Inverse of a matrix problems show up a lot when you need to solve a system of equations quickly. But what happens when the matrix size grows, and recalculating everything from scratch every single time becomes annoying? That’s exactly where the sequential bordering idea helps: we add one row and one column to the matrix, and then update the inverse using a clear formula. Sounds practical, right?
Inverse of a Matrix via Bordering: How We Set Up the Blocks
Let’s start with a simple idea. Suppose we have a square matrix of order \( n \):
\[ A_n = \begin{pmatrix} a_{11} & a_{12} & a_{13} & \dots & a_{1n} \\ a_{21} & a_{22} & a_{23} & \dots & a_{2n} \\ a_{31} & a_{32} & a_{33} & \dots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & a_{n3} & \dots & a_{nn} \end{pmatrix}. \]
We will view it as the result of bordering a matrix of order \( n-1 \). In other words, we write \( A_n \) in block form:
\[ A_n= \begin{pmatrix} A_{n-1} & u_n\\ v_n & a_{nn} \end{pmatrix}, \]
where:
- \( A_{n-1} \) is an \( (n-1)\times(n-1) \) matrix.
- \( u_n \) is an \( (n-1)\times 1 \) column.
- \( v_n \) is a \( 1\times(n-1) \) row.
- \( a_{nn} \) is a scalar (the bottom-right entry).
It’s also helpful to write explicitly what \( u_n \) and \( v_n \) are:
\[ u_n= \begin{pmatrix} a_{1n}\\ a_{2n}\\ a_{3n}\\ \vdots\\ a_{n-1,n} \end{pmatrix}, \qquad v_n= \begin{pmatrix} a_{n1} & a_{n2} & a_{n3} & \dots & a_{n,n-1} \end{pmatrix}. \]
Now the key question is: if we already know \( A_{n-1}^{-1} \), can we quickly obtain \( A_n^{-1} \)? Yes—and that’s exactly why sequential bordering is so convenient in numerical computations.
Sequential Bordering and the Inverse of a Matrix: What Form Are We Looking For?
Next, we proceed symmetrically. If we wrote \( A_n \) as a bordered matrix, then we will also look for the inverse in the same block form:
\[ A_n^{-1}= \begin{pmatrix} P_{n-1} & r_n\\ q_n & \beta_n \end{pmatrix}, \]
where:
- \( P_{n-1} \) is an \( (n-1)\times(n-1) \) matrix.
- \( r_n \) is an \( (n-1)\times 1 \) column.
- \( q_n \) is a \( 1\times(n-1) \) row.
- \( \beta_n \) is a number (a scalar).
Because we are computing the inverse of a matrix, we have the standard condition:
\[ A_n\cdot A_n^{-1}=E_n, \]
where \( E_n \) is the identity matrix of order \( n \). Now multiply the block matrices:
\[ \begin{pmatrix} A_{n-1} & u_n\\ v_n & a_{nn} \end{pmatrix} \cdot \begin{pmatrix} P_{n-1} & r_n\\ q_n & \beta_n \end{pmatrix} = \begin{pmatrix} A_{n-1} \cdot P_{n-1} + u_n \cdot q_n & A_{n-1} \cdot r_n + u_n \cdot \beta_n\\ v_n \cdot P_{n-1} + a_{nn} \cdot q_n & v_n \cdot r_n + a_{nn} \cdot \beta_n \end{pmatrix} = \begin{pmatrix} E_{n-1} & 0\\ 0 & 1 \end{pmatrix}. \]
So we obtain a system of block equalities:
\[ \begin{cases} A_{n-1} \cdot P_{n-1} + u_n \cdot q_n = E_{n-1},\\ A_{n-1} \cdot r_n + u_n \cdot \beta_n = 0,\\ v_n \cdot P_{n-1} + a_{nn} \cdot q_n = 0,\\ v_n \cdot r_n + a_{nn} \cdot \beta_n = 1. \end{cases} \]
An Important Point for Using the Method
Here we rely on two conditions:
- The matrix \( A_{n-1} \) must be invertible, meaning \( A_{n-1}^{-1} \) must exist.
- During the process, a number \( \alpha_n \) will appear, and we will need \( \alpha_n \neq 0 \).
If at least one of these conditions fails, the formulas below cannot be applied directly. In that case, you typically change the bordering order (for example, by swapping rows/columns) or switch to another method for finding the inverse.
