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These are the notes of Solved Exam of Linear Algebra which includes General Solution, Linear Systems, Homogeneous System, Solution Sets, Particular Solution, Nonhomogeneous, Coefficient Matrix etc. Key important points are: Given Matrix, Properties, Determinants, Determinant, Inverse, Upper Triangular Matrix, Operations, Matrix Is Invertible, Linear System, Linear Transformations
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Find the determinant of the matrix below. Specify whether the matrix has an inverse without trying to
compute the inverse.
Row reduce the given matrix to an upper triangular matrix and keep track of the operations you perform:
1 · R 1 + R 2 ↔ R 2 , 1 · R 1 + R 3 ↔ R 3 , −
1 2
· R 1 + R 4 ↔ R 4. Then perform 2 · R 2 + R 4 ↔ R 4 and R 2 ↔ R 3.
Note that there is only one switch. As a result, the determinant is -32 which is not 0. Therefore, the given
matrix is invertible.
Solve the linear system using Cramer’s Rule:
2 x 1 + 3 x 2 − x 3 = 2
3 x 1 − 2 x 2 + x 3 = − 1
− 5 x 1 − 4 x 2 + 2 x 3 = 3
x 1 =
x 2 =
x 3 =
The solution is ( − 5 / 11 , 36 / 11 , 76 / 11).
Let v ⃗ 1 = (1 , 1 , 1), v ⃗ 2 = (1 , 2 , 3) and v⃗ (^) 3 = (1 , 1 , 2).
a. Show that the vectors above are linearly independent.
There are three pivot positions and three vectors. Therefore, v⃗ (^) 1 ,⃗v (^) 2 v,⃗ (^) 3 are linearly independent.
b. Find the unique scalars(weights) c 2 , c 2 , c 3 such that v⃗ = (2 , 1 , 3) can be written as v ⃗ = cv 1 ⃗ 2 + cv 2 ⃗ 2 + cv 3 ⃗ 3.
Thus, we get c 1 = 0, c 2 = − 1 , and c 3 = 3 or v⃗ = 0 ·v⃗ (^) 1 + ( − 1) v ⃗ 2 + 3 v⃗ (^) 3.
Define a linear transformation T : P 2 Ï P 2 by T (p( x )) = p ′ ( x ).
a. Describe the range of T
Consider the vector space P 2 with the standard basis { 1 , x, x 2 }. Then, the transformation T is given as
T ( ax 2
Consequently,
Range of T = {s + tx : s, t ∈ R }.
b. Find dim ( R ( T )).
Since the basis C has 2 elements, dim ( R ( T )) = 2.
c. Find dim ( N ( T )).
By the RANK THEOREM, we have dim ( N ( T )) + dim ( R ( T )) = dim P 2 , i.e., dim ( N ( T )) = 3 − 2 = 1.