Use Cramer’s rule to solve the linear system
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and also solve the system using the inverse
A-1 of the coefficient matrix A.
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Use Cramer’s rule to solve the linear system

and verify your answer by substituting it into
the left side of the linear system.


OR use a cofactor expansion down
column 1:

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For A1 use a cofactor expansion
along row 3:

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For A2 use a cofactor expansion
down column 1:

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For A3 use a cofactor expansion
along row 3:

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Verification of this solution:
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Find the adjugate [the transpose of the matrix of cofactors]
and hence the inverse
(A-1) of the matrix
![A = [ 1 3 2 ; 2 -2 5 ; 3 1 -4 ]](a6w09/q3a.gif)
and use this inverse matrix to verify your
answer to question (2) above.
The sub-matrix Aij is what
remains of matrix A after the deletion
of row i and column j.
The cofactors are
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![adjugate A =
[ 3 14 19 ]
[ 23 -10 -1 ]
[ 8 8 -8 ]](a6w09/q3adj.gif)
From question 2, det A = 88


![X = [ 1 2 1 ]T](a6w09/q3sol2.gif)
Find the eigenvalues and corresponding set of basic
eigenvectors for
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Write down the matrix P that diagonalizes
A
and verify by matrix
multiplication that
P -1AP = D.
Hence find A 43.
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l = –1:

The (–1)-eigenvector is therefore any non-zero multiple of
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l = +1:

The (1)-eigenvector is therefore any non-zero multiple of
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![{ [ 1 1 ], [ 1 2 ] }, leading to P](a6w09/q4basis.gif)
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Find the eigenvalues and corresponding set of basic
eigenvectors for
![A = [ -1 2 3 ; 0 1 0 ; 0 0 2 ]](a6w09/q5a.gif)
Write down the matrix P that diagonalizes
A .
The matrix A is upper triangular
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All eigenvalues have multiplicity 1, which guarantees the
existence of a matrix P that diagonalizes
A

l = –1:

The (–1)-eigenvector is therefore any non-zero multiple of
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l = +1:

The (1)-eigenvector is therefore any non-zero multiple of
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l = +2:

The (2)-eigenvector is therefore any non-zero multiple of
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Therefore the matrix P that diagonalizes A is
![P =
[ 1 1 1 ]
[ 0 1 0 ]
[ 0 0 1 ]](a6w09/q5p.gif)
and the set of basic eigenvectors is just the three columns of P .