1. If the sum of all the elements of a $$ 3 imes 3 $$ scalar matrix is 9 , then the product of all its elements i $$ \begin{array}{llll} ext { (a) } 0 & ext { (b) } 9 & ext { (c) } 27\end{array} $$ 2. Let $$ f: R, ightarrow[-5, a c) $$ be defined $$ a s f(x)=9 x^{2}+6 x-5 $$, where $$ R $$, is the set of all non-negative re.

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1. If the sum of all the elements of a $$ 3 	imes 3 $$ scalar matrix is 9 , then the product of all its elements i $$ \begin{array}{llll}	ext { (a) } 0 & 	ext { (b) } 9 & 	ext { (c) } 27\end{array} $$ 2. Let $$ f: R, ightarrow[-5, a c) $$ be defined $$ a s f(x)=9 x^{2}+6 x-5 $$, where $$ R $$, is the set of all non-negative re.

UpStudy AI · Accepted Answer

Let's address each problem one by one.

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### **Problem 1**

**Statement:**
If the sum of all the elements of a $$ 3 \times 3 $$ scalar matrix is 9, then the product of all its elements is:
$$
\begin{array}{llll}
\text{(a) } 0 & \text{(b) } 9 & \text{(c) } 27
\end{array}
$$

**Solution:**

A **scalar matrix** is a diagonal matrix where all the diagonal elements are equal, and the off-diagonal elements are zero. A general form of a $$ 3 \times 3 $$ scalar matrix is:

$$
\begin{bmatrix}
k & 0 & 0 \
0 & k & 0 \
0 & 0 & k \
\end{bmatrix}
$$

**Sum of All Elements:**
$$
\text{Sum} = k + 0 + 0 + 0 + k + 0 + 0 + 0 + k = 3k
$$
Given that the sum is 9:
$$
3k = 9 \implies k = 3
$$

**Product of All Elements:**
$$
\text{Product} = k \times 0 \times 0 \times 0 \times k \times 0 \times 0 \times 0 \times k = 0
$$
Since the off-diagonal elements are zero, the entire product becomes zero regardless of the value of $$ k $$.

**Answer:** **(a) 0**

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### **Problem 2**

**Statement:**
Let $$ f: \mathbb{R} \rightarrow [-5, a) $$ be defined as $$ f(x) = 9x^{2} + 6x - 5 $$, where $$ \mathbb{R} $$ is the set of all non-negative real numbers.

**Solution:**

First, let's determine the range of the function $$ f(x) = 9x^{2} + 6x - 5 $$ for $$ x \geq 0 $$.

1. **Finding the Minimum Value:**
   - The function is a quadratic equation that opens upwards (since the coefficient of $$ x^2 $$ is positive).
   - The vertex of the parabola gives the minimum value. The vertex occurs at:
     $$
     x = -\frac{b}{2a} = -\frac{6}{2 \times 9} = -\frac{1}{3}
     $$
   - However, since $$ x \geq 0 $$, the minimum value within the domain occurs at $$ x = 0 $$:
     $$
     f(0) = 9(0)^2 + 6(0) - 5 = -5
     $$

2. **Behavior as $$ x $$ Increases:**
   - As $$ x $$ increases beyond 0, $$ f(x) $$ increases without bound because it's a quadratic function opening upwards.
   - Therefore, $$ f(x) $$ approaches infinity as $$ x $$ approaches infinity.

**Conclusion:**
The range of $$ f(x) $$ for $$ x \geq 0 $$ is:
$$
[-5, \infty)
$$
This implies that $$ a = \infty $$.

**Answer:** The range of $$ f(x) $$ is $$[-5, \infty)$$, so $$ a = \infty $$.

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### **Summary of Answers**

1. **Problem 1 Answer:** (a) 0
2. **Problem 2 Answer:** The range of $$ f(x) $$ is $$[-5, \infty)$$, so $$ a = \infty $$.
**Problem 1 Answer:** (a) 0 **Problem 2 Answer:** The range of $$ f(x) $$ is $$[-5, \infty)$$, so $$ a = \infty $$.

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