Calculate the determinant of the 2 × 2 matrix.

Question

ScanSolve AI Tutor · Accepted Answer

Here are the solutions to questions 17, 18, and 19 based on the image provided:

17. Find the value of $x$ if $\begin{vmatrix} x & 3 \\ 4 & 7 \end{vmatrix} = 9$

Calculate the determinant of the $2 \times 2$ matrix.
The determinant of a matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $ad - bc$ .
For the given matrix $\begin{bmatrix} x & 3 \\ 4 & 7 \end{bmatrix}$ , the determinant is $(x)(7) - (3)(4)$ .
$7x - 12$
Set the determinant equal to 9 as given in the problem.
$7x - 12 = 9$
Solve the linear equation for $x$ .
Add 12 to both sides:
$7x = 9 + 12$ $7x = 21$
Divide by 7:
$x = \frac{21}{7}$ $x = 3$
The correct answer is (b) 3.
$\boxed{3}$
18. Rationalise $\frac{2\sqrt{3} - \sqrt{5}}{\sqrt{5} - \sqrt{3}}$
Identify the conjugate of the denominator.
The denominator is $\sqrt{5} - \sqrt{3}$ . Its conjugate is $\sqrt{5} + \sqrt{3}$ .
Multiply the numerator and the denominator by the conjugate.
$\frac{2\sqrt{3} - \sqrt{5}}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}}$
Expand the numerator
$(2\sqrt{3} - \sqrt{5})(\sqrt{5} + \sqrt{3})$ .
$= (2\sqrt{3})(\sqrt{5}) + (2\sqrt{3})(\sqrt{3}) - (\sqrt{5})(\sqrt{5}) - (\sqrt{5})(\sqrt{3})$ $= 2\sqrt{15} + 2(3) - 5 - \sqrt{15}$ $= 2\sqrt{15} + 6 - 5 - \sqrt{15}$ $= (2\sqrt{15} - \sqrt{15}) + (6 - 5)$ $= \sqrt{15} + 1$
Expand the denominator using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$
$(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})$ .
$= (\sqrt{5})^2 - (\sqrt{3})^2$ $= 5 - 3$ $= 2$
Combine the simplified numerator and denominator.
$\frac{\sqrt{15} + 1}{2}$
Upon reviewing the provided options, none of them match the calculated answer $\frac{\sqrt{15} + 1}{2}$ . There might be a typo in the question or the options.
$\boxed{\frac{\sqrt{15} + 1}{2}}$
19. If $y=x$ , find $dy/dx$ .
The given function is $y=x$ .
Differentiate $y$ with respect to $x$ .
Using the power rule for differentiation, if $y = x^n$ , then $\frac{dy}{dx} = nx^{n-1}$ .
For $y=x$ , we have $n=1$ .
$\frac{dy}{dx} = 1 \cdot x^{1-1}$ $\frac{dy}{dx} = 1 \cdot x^0$
Since any non-zero number raised to the power of 0 is 1 ( $x^0=1$ for $x \neq 0$ ),
$\frac{dy}{dx} = 1 \cdot 1$ $\frac{dy}{dx} = 1$
The correct answer is (a) 1.
$\boxed{1}$

Calculate the determinant of the 2 x 2 matrix.

Answer

Step-by-step solution

Calculate the determinant of the $2 \times 2$ matrix.

Set the determinant equal to 9 as given in the problem.

Solve the linear equation for $x$ .

Identify the conjugate of the denominator.

Multiply the numerator and the denominator by the conjugate.

Expand the numerator

Expand the denominator using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$

Combine the simplified numerator and denominator.

The given function is $y=x$ .

Differentiate $y$ with respect to $x$ .

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