Find the inverse of \(\begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}\)
\(\begin{pmatrix} 5 & 1 \\ -3 & 2 \end{pmatrix}\)
\(\begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}\)
\(\begin{pmatrix} -5 & 2 \\ -1 & 3 \end{pmatrix}\)
\(\begin{pmatrix} 5 & 1 \\ 2 & 3 \end{pmatrix}\)
Correct answer is B
Let A = \(\begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}\)
|A| = (3 x 2 - 5 x 1)
= 6 - 5
= 1
A\(^{-1}\) = \(\begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}\)
\({\color{red}2x} \times 3\)
1 - \(\sqrt{6}\)
\(\sqrt{6}\) - 1
\(\sqrt{6}\)
1 + \(\sqrt{6}\)
Correct answer is D
x*y = x\(^2\) - y\(^2\) + xy
(\(\sqrt{3}\))*(\(\sqrt{2}\)) = (\(\sqrt{3}\))\(^2\) - (\(\sqrt{2}\))\(^2\) + \(\sqrt{3}\) x \(\sqrt{2}\)
= 3 - 2 + \(\sqrt{6}\)
= 1 + \(\sqrt{6}\)
Evaluate: \(^{lim}_{x \to 1} \begin{pmatrix} \frac{1 - x}{x^2 - 3x + 2} \end {pmatrix}\)
-1
- \(\frac{1}{2}\)
\(\frac{1}{2}\)
1
Correct answer is D
Lim(\(\frac{1 - x}{x^2 - 3x + 2}\))
= \(^{lim}_{x \to 1} \begin{pmatrix} \frac{1 - x}{x^2 - 3x + 2} \end {pmatrix}\)
= \(^{lim}_{x \to 1} \begin{pmatrix} \frac{x - 1}{(x - 2)(x + 1)} \end {pmatrix}\)
= \(\frac{-1}{1 - 2}\)
= \(\frac{-1}{-1}\)
= 1
69
60
35
30
Correct answer is D
54 x 40 = 35 x 10 + 60F
160 = 350 + 60F
F = \(\frac{1810}{60}\)
= 30N
0.6
0.5
0.4
0.3
Correct answer is C
P(X \(\cup\) Y) = P(X)P(Y)
0.15 = 0.5m
m = \(\frac{0.015}{0.5}\)
= 0.3