Q. If \( A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \), find \( |A| \).
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Solution
The determinant is \( 1*4 - 2*3 = 4 - 6 = -2 \).
Correct Answer: A — -2
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Q. If \( A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \), what is \( |2A| \)?
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Solution
The determinant of \( 2A \) is \( 2^2 * |A| = 4 * (-2) = -8 \).
Correct Answer: B — 8
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Q. If \( A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \), what is \( |A| \)?
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Solution
The determinant is calculated as (1*4) - (2*3) = 4 - 6 = -2.
Correct Answer: A — -2
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Q. If \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), what is the determinant of A?
A.
ad - bc
B.
bc - ad
C.
a + d
D.
b + c
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Solution
The determinant of matrix A is given by the formula \( ad - bc \).
Correct Answer: A — ad - bc
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Q. If \( B = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 0 \\ 2 & 3 & 1 \end{pmatrix} \), what is \( |B| \)?
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Solution
The determinant is calculated as 1(1*1 - 0*3) - 2(0*1 - 1*2) + 1(0*3 - 1*2) = 1 - 4 - 2 = -5.
Correct Answer: B — 2
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Q. If \( B = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} \), what is \( \det(B) \)?
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Solution
Using the determinant formula, we find \( \det(B) = 1(1*0 - 4*6) - 2(0 - 4*5) + 3(0 - 1*5) = -24 \).
Correct Answer: A — -24
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Q. If \( B = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix} \), what is \( |B| \)?
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Solution
The determinant is 0 because the rows are linearly dependent.
Correct Answer: A — 0
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Q. If \( B = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \), what is \( |B| \)?
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Solution
The determinant is \( 1*4 - 2*3 = 4 - 6 = -2 \).
Correct Answer: A — -2
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Q. If \( B = \begin{pmatrix} 1 & 2 \\ 3 & 5 \end{pmatrix} \), what is \( |B| \)?
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Solution
The determinant is \( 1*5 - 2*3 = 5 - 6 = -1 \).
Correct Answer: B — 1
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Q. If \( B = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} \), find \( |B| \).
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Solution
The determinant is \( 2*4 - 3*1 = 8 - 3 = 5 \).
Correct Answer: A — 5
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Q. If \( B = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} \), what is \( |B| \)? (2022)
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Solution
The determinant of \( B \) is \( 2*4 - 3*1 = 8 - 3 = 5 \).
Correct Answer: A — 5
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Q. If \( C = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 6 \end{pmatrix} \), find \( \det(C) \).
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Solution
The determinant is 0 because the first column is a linear combination of the other columns.
Correct Answer: A — 0
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Q. If \( C = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 6 \end{pmatrix} \), find \( |C| \).
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Solution
The determinant is 0 because the first column is a linear combination of the others.
Correct Answer: A — 0
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Q. If \( C = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), what is the determinant of C?
A.
ad - bc
B.
bc - ad
C.
a + d
D.
b + c
Show solution
Solution
The determinant of C is given by the formula \( ad - bc \).
Correct Answer: A — ad - bc
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Q. If \( D = \begin{vmatrix} 2 & 3 & 1 \\ 1 & 0 & 2 \\ 4 & 1 & 0 \end{vmatrix} \), find \( D \).
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Solution
Calculating gives \( 2(0*0 - 2*1) - 3(1*0 - 2*4) + 1(1*1 - 0*4) = -4 + 24 + 1 = 21 \).
Correct Answer: A — -10
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Q. If \( E = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \), what is \( |E| \)? (2023)
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Solution
The determinant is calculated as \( 0*0 - 1*1 = 0 - 1 = -1 \).
Correct Answer: C — -1
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Q. If \( E = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \), what is \( |E| \)? (2022)
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Solution
The determinant is \( 1*1 - 1*1 = 1 - 1 = 0 \).
Correct Answer: A — 0
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Q. If \( F = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 6 \end{pmatrix} \), what is the value of the determinant?
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Solution
The determinant is 0 because the first column is a linear combination of the other columns.
Correct Answer: A — 0
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Q. If \( I = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \), what is \( |I| \)? (2021)
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Solution
The determinant is \( 1*1 - 1*1 = 1 - 1 = 0 \).
Correct Answer: A — 0
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Q. If \( J = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 0 \\ 2 & 1 & 3 \end{pmatrix} \), what is the value of the determinant?
