Engineering
Mathematics
Inverse of a Matrix
Question

Let P=[3−1−220α3−50] , where a ∈ R. Suppose Q = [qij] is a matrix such that PQ = kI, where k ∈ R, k ≠ 0 and I is the identity matrix of order 3. If q23 = −k8  and det (Q) = k22, then

α = 0, k = 8

det (P adj (Q)) = 29

det (Q adj (P)) = 213

4α – k + 8 = 0

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Solution

Q = P–1k

| Q | = k3|P|=k22

 | P | = 2k

Q=k (adj. P)|P|=k|P| adj.(P)=−k|P|   (3α+4)=−k8

⇒ 24α + 32 = | P |

| P | = 12α + 20

24α + 32 = 12α + 20 ⇒ α = – 1

| P | = 8, k = 4

(B)        4α – k + 8 = 0

(C)        | P adj. (Q) | = | P | | Q |2 = 8 · 82 = 29

(D)        | Q (adj. P) | = | Q | | adj. P | = | Q | · | P |2 = 83 = 29.