If , then show that A2 = 2A & A3 = 4A.

Engineering

Mathematics

Introduction to Determinants

Question

If $A = |\begin{array}{cc} 1 & - 1 \\ - 1 & 1 \end{array}|$ , then show that A² = 2A & A³ = 4A.

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Given that:

A = ∣ \begin{array}{cc} 1 & - 1 \\ - 1 & 1 \end{array} ∣

Now,

A² = A × A

A \times A = ∣ \begin{array}{cc} 1 & - 1 \\ - 1 & 1 \end{array} ∣ \times ∣ \begin{array}{cc} 1 & - 1 \\ - 1 & 1 \end{array} ∣

A \times A = ∣ \begin{array}{cc} 1 \times 1 + (- 1) \times (- 1) & - 1 \times 1 + 1 \times (- 1) \\ - 1 \times 1 + 1 \times (- 1) & 1 \times 1 + (- 1) \times (- 1) \end{array} ∣

A^{2} = ∣ \begin{array}{cc} 2 & - 2 \\ - 2 & 2 \end{array} ∣

A^{2} = 2 ∣ \begin{array}{cc} 1 & - 1 \\ - 1 & 1 \end{array} ∣

A² = 2A

Hence proved.

Now,

A³ = A² × A

= 2A × A (Since we proved A² = 2A)

= 2A²

= 2 × 2A

A³ = 4A

Hence proved.

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