a) In an examination paper on Business Mathematics, 18 questions are set. In how many different ways can you choose 15 questions to answer? b) If A = pmatrix 1 & 2 \\ 3 & 1 pmatrix show that A^2 2A…

a) Number of ways to choose 15 questions out of 18

This is a combination problem, where the order of selection does not matter. The formula for combinations is:

$$C(n,r)=\frac{n!}{r!(n-r)!}$$

where:

• $n=18$ (total number of questions),
• $r=15$ (number of questions to choose).

Step 1: Write the combination formula

$$C(18,15)=\frac{18!}{15!(18-15)!}=\frac{18!}{15!\cdot 3!}$$

Step 2: Simplify the factorials

Notice that $18!=18\times 17\times 16\times 15!$. Thus, the $15!$ terms cancel out:

$$C(18,15)=\frac{18\times 17\times 16\times 15!}{15!\cdot 3!}=\frac{18\times 17\times 16}{3!}$$

Step 3: Compute $3!$

$$3!=3\times 2\times 1=6$$

Step 4: Compute the numerator

$$18\times 17=306$$ $$306\times 16=4896$$

Step 5: Divide by $3!$

$$C(18,15)=\frac{4896}{6}=816$$

Final Answer: There are 816 different ways to choose 15 questions out of 18.

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b) Show that $A^2-2A-5I=0$, where $A=\left(\begin{array}{cc}1& 2\\ 3& 1\end{array}\right)$

Step 1: Compute $A^2$

First, multiply matrix $A$ by itself:

$$A^2=A\times A=\left(\begin{array}{cc}1& 2\\ 3& 1\end{array}\right)\times \left(\begin{array}{cc}1& 2\\ 3& 1\end{array}\right)$$

Using matrix multiplication rules:

$$A_{11}^2=(1\times 1)+(2\times 3)=1+6=7$$ $$A_{12}^2=(1\times 2)+(2\times 1)=2+2=4$$ $$A_{21}^2=(3\times 1)+(1\times 3)=3+3=6$$ $$A_{22}^2=(3\times 2)+(1\times 1)=6+1=7$$

Thus,

$$A^2=\left(\begin{array}{cc}7& 4\\ 6& 7\end{array}\right)$$

Step 2: Compute $2A$

Multiply matrix $A$ by 2:

$$2A=2\times \left(\begin{array}{cc}1& 2\\ 3& 1\end{array}\right)=\left(\begin{array}{cc}2& 4\\ 6& 2\end{array}\right)$$

Step 3: Compute $5I$

The identity matrix $I$ is:

$$I=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$$

Multiply by 5:

$$5I=5\times \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}5& 0\\ 0& 5\end{array}\right)$$

Step 4: Compute $A^2-2A-5I$

Subtract $2A$ and $5I$ from $A^2$:

$$A^2-2A-5I=\left(\begin{array}{cc}7& 4\\ 6& 7\end{array}\right)-\left(\begin{array}{cc}2& 4\\ 6& 2\end{array}\right)-\left(\begin{array}{cc}5& 0\\ 0& 5\end{array}\right)$$

Perform the subtraction element-wise:

$$=\left(\begin{array}{cc}7-2-5& 4-4-0\\ 6-6-0& 7-2-5\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$$

Since the result is the null matrix (0), we have shown that:

$$A^2-2A-5I=0$$

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Verification (Optional but Good Practice)

To ensure correctness, let's verify by solving the matrix equation $A^2-2A-5I=0$ for $A$:

$$A^2-2A=5I$$

This resembles a quadratic equation in matrix form. The characteristic equation of $A$ can be found by solving $\det (A-\lambda I)=0$:

$$A-\lambda I=\left(\begin{array}{cc}1-\lambda & 2\\ 3& 1-\lambda \end{array}\right)$$

$$\det (A-\lambda I)=(1-\lambda {)}^2-6={\lambda }^2-2\lambda -5=0$$

This matches the given equation $A^2-2A-5I=0$, confirming our result.