Theory of Intelligence
11TH•Mathematics•Matrices And Determinants•Exercise 4.1

Introduction

Practice set 4.1 (identity matrix, matrix operations, transpose, and basic properties).

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Question 1

If A=[aij]3×3A=[a_{ij}]_{3\times 3}, then show that:

Select Part:Part (i) of 2
(i)

I3A=AI_3A=A

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Step-by-Step Solution

Question 1 • Part (i)

verifiedVerified Answer

Part (i): Proof that I3A=AI_3 A = A

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Matrix Lab•Interactive

Matrix Identity Left-Multiplication (I₃ A = A)

Lab Mode:
Target Scanner:
×
Identity Matrix (I₃)Row 1
1
0
0
0
1
0
0
0
1
×
Arbitrary Matrix (A)Col 1
2
-1
4
0
3
-2
5
1
6
=
Product (I₃A)Exact Match A
2
-1
4
0
3
-2
5
1
6
Row-by-Column Kronecker Delta Dot Product: (1, 1)Formula: (I₃A)_{11} = ∑ δ_{1k} a_{k1}
Cell (1, 1) = (1 × 2)+(0 × 0)+(0 × 5)=2
✓ Mathematical Invariant: The Kronecker delta δ_ik is zero everywhere except when k = i. Hence, exactly one non-zero product survives, reproducing entry a_{11} = 2 identically!

Step-by-Step Proof

Let A=[aij]3×3A = [a_{ij}]_{3 \times 3} and I3=[δij]3×3I_3 = [\delta_{ij}]_{3 \times 3}:

I3A=[100010001][a11a12a13a21a22a23a31a32a33]I_3 A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}

For each element (i,j)(i, j):

(I3A)ij=∑k=13δikakj=δi1a1j+δi2a2j+δi3a3j=aij(I_3 A)_{ij} = \sum_{k=1}^3 \delta_{ik} a_{kj} = \delta_{i1} a_{1j} + \delta_{i2} a_{2j} + \delta_{i3} a_{3j} = a_{ij}

Since each corresponding entry equals aija_{ij}:

I3A=A\boxed{I_3 A = A}
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