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Solution
The determinant is calculated as \( 1(1*3 - 0*1) - 2(0*3 - 1*2) + 1(0*1 - 1*2) = 3 + 4 - 2 = 5 \).
Correct Answer: A — 0
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Q. If \( y = \cot^{-1}(x) \), what is \( \frac{dy}{dx} \)?
A.
\( -\frac{1}{1+x^2} \)
B.
\( \frac{1}{1+x^2} \)
C.
0
D.
undefined
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Solution
The derivative of \( y = \cot^{-1}(x) \) is \( -\frac{1}{1+x^2} \).
Correct Answer: A — \( -\frac{1}{1+x^2} \)
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Q. If \( y = \sec^{-1}(x) \), what is \( \frac{dy}{dx} \)?
A.
\( \frac{1}{
B.
x
C.
\sqrt{x^2-1}} \)
D.
\( \frac{1}{x\sqrt{x^2-1}} \)
.
0
.
undefined
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Solution
The derivative of \( y = \sec^{-1}(x) \) is \( \frac{1}{|x|\sqrt{x^2-1}} \).
Correct Answer: B — x
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Q. If \( y = \sin^{-1}(x) + \cos^{-1}(x) \), what is the value of \( y \)?
A.
0
B.
1
C.
\( \frac{\pi}{2} \)
D.
undefined
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Solution
Since \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \) for all \( x \) in the domain of \( \sin^{-1} \) and \( \cos^{-1} \), the answer is \( \frac{\pi}{2} \).
Correct Answer: C — \( \frac{\pi}{2} \)
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Q. If \( y = \tan^{-1}(x) + \tan^{-1}(y) \), what is the value of \( y \) when \( x = 1 \)?
A.
0
B.
1
C.
\( \frac{\pi}{4} \)
D.
undefined
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Solution
When \( x = 1 \), \( y = \tan^{-1}(1) + \tan^{-1}(y) \) leads to \( y = \frac{\pi}{4} \).
Correct Answer: C — \( \frac{\pi}{4} \)
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Q. If ΔG is negative for a reaction, what can be inferred about the reaction?
A.
The reaction is at equilibrium.
B.
The reaction is spontaneous.
C.
The reaction is non-spontaneous.
D.
The reaction requires energy input.
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Solution
A negative ΔG indicates that the reaction is spontaneous.
Correct Answer: B — The reaction is spontaneous.
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Q. If ΔG is negative for a reaction, what can be inferred?
A.
The reaction is non-spontaneous.
B.
The reaction is at equilibrium.
C.
The reaction is spontaneous.
D.
The reaction requires energy input.
Show solution
Solution
A negative ΔG indicates that the reaction is spontaneous under the given conditions.
Correct Answer: C — The reaction is spontaneous.
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Q. If ΔG is negative, what does it indicate about the reaction?
A.
Reaction is at equilibrium
B.
Reaction is spontaneous
C.
Reaction is non-spontaneous
D.
Reaction requires energy input
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Solution
A negative ΔG indicates that the reaction is spontaneous.
Correct Answer: B — Reaction is spontaneous
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Q. If ΔG is positive, what can be inferred about the reaction?
A.
The reaction is spontaneous.
B.
The reaction is at equilibrium.
C.
The reaction is non-spontaneous.
D.
The reaction will proceed in reverse.
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Solution
If ΔG is positive, the reaction is non-spontaneous under the given conditions.
Correct Answer: C — The reaction is non-spontaneous.
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Q. If ΔH = 100 kJ and ΔS = 0.2 kJ/K, what is ΔG at 298 K?
A.
100 kJ
B.
96 kJ
C.
104 kJ
D.
90 kJ
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Solution
ΔG = ΔH - TΔS = 100 kJ - 298 K * 0.2 kJ/K = 100 kJ - 59.6 kJ = 40.4 kJ.
Correct Answer: B — 96 kJ
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Q. If ΔH is negative and ΔS is positive, what can be said about ΔG?
A.
ΔG is always positive.
B.
ΔG is always negative.
C.
ΔG can be positive or negative depending on temperature.
D.
ΔG is zero.
Show solution
Solution
If ΔH is negative and ΔS is positive, ΔG will always be negative, indicating that the reaction is spontaneous at all temperatures.
Correct Answer: B — ΔG is always negative.
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