Inverse Matrices and Determinants — Solutions

← Back to Questions›Matrices

1. Fluency Show Solution

   A =

   3	2
   1	4

   → a = 3, b = 2, c = 1, d = 4

   det(A) = ad − bc = (3)(4) − (2)(1) = 12 − 2 = 10

   Since det(A) = 10 ≠ 0, this matrix is non-singular (invertible).

2. Fluency Show Solution

   B =

   −5	3
   2	−1

   → a = −5, b = 3, c = 2, d = −1

   det(B) = ad − bc = (−5)(−1) − (3)(2) = 5 − 6 = −1

   Since det(B) = −1 ≠ 0, this matrix is invertible.

3. Fluency Show Solution

   Step 1: Find det(A).

   det(A) = (4)(1) − (1)(3) = 4 − 3 = 1

   Step 2: Apply inverse formula. Swap 4 and 1 on the main diagonal; negate 1 and 3.

   A⁻¹ = ¹⁄₁

   1	−1
   −3	4

   =

   1	−1
   −3	4

   Quick verify: (4)(1)+(1)(−3) = 1 and (4)(−1)+(1)(4) = 0 → row 1 of AA⁻¹ = [1, 0] ✓

4. Fluency Show Solution

   Step 1: det(B) = (2)(−1) − (−3)(1) = −2 + 3 = 1

   Step 2: Apply inverse formula. Swap −1 and 2; negate −3 (becomes +3) and 1 (becomes −1).

   B⁻¹ = ¹⁄₁

   −1	3
   −1	2

   =

   −1	3
   −1	2

5. Understanding Show Solution

   Computing the determinant:

   det = (6)(1) − (2)(3) = 6 − 6 = 0

   Since det = 0, the matrix is singular.

   Algebraic meaning: No inverse exists. If you try to apply the formula, you would be dividing by zero, which is undefined.

   Geometric meaning: The transformation represented by this matrix collapses 2D space onto a line (the area scaling factor is |det| = 0). All regions are mapped to zero area — information is irreversibly lost, so the transformation cannot be undone (no inverse).

   Note: Observe that row 2 = ½ × row 1 (since [3, 1] = ½[6, 2]). When one row is a scalar multiple of the other, the determinant is always zero.

6. Understanding Show Solution

   For an n×n matrix: det(kA) = kⁿ × det(A).

   Here n = 2 (a 2×2 matrix) and k = 3.

   det(3A) = 3² × det(A) = 9 × 5 = 45

   Why k²? Multiplying A by scalar k multiplies every element by k. The determinant formula ad − bc involves products of two elements. Each product picks up k twice (once from each factor), giving k² overall. In general, for n×n matrices, every term in the determinant is a product of n elements, giving kⁿ.

7. Understanding Show Solution

   For the matrix to be singular: det = 0.

   det =

   k	2
   3	4

   = (k)(4) − (2)(3) = 4k − 6

   Set equal to zero: 4k − 6 = 0 → 4k = 6 → k = 3/2

   Check: det = 4(3/2) − 6 = 6 − 6 = 0 ✓

8. Understanding Show Solution

   Step 1: Find det(A).

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

   Step 2: Find A⁻¹.

   A⁻¹ = ¹⁄₁

   3	−1
   −5	2

   =

   3	−1
   −5	2

   Step 3: Compute AA⁻¹.

   c₁₁ = (2)(3)+(1)(−5) = 6−5 = 1

   c₁₂ = (2)(−1)+(1)(2) = −2+2 = 0

   c₂₁ = (5)(3)+(3)(−5) = 15−15 = 0

   c₂₂ = (5)(−1)+(3)(2) = −5+6 = 1

   AA⁻¹ =

   1	0
   0	1

   = I ✓

9. Problem Solving Show Solution

   Find det(T):

   det(T) = (1)(1) − (2)(0) = 1 − 0 = 1

   Apply the area scaling principle:

   Area of R′ = |det(T)| × Area of R = |1| × 8 = 8 cm²

   Interpretation: Since |det(T)| = 1, this transformation (a horizontal shear) preserves area. The region R′ has the same area as R, namely 8 cm², even though the shape is distorted. A shear transformation does not expand or shrink space — it only slides layers parallel to an axis.

10. Problem Solving Show Solution

    (a) Example: The matrix

    1	2
    2	4

    is not the zero matrix (it has non-zero entries).

    (b) Verification: det = (1)(4) − (2)(2) = 4 − 4 = 0 ✓

    (c) Problems with AX = B:

    The matrix inverse method requires X = A⁻¹B. But if det(A) = 0, then A⁻¹ does not exist (division by zero in the formula). We cannot find a unique solution this way.

    Looking at the system represented by this matrix:

    Row 1: x + 2y = b₁

    Row 2: 2x + 4y = b₂

    Row 2 is exactly twice Row 1. So either b₂ = 2b₁ (equations are dependent → infinitely many solutions) or b₂ ≠ 2b₁ (equations are contradictory → no solution). In neither case is there a unique solution. The singular matrix makes unique solution impossible.