Given that A is a square matrix and A² = A, we must reduce (I + A)² - 3A. Step 1: Expand (I + A)² (I + A)² = I² + 2IA + A² Given that I² = I and A² = A, we get: (I + A)² = I + 2A + A (I + A)² = I + 3A Step 2: Subtract 3A Substitute this back into the expression: (I + A)² - 3A = (I + 3A) - 3A (I + A)Read more
Given that A is a square matrix and A² = A, we must reduce (I + A)² – 3A.
Step 1: Expand (I + A)²
(I + A)² = I² + 2IA + A²
Given that I² = I and A² = A, we get:
(I + A)² = I + 2A + A
(I + A)² = I + 3A
Step 2: Subtract 3A
Substitute this back into the expression:
(I + A)² – 3A = (I + 3A) – 3A
(I + A)² – 3A = I
If A is a square matrix and A² = A, then (I + A)² – 3A is equal to:
Given that A is a square matrix and A² = A, we must reduce (I + A)² - 3A. Step 1: Expand (I + A)² (I + A)² = I² + 2IA + A² Given that I² = I and A² = A, we get: (I + A)² = I + 2A + A (I + A)² = I + 3A Step 2: Subtract 3A Substitute this back into the expression: (I + A)² - 3A = (I + 3A) - 3A (I + A)Read more
Given that A is a square matrix and A² = A, we must reduce (I + A)² – 3A.
Step 1: Expand (I + A)²
(I + A)² = I² + 2IA + A²
Given that I² = I and A² = A, we get:
(I + A)² = I + 2A + A
(I + A)² = I + 3A
Step 2: Subtract 3A
Substitute this back into the expression:
(I + A)² – 3A = (I + 3A) – 3A
(I + A)² – 3A = I
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