Linear Algebra

Running head: LINEAR ALGEBRA 1
Linear Algebra
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LINEAR ALGEBRA 2
Question 1
A=
[ 1 0 0 3 2 4
0 2 1 2 1 2
0 0 1 1 2 1
1 0 0 4 3 1
0 0 0 0 2 1
0 0 0 0 0 7 ]First reduce the given matrix to row echelon form using matrix transformation
To eliminate the leading coefficient of the forth row, subtract row 1 from row 4 and
replace row 4 with the difference
R4 R4 – R1
A=
[1 0 0 3 2 4
0 2 1 2 1 2
0 0 1 1 2 1
0 0 0 1 1 −3
0 0 0 0 2 1
0 0 0 0 0 7
]The determinant of the matrix is the diagonal product of its row echelon form
Determinant = 1 ×2 ×1× 1× 2× 7=28
Determinant of A = 28
Question 2
The characteristic polynomial of B is defined as Det (B-λI) where I is the identity
matrix of the same dimension as B.
In this case, I =
[1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1 ]
λI =
[ λ 0 0 0
0 λ 0 0
0 0 λ 0
0 0 0 λ ]

LINEAR ALGEBRA 3
B-λI =
[2 5 1 1
1 4 2 2
0 0 6 −5
0 0 2 3 ]−
[ λ 0 0 0
0 λ 0 0
0 0 λ 0
0 0 0 λ ]=
[2−λ 5 1 1
1 4− λ 2 2
0 0 6−λ −5
0 0 2 3−λ ]
To get the determinant, first convert the matrix to its row echelon form
To eliminate the leading coefficient in row 2, multiply R1 by 1
2−λ and subtract the
product from R2. Replace R2 with the difference
R2 R 2− 1
2−λ ∙ R 1
[ 2−λ 5 1 1
0 λ2−6 λ +3
2−λ
−2 λ+3
2−λ
−2 λ+3
2−λ
0 0 6−λ −5
0 0 2 3−λ ]
To eliminate the leading coefficient in row 4, multiply R3 by 2
6− λ and subtract the
product from R4. Replace R4 with the difference
R4R 4− 2
6−λ ∙ R 3
[2−λ 5 1 1
0 λ2−6 λ +3
2−λ
−2 λ+3
2−λ
−2 λ+3
2−λ
0 0 6−λ −5
0 0 0 λ2−9 λ+28
6− λ
]The determinant of the matrix is the diagonal product of its row echelon form
Determinant = 2− λ × λ2−6 λ +3
2−λ × 6− λ× λ2−9 λ +28
6− λ
¿ ( λ2−6 λ+3 ) ( λ2−9 λ+28 )
¿ λ4−15 λ3+ 85 λ2−195 λ+ 84