The graded gate, same seven questions, same order, new numbers
Gate 3 is the only graded work this block, and it is the single best predictor of WPR 2. This page rebuilds it exactly: the same seven questions in the same order, carrying the same point values, with every number changed. Work it closed-book, then check.
What Gate 3 actually was: 25 points, seven questions, 100% computation: no caption, no interpretation item, nothing to write a paragraph about. That is a hard break from Gates 1 and 2, and it matches the course director's warning that WPR 2 is much more computational.
Use these vectors for problems 1–3.
|u| = √((−8)2 + 152) = √(64 + 225) = √289 = 17
Squaring kills the minus sign: (−8)2 = +64, not −64. That single slip turns 289 into 161 and the answer into √161.
û = u|u| = ⟨−8, 15⟩17 = ⟨−8/17, 15/17⟩
Divide every component by the magnitude. Leave it as fractions. The gate asked for exact values, so ⟨−0.47, 0.88⟩ is not the answer it wants.
Check your work: a unit vector has magnitude 1. (−8/17)2 + (15/17)2 = (64 + 225)/289 = 289/289 = 1 ✓
u · v = (−8)(2) + (15)(1) = −16 + 15 = −1
A scalar, not a vector. The negative sign is real and means the angle between them is obtuse, just barely.
cos θ = 8 / (3 × 7) = 8/21 = 0.3810
θ = cos−1(0.3810) = 67.61°
Note the gate hands you the dot product and both magnitudes. The only work being tested is dividing and taking the inverse cosine, so there is no excuse for losing these four points.
Use the following matrices for problems 5–6.
| A = | [ 2 −1 ] [ 3 5 ] |
B = | [ 4 0 ] [ 1 −2 ] |
Each entry is a row of A dotted with a column of B:
(1,1): 2(4) + (−1)(1) = 8 − 1 = 7
(1,2): 2(0) + (−1)(−2) = 0 + 2 = 2
(2,1): 3(4) + 5(1) = 12 + 5 = 17
(2,2): 3(0) + 5(−2) = 0 − 10 = −10
AB = [ 7 2 ; 17 −10 ]
Two sign traps live in this one: (−1)(−2) = +2, and 5(−2) = −10.
det(A) = (2)(5) − (−1)(3) = 10 − (−3) = 10 + 3 = 13
13 ≠ 0, so yes, A−1 exists.
The subtraction of a negative is the whole difficulty. A cadet who writes 10 − 3 = 7 has the right formula and the wrong sign.
Answer both halves. Gate 3's rubric drops you from 4 to 2 if the determinant is right but the existence conclusion does not follow.
Augmented matrix: [ 2 3 | 8 ; 1 −1 | 1 ]
R2 ← 2R2 − R1: [ 2 3 | 8 ; 0 −5 | −6 ] ← this is a Row Echelon Form
Row 2: −5x2 = −6, so x2 = 6/5
Row 1: 2x1 + 3(6/5) = 8 → 2x1 = 8 − 18/5 = 22/5 → x1 = 11/5
Solution type: unique. Every column has a pivot and there is no contradictory row.
Check: 2(11/5) + 3(6/5) = 22/5 + 18/5 = 40/5 = 8 ✓ and 11/5 − 6/5 = 5/5 = 1 ✓
Expect fractions. The real Gate 3 answers were 7/3 and 4/3. If your answer comes out ugly, that is not evidence you made a mistake. Check it by substitution and move on. 11/5 = 2.2 and 6/5 = 1.2 are both acceptable forms.