If A = bmatrix 1 & 2 \\ 4 & 2 bmatrix, then show that 2A = 4 A

[object Object] — This question tests the property that for a square matrix A of order n, kA = kⁿ A . Since A is a 2×2 matrix (order 2), we have 2A = 2² A = 4 A . The student calculates both sides independently to verify this property.

If A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}, then show that |…

Question 29 of 58intermediate🔧 ApplyLong Answer2 marks

If A = $\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$ , then show that |2A| = 4 |A|

Correct Answer

Given matrix A = $\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$

Step 1: Calculate |A| |A| = (1)(2) - (2)(4) = 2 - 8 = -6

Step 2: Find matrix 2A 2A = $\begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}$

Step 3: Calculate |2A| |2A| = (2)(4) - (4)(8) = 8 - 32 = -24

Step 4: Verify the result 4|A| = 4 × (-6) = -24

Therefore, |2A| = -24 = 4|A| (Hence proved)

Exercise: EXERCISE 4.1 | Q: 3 | (Chapter: 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 $2 \times 2$ matrix (order 2), we have |2A| = 2²|A| = 4|A|. The student calculates both sides independently to verify this property.

Solution Steps

Step 1: Calculate determinant of A using formula |A| = a₁₁a₂₂ - a₁₂a₂₁
Step 2: Multiply each element of A by 2 to get matrix 2A
Step 3: Calculate determinant of 2A
Step 4: Compare |2A| with 4|A| to verify equality

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