Invertible Matrix Theorem: Detailed Analysis and Application Examples

This assignment provides a detailed analysis of the Invertible Matrix Theorem, covering key aspects such as pivot positions, linear independence of columns, eigenvalues, and determinants. The solution includes step-by-step calculations to verify the theorem's statements using a specific matrix example. The assignment demonstrates how to compute the row echelon form, analyze linear independence, and determine eigenvalues to confirm the theorem's validity. References to relevant research papers are also included, offering a comprehensive understanding of the topic. This resource, available on Desklib, helps students grasp the fundamental concepts of linear algebra and the applications of the Invertible Matrix Theorem.

Running head: THE INVERTIBLE MATRIX THEOREM 1
The Invertible Matrix Theorem
Name
Institution

THE INVERTIBLE MATRIX THEOREM 2
The Invertible Matrix Theorem
Let then A be a 3 ×3 ¿ where n=3. A=
[ 1 −3 0
−4 11 1
0 7 3 ]
We consider four statements from the invertible Matrix Theorem as shown.
The statements extracted from the Invertible Matrix Theorem include: “Matrix A has n pivot
positions, the columns of the matrix form a linearly independent set, zero is not an eigenvalue of
matrix A, and the determinant of A, det ( A)≠0 .”
Part a
“Matrix A has n pivot positions”
The row echelon form of the Matrix A can be computed as:
R1: “add 4 times the 1st row to the 2nd row” to obtain [1 −3 0
0 −1 1
2 7 3 ]
R2: “add 2 times the 1st row to the 3rd row” to obtain [1 −3 0
0 −1 1
0 13 3 ]
R3: “multiply the 2nd row by -1” to obtain [1 −3 0
0 1 1
0 13 3 ]
R4: add -13 times the 2nd row to 3rd row to obtain [1 −3 0
0 1 −1
0 0 16 ]

THE INVERTIBLE MATRIX THEOREM 3
R5: multiply the third row by 1/16 to obtain [1 −3 0
0 1 −1
0 0 1 ]
Therefore, 3(=n) pivot positions and the statement uphold.
Part b
“The columns of the matrix form a linearly independent set.”
Here, we write the matrix A into a homogenous system of equations
1 c1−4 c2+2 c3=0
−3 c1 +11 c2 +7 c3=0
0 c1 +1 c2 +3 c3=0
Then, transforming the coefficient matrix into the “reduced row echelon form” to obtain
[1 0 0
0 1 0
0 0 1 ], which corresponds to 1 c1=1 c2 =1c3=0. Evidently, each column contains a leading
entry and the matrix has the trivial solution (0,0,0). Therefore, the set is linearly independent and
the statement upholds.
Part c
“Zero is not an eigenvalue of matrix A”
|A|= [ 1 −3 0
−4 11 1
0 7 3 ]

THE INVERTIBLE MATRIX THEOREM 4
The singular matrix, A−λI = [1−λ −3 0
−4 11−λ 1
0 7 3−λ ]=−λ3 +15 λ2−28 λ−16=0
The corresponding eigen values are ( 12.695
2.7613
−0.4564 ) which are non-zero.
Part d
“The determinant of A, det ( A)≠0”
det |A |=
[ 1 −3 0
−4 11 1
0 7 3 ]=1 ( 33−7 ) + 3 ( −12−2 ) +0=−16(≠ 0)
Hence, the four statements (part a through d) obtained from the Invertible Matrix Theorem are all
true.

THE INVERTIBLE MATRIX THEOREM 5
References
Rodman, L. X., & Spitkovsky, I. M. (2005). Analytic roots of invertible matrix
functions. Electronic Journal of Linear Algebra, 13(1). doi:10.13001/1081-3810.1161
Wawro, M. (2015). Reasoning About Solutions in Linear Algebra: the Case of Abraham and the
Invertible Matrix Theorem. International Journal of Research in Undergraduate
Mathematics Education, 1(3), 315-338. doi:10.1007/s40753-015-0017-7