Let A be a 3×3 matrix such that XTAX=0 for all nonzero 3×1 matrices X=xyz.A=111,A=14−5,A=121,A=04−8Ifdet(adj(2(A+I)))=2α3β5γ, α,β,γ∈N, then α2+β2+γ2 is:
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When working with determinants of matrices, be sure to use the properties of the adjugate matrix and the fact that the determinant of the adjugate is related to the determinant of the original matrix. Pay attention to the exponentiation for power terms when dealing with matrices.
Step 1: Understand the condition XTAX=0. If XTAX=0 for all nonzero vectors X, then A must be a skew-symmetric matrix. Therefore, AT=−A.
Step 2: Use the given information to find A. Let v1=111 and v2=121. We are given:
Av1=14−5andAv2=04−8. Also, let v3=xyz be an arbitrary vector. Since A is skew-symmetric, the diagonal entries must be zero.
Let us compute Av1+Av2:
Av1+Av2=A(v1+v2)=18−13. Since we do not know enough about A, let us proceed. The relation XTAX=0 for all X implies A is skew-symmetric. Let v1=111,v2=121, then: Av1=14−5,Av2=04−8. From AT=−A, we know aii=0 for all i. Let A=0−a−ba0−cbc0. Then: Av1=a+b−a+c−b−c=14−5 and Av2=2a+b−a+2c−2b−c=04−8. So we have the following system of equations: a+b=1,−a+c=4,−b−c=−5,2a+b=0,−a+2c=4,−2b−c=−8. From a+b=1 and 2a+b=0, we get a=−1 and b=2. From −a+c=4, we get c=3. Therefore, the matrix A is: A=01−2−10−3230.
Step 3: Calculate det(2(A+I))
We first compute A+I, where I is the identity matrix: A+I=11−2−11−3231. Then: 2(A+I)=22−4−22−6462. The determinant of 2(A+I) is calculated as: det(2(A+I))=2(4+36)+2(4+24)+4(−12+8)=80+56−16=120.
Step 4: Calculate det(adj(2(A+I)))
We know that: det(adj(M))=(detM)n−1, where n is the size of the matrix. In our case, n=3, so: det(adj(2(A+I)))=(det(2(A+I)))3−1=(120)2=26⋅32⋅52.
Step 5: Find α,β,γ and α2+β2+γ2
We have: det(adj(2(A+I)))=26⋅32⋅52, so α=6, β=2, and γ=2. Thus: α2+β2+γ2=62+22+22=36+4+4=44.