How do I solve matrix for #a in CC#?

Question

#det A=|[1,1,1,-a],[1,1,-a,1],[1,-a,1,1],[-a,1,1,1]|!=0#

HIX Tutor · Accepted Answer

undefined To solve for matrix a in CC, you can use the following steps:

Step 1: Write down the given matrix equation. Let's say it is A * X = B, where A represents the coefficient matrix, X represents the unknown matrix a, and B represents the result matrix.

Step 2: If the coefficient matrix A has an inverse (a matrix that "undoes" its effect), find the inverse of A and multiply both sides of the equation by it. This step gives you X = A^(-1) * B, where A^(-1) represents the inverse of A.

Step 3: If the coefficient matrix A does not have an inverse, you can use other methods like row operations or matrix factorization to solve the equation. These methods involve manipulating the equations and matrices to isolate the unknown matrix a.

Step 4: Once you have simplified the equation to X = [values], you have found the solution for matrix a in CC.

Julia Adams · Answer

#det A = (x+1)^3(x-3)#, so #det A != 0# if and only if #a != -1# and #a != 3#

#abs((1,1,1,-a),(1,1,-a,1),(1,-a,1,1),(-a,1,1,1))#

#= abs((-a,1,1,1),(1,-a,1,1),(1,1,-a,1),(1,1,1,-a))#

#= 1/(a-2) abs((-a,1,1,1),(1,-a,1,1),(1,1,-a,1),(a-2,a-2,a-2,2a-a^2(-a)))#

#= 1/(a-2) abs((-a,1,1,1),(1,-a,1,1),(1,1,-a,1),(0,0,0,3+2a-a^2))#

#= (3+2a-a^2)/(a-2) abs((-a,1,1),(1,-a,1),(1,1,-a))#

#= (3+2a-a^2)/((a-1)(a-2)) abs((-a,1,1),(1,-a,1),(a-1,a-1,a-a^2))#

#= (3+2a-a^2)/((a-1)(a-2)) abs((-a,1,1),(1,-a,1),(0,0,2+a-a^2))#

#= ((3+2a-a^2)(2+a-a^2))/((a-1)(a-2)) abs((-a,1),(1,-a))#

#= ((3+2a-a^2)(2+a-a^2)(a^2-1))/((a-1)(a-2))#

#= ((a^2-2a-3)(a^2-a-2)color(red)(cancel(color(black)((a-1))))(a+1))/(color(red)(cancel(color(black)((a-1))))(a-2))#

#= ((a-3)(a+1)color(red)(cancel(color(black)((a-2))))(a+1)(a+1))/color(red)(cancel(color(black)((a-2))))#

#= (a+1)^3(a-3)#

So:

#det A != 0# effectively means #a != -1# and #a != 3#