be repeated in a single license plate.

1. First digit: We have 6 choices (1, 2, 3, 4, 5, 6).

2. Second digit: Since digits must be different, we now have 5 remaining choices.

3. Third digit: After choosing the first two digits, we have 4 remaining choices.

By the Fundamental Counting Principle, the total number of possible license plates is: $$6\times 5\times 4=120$$

Answer: 120 license plates can be made.

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b) Finding $|AB|$ for Given Matrices

Given: $$A=\left(\begin{array}{cc}-3& -2\\ 5& 3\end{array}\right),B=\left(\begin{array}{cc}3& 2\\ -5& 3\end{array}\right)$$

Objective: Find the determinant of the product $AB$, i.e., $|AB|$.

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Step 1: Compute the Product $AB$

The product of two $2\times 2$ matrices $A$ and $B$ is given by: $$AB=\left(\begin{array}{cc}{a}_{11}{b}_{11}+a_{12}{b}_{21}& a_{11}{b}_{12}+a_{12}{b}_{22}\\ a_{21}{b}_{11}+a_{22}{b}_{21}& a_{21}{b}_{12}+a_{22}{b}_{22}\end{array}\right)$$

Substituting the values: $$AB=\left(\begin{array}{cc}(-3)(3)+(-2)(-5)& (-3)(2)+(-2)(3)\\ (5)(3)+(3)(-5)& (5)(2)+(3)(3)\end{array}\right)$$

Calculate each element:

1. $(AB{)}_{11}=(-3)(3)+(-2)(-5)=-9+10=1$
2. $(AB{)}_{12}=(-3)(2)+(-2)(3)=-6-6=-12$
3. $(AB{)}_{21}=(5)(3)+(3)(-5)=15-15=0$
4. $(AB{)}_{22}=(5)(2)+(3)(3)=10+9=19$

Thus, $$AB=\left(\begin{array}{cc}1& -12\\ 0& 19\end{array}\right)$$

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Step 2: Compute $|AB|$

The determinant of a $2\times 2$ matrix $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ is: $$|AB|=ad-bc$$

For $AB$: $$|AB|=(1)(19)-(-12)(0)=19-0=19$$

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Alternative Approach: Using Determinant Properties

We know that for any two square matrices $A$ and $B$ of the same size: $$|AB|=|A|\cdot |B|$$

Let's verify this:

1. Compute $|A|$: $$|A|=(-3)(3)-(-2)(5)=-9+10=1$$

2. Compute $|B|$: $$|B|=(3)(3)-(2)(-5)=9+10=19$$

3. Multiply $|A|$ and $|B|$: $$|AB|=|A|\cdot |B|=1\times 19=19$$

This matches our previous result.

Answer: 19