Solutions — Determinants and Inverses

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1. Q1 — Computing Determinants

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   Formula: det([[a,b],[c,d]]) = ad − bc.

   (a) A = [[5,2],[3,4]]: det = 5(4) − 2(3) = 20 − 6 = 14. Invertible.

   (b) A = [[−1,3],[2,−6]]: det = (−1)(−6) − 3(2) = 6 − 6 = 0. Singular! Notice row 2 = −2 × row 1, confirming linear dependence.

   (c) A = [[7,0],[4,−2]]: det = 7(−2) − 0(4) = −14. The negative determinant tells us the transformation reverses orientation. det = −14. Invertible.

2. Q2 — Finding an Inverse

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   A = [[3,1],[5,2]].

   Step 1: det(A) = 3(2) − 1(5) = 6 − 5 = 1

   Step 2: Apply the inverse formula: swap a and d, negate b and c, divide by det.

   A⁻¹ = (1/1) [[2, −1], [−5, 3]] = [[2, −1], [−5, 3]]

   Verification: AA⁻¹

   (1,1): 3(2)+1(−5)=6−5=1. (1,2): 3(−1)+1(3)=−3+3=0.

   (2,1): 5(2)+2(−5)=10−10=0. (2,2): 5(−1)+2(3)=−5+6=1.

   AA⁻¹ = [[1,0],[0,1]] = I ✓

3. Q3 — Identifying Singular Matrices

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   (a) [[4,2],[6,3]]: det = 4(3) − 2(6) = 12 − 12 = 0. Singular. Row 2 = 1.5 × Row 1 (linearly dependent rows).

   (b) [[1,0],[0,5]]: det = 1(5) − 0(0) = 5 ≠ 0. Invertible. This is a diagonal matrix (scales x by 1, y by 5); its inverse scales x by 1 and y by 1/5.

   (c) [[3,−6],[−1,2]]: det = 3(2) − (−6)(−1) = 6 − 6 = 0. Singular. Row 2 = −(1/3) × Row 1.

4. Q4 — Determinant of a Product

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   Given: det(A) = 4, det(B) = −3.

   (a) det(AB) = det(A) × det(B) = 4 × (−3) = −12

   The combined transformation scales area by a factor of 12 and reverses orientation.

   (b) Since det(A) × det(A⁻¹) = det(I) = 1:

   det(A⁻¹) = 1/det(A) = 1/4

   (c) det(A²) = det(A·A) = det(A) × det(A) = 4 × 4 = 16

5. Q5 — Finding k for a Singular Matrix

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   M = [[k,2],[3,k]]. For M to be singular: det(M) = 0.

   det(M) = k × k − 2 × 3 = k² − 6

   Set equal to zero: k² − 6 = 0 → k² = 6 → k = √6 or k = −√6

   Check: for k = √6, det = 6 − 6 = 0 ✓

6. Q6 — Using A⁻¹ to Solve a System

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   System: 2x + 3y = 8 and x + 2y = 5.

   Matrix form: [[2,3],[1,2]] [[x],[y]] = [[8],[5]].

   det(A) = 2(2) − 3(1) = 4 − 3 = 1

   A⁻¹ = (1/1)[[2,−3],[−1,2]] = [[2,−3],[−1,2]]

   [[x],[y]] = A⁻¹B = [[2,−3],[−1,2]] [[8],[5]]

   x = 2(8) + (−3)(5) = 16 − 15 = 1

   y = (−1)(8) + 2(5) = −8 + 10 = 2

   Verify: 2(1)+3(2) = 2+6 = 8 ✓   1+2(2) = 1+4 = 5 ✓

7. Q7 — Inverse of a Product

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   A = [[2,1],[1,1]], B = [[3,0],[1,2]].

   Method (i): Compute AB, then invert.

   AB: (1,1)=6+1=7, (1,2)=0+2=2, (2,1)=3+1=4, (2,2)=0+2=2.

   AB = [[7,2],[4,2]]. det(AB) = 14−8 = 6.

   (AB)⁻¹ = (1/6)[[2,−2],[−4,7]]

   Method (ii): Use (AB)⁻¹ = B⁻¹A⁻¹.

   det(A) = 2−1 = 1. A⁻¹ = [[1,−1],[−1,2]].

   det(B) = 6−0 = 6. B⁻¹ = (1/6)[[2,0],[−1,3]].

   B⁻¹A⁻¹ = (1/6)[[2,0],[−1,3]][[1,−1],[−1,2]]

   (1/6)[(1,1): 2+0=2, (1,2): −2+0=−2, (2,1): −1−3=−4, (2,2): 1+6=7]

   B⁻¹A⁻¹ = (1/6)[[2,−2],[−4,7]] ✓ Both methods agree.

8. Q8 — Finding a Matrix from its Inverse

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   M⁻¹ = [[3,−1],[−5,2]].

   Since (M⁻¹)⁻¹ = M, we find the inverse of M⁻¹.

   det(M⁻¹) = 3(2) − (−1)(−5) = 6 − 5 = 1

   Apply the inverse formula to M⁻¹: swap main diagonal, negate off-diagonal, divide by det.

   M = (1/1)[[2, 1], [5, 3]] = [[2, 1], [5, 3]]

   Verify: M × M⁻¹: (1,1)=6−5=1, (1,2)=−2+2=0, (2,1)=15−15=0, (2,2)=−5+6=1. = I ✓

9. Q9 — Determining Solution Type Using the Determinant

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   (a) System: 3x − y = 2, 6x − 2y = 4. A = [[3,−1],[6,−2]].

   det = 3(−2)−(−1)(6) = −6+6 = 0. Singular.

   Equation 2 = 2 × Equation 1 (both equations represent the same line). Infinitely many solutions.

   (b) System: 2x + y = 5, x − y = 1. A = [[2,1],[1,−1]].

   det = 2(−1)−1(1) = −2−1 = −3 ≠ 0. Unique solution. (Solution: x=2, y=1)

   (c) System: 4x + 6y = 3, 6x + 9y = 7. A = [[4,6],[6,9]].

   det = 4(9)−6(6) = 36−36 = 0. Singular.

   Row 2 = (3/2) × Row 1 for the coefficient matrix. But RHS: 7 ≠ (3/2)(3) = 4.5. The lines are parallel but distinct. No solution.

10. Q10 — Proof: If det(A) ≠ 0, then AX = 0 has only the trivial solution

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    Given: A is an invertible 2×2 matrix (det(A) ≠ 0), and AX = 0 for some column vector X.

    To prove: X = 0 (the zero vector).

    Proof:

    Since A is invertible, A⁻¹ exists. Multiply both sides of AX = 0 on the left by A⁻¹:

    A⁻¹(AX) = A⁻¹ × 0

    (A⁻¹A)X = 0     [matrix multiplication is associative]

    IX = 0           [since A⁻¹A = I]

    X = 0            [since IX = X for any X]

    Therefore, the only solution to AX = 0 is X = 0, the trivial solution. □

    Significance: Geometrically, an invertible transformation cannot collapse a non-zero vector to the origin. This property (called “injectivity” or “one-to-one”) is equivalent to the matrix being invertible.