2. Determinant of block matrix

Definition 7.2.3

$det (A) = \sum_{σ \in \prod_{n}} s g n (σ) \prod_{i = 1}^{n} A_{i, σ (i)}$

a)

Using the definition 7.2.3 $det M = \sum_{σ \in \prod_{n}} s g n (σ) \prod_{i = 1}^{n} M_{i, σ (i)}$

Because the lower-left block is zero, any term with some $i > m$ and $σ (i) \leq m$ contains a factor $M_{i, σ (i)} = 0$ and vanishes. Hence the only permutations that contribute satisfy

$i \in {m + 1 \dots n} ⟹ σ (i) \in {m + 1 \dots n}$ and $i \in {1 \dots m} ⟹ σ (i) \in {1 \dots m}$ (since $σ$ is bijective) This means we can split $σ$ into $σ_{1}$ and $σ_{2}$ where $σ_{1}$ maps ${1 \dots m} \to {1 \dots m}$ and $σ_{2}$ maps ${m + 1 \dots n} \to {m + 1 \dots n}$ with $σ = σ_{1} \circ σ_{2}$ and $s g n (σ) = s g n (σ_{1}) s g n (σ_{2})$ The product splits:

i = 1 \prod n M_{i, σ (i)} = (i = 1 \prod m A_{i, σ_{1} (i)}) (i = m + 1 \prod n C_{i - m, σ_{2} (i) - m})

Therefore, $det M = \sum_{σ_{1} \in \prod_{m}} \sum_{σ_{2} \in \prod_{n = m}} s g n (σ_{1}) s g n (σ_{2}) (\prod_{i = 1}^{m} A_{i, σ_{1} (i)}) (\prod_{i = m + 1}^{n} C_{i - m, σ_{2} (i) - m}) = det A \cdot det C$

b)

$M = 291328000238000017400553000002740721$ $det (M) = det (M^{⊤}) = det (200407900004100000320507231502887321)$ swap row 2 with row 6, swap row 3 with row 4 (swap twice makes determinant unchanged) $= det (274000940000100000375002225103813728)$ Let $A = 274940100, B = 375225813, C = 002103728$ $det (M) = det ([A 0 B C]) = det (A) det (C)$ (result from previous subtask) $det (A) = - 16, det (C) = 4$ $det (M) = - 16 \cdot 4 = - 64$

Thomas Second Brain

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LinArg assignment 11

2. Determinant of block matrix

a)

b)

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