የሒሳብ መግቢያ ፈተና፣ 2013 ዓ.ም. (2021)

For a 2 by 2 matrix the determinant is \(ad - bc\), where \(a\) and \(b\) are the top row and \(c\) and \(d\) are the bottom row.

Here \(a = 1\), \(b = -7\), \(c = 6\), \(d = 5\).

Step 1: \(ad = 1 \times 5 = 5\)
Step 2: \(bc = (-7) \times 6 = -42\)
Step 3: \(ad - bc = 5 - (-42) = 5 + 42 = 47\)

The determinant is 47, so the answer is A.

Watch the sign: if you drop the minus on \(-7\) you get \(5 - 42 = -37\), which is the trap in option C.

Textbook: Mathematics Grade 11, Unit 4 Determinants and Their Properties, section 4.1 Determinants of Matrices of Order 2 (page 209).

The additive inverse of a number \(z\) is the number you add to \(z\) to get 0. It is \(-z\).

Step 1: write the inverse: \(-(3 - 4i)\)
Step 2: change the sign of both parts: \(-3 + 4i\)

Check: \((3 - 4i) + (-3 + 4i) = 0\). So the answer is B.

Do not mix this up with the conjugate. The conjugate of \(3 - 4i\) is \(3 + 4i\), which is the trap in option C.

The modulus of a complex number \(a + bi\) is its distance from the origin: \(|a + bi| = \sqrt{a^2 + b^2}\).

Here \(a = -6\) and \(b = 8\).

Step 1: \(a^2 = (-6)^2 = 36\)
Step 2: \(b^2 = 8^2 = 64\)
Step 3: \(|z| = \sqrt{36 + 64} = \sqrt{100} = 10\)

So the answer is A.

Remember that squaring removes the minus sign, so the \(-6\) contributes \(+36\), never \(-36\).

To find \(6u + 2v\), multiply each vector by its number, then add the matching coordinates.

Step 1: \(6u = 6(1, 3) = (6, 18)\)
Step 2: \(2v = 2(-3, 5) = (-6, 10)\)
Step 3: add: \((6 + (-6),\ 18 + 10) = (0, 28)\)

So the answer is B.

The trap in option A is adding \(6 + (-6)\) wrongly; the first coordinates cancel to 0 exactly.

Textbook: Mathematics Grade 11, Unit 5 Vectors, section 5.1 Revision on Vectors and Scalars (page 252).

The standard form of a circle with center \((h, k)\) and radius \(r\) is \((x - h)^2 + (y - k)^2 = r^2\).

Here \(h = 2\), \(k = -3\), \(r = 5\).

Step 1: \(x - h = x - 2\)
Step 2: \(y - k = y - (-3) = y + 3\)
Step 3: \(r^2 = 5^2 = 25\)

So the equation is \((x - 2)^2 + (y + 3)^2 = 25\), which is C.

Two traps here: option A forgets to square the radius, and option D keeps \(y - 3\) instead of flipping the sign of \(-3\).

A sphere with center \((h, k, l)\) and radius \(r\) has the equation \((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\). It is the circle formula with one more coordinate.

Here the center is \((0, -1, 2)\) and \(r = 6\).

Step 1: \(x - 0 = x\), so the first term is \(x^2\)
Step 2: \(y - (-1) = y + 1\), giving \((y + 1)^2\)
Step 3: \(z - 2\) gives \((z - 2)^2\)
Step 4: \(r^2 = 6^2 = 36\)

So the equation is \(x^2 + (y + 1)^2 + (z - 2)^2 = 36\), which is C.

Option A uses \(2 \times 6 = 12\) instead of \(6^2 = 36\), and option B flips the sign on \(z\). Both are common slips.