Question 4 of 58intermediate🔧 ApplyLong Answer2 marks

If A = [101012004]\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}, then show that |3A| = 27 |A|

Correct Answer

Step 1: Finding |A|

Given A = [101012004]\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}

Expanding along R₁ (as it contains maximum zeros for easier calculation):

|A| = 1 × ∣1204∣−0×∣0204∣+1×∣0100∣\begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} - 0 \times \begin{vmatrix} 0 & 2 \\ 0 & 4 \end{vmatrix} + 1 \times \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix}

= 1 × (4 - 0) - 0 + 1 × (0 - 0) = 4

Step 2: Finding |3A|

3A = [3030360012]\begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{bmatrix}

|3A| = 3 × ∣36012∣\begin{vmatrix} 3 & 6 \\ 0 & 12 \end{vmatrix} - 0 + 3 × ∣0300∣\begin{vmatrix} 0 & 3 \\ 0 & 0 \end{vmatrix}

= 3 × (36 - 0) + 0 = 3×363 \times 36 = 108

Step 3: Verification

27|A| = 27×427 \times 4 = 108

Since |3A| = 108 = 27|A|, hence verified.

Exercise: EXERCISE 4.1 | Q: 4 | (Page 6)
For More Understanding

Explanation

This question tests the property that for a square matrix A of order n, |kA| = kⁿ|A|. Since A is a 3×33 \times 3 matrix, |3A| = 3³|A| = 27|A|. The solution demonstrates this by directly calculating both determinants using expansion along R₁, which is efficient as it contains two zeros. The context provides the expansion formula: |A| = a₁₁(minor) - a₁₂(minor) + a₁₃(minor), which is applied correctly.

Solution Steps

  1. Step 1: Calculate |A| by expanding along R₁ to get |A| = 4

  2. Step 2: Calculate |3A| by expanding along R₁ to get |3A| = 108

  3. Step 3: Verify that 27|A| = 27×427 \times 4 = 108 = |3A|

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