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Row echelon form ➔ Identity matrix
𝐊₆ includes 6 elementary row operations that transform the matrix 𝐀 to row echelon form.
$$ 𝐀₆ = 𝐊₆𝐀 \\\ \begin{bmatrix} 1 & 2 & 3 \\\ 0 & 1 & 2 \\\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1/2 & 0 \\\ 1 & -1/2 & 0 \\\ -1 & 1/4 & 1/2 \end{bmatrix} \begin{bmatrix} 1 & 3 & 5 \\\ 2 & 4 & 6 \\\ 1 & 4 & 9 \end{bmatrix} $$𝐀₆ can further perform elementary row operations to make the elements at the top right of the main diagonal 0, i.e., reaching an identity matrix.
𝐀₆: $[^{\_{1\ 2\ 3}} \_{^{0\ 1\ 2} \_{0\ 0\ 1}}]$ $\overset{-2×r3+r2:\ [^{\_{1\ 0\ 0}} \_{^{0\ 1\ -2} \_{0\ 0\ 1}}]}{{⇢}}$ 𝐀₇: $[^{\_{1\ 2\ 3}} \_{^{0\ 1\ 0} \_{0\ 0\ 1}}]$ $\overset{-3×r3+r1:\ [^{\_{1\ 0\ -3}} \_{^{0\ 1\ 0} \_{0\ 0\ 1}}]}{{⇢}}$ 𝐀₈: $[^{\_{1\ 2\ 0}} \_{^{0\ 1\ 0} \_{0\ 0\ 1}}]$ $\overset{-2×r2+r1:\ [^{\_{1\ -2\ 0}} \_{^{0\ 1\ 0} \_{0\ 0\ 1}}]}{{⇢}}$ 𝐀₉: $[^{\_{1\ 0\ 0}} \_{^{0\ 1\ 0} \_{0\ 0\ 1}}]$
where,
$$ [^{\_{1\ -2\ 0}} \_{^{0\ 1\ 0} \_{0\ 0\ 1}}] [^{\_{1\ 0\ -3}} \_{^{0\ 1\ 0} \_{0\ 0\ 1}}] [^{\_{1\ 0\ 0}} \_{^{0\ 1\ -2} \_{0\ 0\ 1}}] 𝐊₆𝐀 = 𝐊₉𝐀 = 𝐈 \\\ 𝐊₉ = [^{\_{-3\ 7/4\ 1⁄2}} \_{^{3\ -1\ -1}\_{-1\ ¼\ 1⁄2}}] $$The above example indicates that there exist a matrix that can modify the matrix 𝐀 to 𝐈.
Let 𝐀⁻¹ denote the 𝐊₉, and 𝐀⁻¹ is called the inverse matrix of 𝐀.
Such that there is 𝐀⁻¹𝐀 = 𝐈. Since the matrix multiplication doesn’t has commutative property, what does 𝐀𝐀⁻¹ equal?
𝐀𝐀⁻¹ = [^{_{1\ 3\ 5} _{^{2\ 4\ 6}_{1\ 4\ 9}}}] [^{_{-3\ 7/4\ 1⁄2}} _{^{3\ -1\ -1}_{-1\ ¼\ 1⁄2}}] = 𝐈
In fact, 𝐀 and 𝐀⁻¹ are a pair of commutative matrices: 𝐀⁻¹𝐀 = 𝐀𝐀⁻¹ = 𝐈. Also, 𝐀 and 𝐀⁻¹ are the inverse matrix of each other: (𝐀⁻¹)⁻¹ = 𝐀
Analogy to reciprocal
Given two numbers a, b, if ab = ba = 1, then b = a⁻¹, a≠0.
And the identity matrix 𝐈 has the same effect as the number 1.
Therefore, given two matrices 𝐀, 𝐁, if 𝐀𝐁 = 𝐁𝐀 = 𝐈, then 𝐁 = 𝐀⁻¹
No all matrix has its inverse matrix, which is like the number 0 doesn’t has its reciprocal.
So what kind of matrix has an inverse? And is the inverse unique? How to solve the inverse matrix?
Invertible matrix
An invertible matrix can perform multiple elementary row operations to become an identity matrix.
Identity matrix is a square matrix (#row = #colmuns) and its rank = #rows (= #cols). And the elementary row operations do not change the size and rank note4.
Therefore, an invertible matrix must be an square matrix and its rank = #rows.
Based on the commutative property that 𝐀⁻¹ and 𝐀 are the inverse matrices of each other, the 𝐀⁻¹ must also be a square matrix with the same rank of 𝐈.
可逆的2个条件：1-方阵， 2-秩=列(行)数
Since elementary row operation is that 𝐀 is multiplied by elementary matrix on the left. And the product of the series of elementary row operations that transformed 𝐀 to 𝐈 is represented as 𝐀⁻¹.
Therefore, an invertible matrix and its inverse are both able to represented as the multiplication of multiple elementary matrix.
Since the product of a series of elementary matrix must be invertible, a special case is that there is only a single elementary matrix. So, any of the elementary matrix is invertible.
Inverse is unique
Let 𝐀𝐁 = 𝐁𝐀 = 𝐈 and 𝐀𝐂 = 𝐂𝐀 = 𝐈, then 𝐁 = 𝐁𝐈 = 𝐁(𝐀𝐂) = (𝐁𝐀)𝐂 = 𝐈𝐂 = 𝐂