If we add the second and fourth equations, we get as before that 2x3 = 2, so we replace the third equation with x3= 1. If we subtract four times the second equation from the third equation and the first equation from the second equation, we get an equivalent system. If we subtract the second from the third equation and the first from the second equation, we get an equivalent system.
Using the first equation to eliminate this group from the second and third equations, we get the equivalent system. For each a∈R, find the rank of the matrix of coefficients and the total matrix of the system of equations.
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Since the coefficient matrix is contained in the common matrix, the best strategy is to discuss the common matrix first.
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Whena=−1, then the matrix of coefficients is of rank 1, while the total matrix is of rank 2. Whena={0,1,2}, then both the matrix of coefficients and the total matrix are of rank 3 , so the solution exists and is unique. When a = 0, then both the matrix of coefficients and the total matrix are of rank 1, so we have infinitely many solutions.
When a = 1, then both the coefficient matrix and the total matrix are of order 2, so we have infinitely many solutions. Ift= 9, then both the coefficient matrix and the total matrix are of rank 2, so we have infinitely many solutions.
2 Matrices
Of course, we can calculate the matrix product AB with the first calculator strip and then use the matrix product definition. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. on the ad to read more Click on the ad to read more.
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Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more. Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more. Since A is a square matrix, the matrix equation AX=A is equivalent to A(X−I) =0, thus AY=0 and X=Y+I. Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more.
On the other hand, this could not be expected because the matrix product is not commutative. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. per ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. If a= 1, then the matrix of coefficients is rank 2 and the entire matrix is rank 3, so there is no solution.
Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more. Ifa=−1, then the matrix of coecients is of rank 2 and the total matrix is of rank 3, and the set of solutions is empty. The condition of exactly one solution is that the matrix of coecients is of rank 3.
It follows that the matrix of coefficients of rank 3 is fora=−3, and therefore there exists exactly one solution, whena=−3 andb∈R. Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more. Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Get help now.
It follows that if fa=−1 then we have no solution because the total matrix is of rank 3 and the coefficient matrix is only of rank 2. Click an ad for more Click an ad for more Click an ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad. to read more Click on ad for more Click on ad for more.
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3 Determinants
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Find a lower triangular unit matrix L and an upper triangular matrix U, such that A = LU
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