The Key Scalar and Update Formulas: How We Compute the Inverse of a Matrix
Let’s start with the second equation:
\[ A_{n-1} \cdot r_n + u_n \cdot \beta_n=0. \]
Multiply on the left by \( A_{n-1}^{-1} \) (and remember: it must exist):
\[ r_n = -A_{n-1}^{-1} \cdot u_n \cdot \beta_n. \]
Now substitute this into the fourth equation:
\[ v_n \cdot r_n + a_{nn} \cdot \beta_n=1. \]
We get:
\[ v_n \cdot \left(-A_{n-1}^{-1} \cdot u_n \cdot \beta_n\right) + a_{nn} \cdot \beta_n = 1, \]
factor out \( \beta_n \):
\[ \left(a_{nn} – v_n \cdot A_{n-1}^{-1} \cdot u_n\right) \cdot \beta_n = 1. \]
Now introduce a very important notation:
\[ \alpha_n = a_{nn} – v_n \cdot A_{n-1}^{-1} \cdot u_n. \]
You can think of \( \alpha_n \) as an “adjusted” version of \( a_{nn} \) that takes into account the effect of the added row and column through \( A_{n-1}^{-1} \).
Тоді:
\[ \beta_n = \frac{1}{\alpha_n}. \]
And here the second key condition becomes obvious: we need \( \alpha_n\neq 0 \). Otherwise \( \beta_n \) is undefined, and the formula for \( A_n^{-1} \) does not work.
Now return to \( r_n \):
\[ r_n = -A_{n-1}^{-1} \cdot u_n \cdot \beta_n = -\frac{A_{n-1}^{-1} \cdot u_n}{\alpha_n}. \]
Before moving on, let’s note the dimensions (this is a great way to avoid mistakes):
- \( A_{n-1}^{-1} \cdot u_n \) has size \( (n-1)\times 1 \).
- \( v_n \cdot A_{n-1}^{-1} \) has size \( 1\times(n-1) \).
- \( v_n \cdot A_{n-1}^{-1} \cdot u_n \) is a scalar (a single number).
Next, work with the third equation:
\[ v_n \cdot P_{n-1} + a_{nn} \cdot q_n=0. \]
From the first equation, express \( P_{n-1} \):
\[ A_{n-1} \cdot P_{n-1} = E_{n-1} – u_n \cdot q_n \quad\Rightarrow\quad P_{n-1} = A_{n-1}^{-1} – A_{n-1}^{-1} \cdot u_n \cdot q_n. \]
Substitute this into \( v_n \cdot P_{n-1} + a_{nn} \cdot q_n=0 \):
\[ v_n \cdot \left(A_{n-1}^{-1} – A_{n-1}^{-1} \cdot u_n \cdot q_n\right) + a_{nn} \cdot q_n = 0, \]
expand the parentheses:
\[ v_n \cdot A_{n-1}^{-1} – v_n \cdot A_{n-1}^{-1} \cdot u_n \cdot q_n + a_{nn} \cdot q_n=0. \]
Group the terms containing \( q_n \):
\[ v_n \cdot A_{n-1}^{-1} + \left(a_{nn} – v_n \cdot A_{n-1}^{-1} \cdot u_n\right) \cdot q_n = 0. \]
But the expression in parentheses is exactly \( \alpha_n \). So:
\[ v_n \cdot A_{n-1}^{-1} + \alpha_n \cdot q_n=0 \quad\Rightarrow\quad q_n = -\frac{v_n \cdot A_{n-1}^{-1}}{\alpha_n}. \]
Finally, return to the formula for \( P_{n-1} \):
\[ P_{n-1} = A_{n-1}^{-1} – A_{n-1}^{-1} \cdot u_n \cdot q_n. \]
Substitute \( q_n \):
\[ P_{n-1} = A_{n-1}^{-1} – A_{n-1}^{-1} \cdot u_n \cdot \left(-\frac{v_n \cdot A_{n-1}^{-1}}{\alpha_n}\right) = A_{n-1}^{-1} + \frac{A_{n-1}^{-1} \cdot u_n \cdot v_n \cdot A_{n-1}^{-1}}{\alpha_n}. \]
Now we can assemble everything into one formula. This is the update for the inverse of a matrix after bordering:
\[ A_n^{-1}= \begin{pmatrix} A_{n-1}^{-1} + \dfrac{A_{n-1}^{-1} \cdot u_n \cdot v_n \cdot A_{n-1}^{-1}}{\alpha_n} & -\dfrac{A_{n-1}^{-1} \cdot u_n}{\alpha_n} \\[10pt] -\dfrac{v_n \cdot A_{n-1}^{-1}}{\alpha_n} & \dfrac{1}{\alpha_n} \end{pmatrix}, \qquad \alpha_n = a_{nn} – v_n \cdot A_{n-1}^{-1} \cdot u_n. \]
One more useful connection: this is the same mechanism as block matrix inversion, just in the special case where one of the blocks has size \( 1\times 1 \). That’s why many ideas from “inverting block matrices” carry over here almost directly.
And that’s the core of the method: instead of fully inverting an \( n \times n \) matrix, we do a few multiplications and one division by \( \alpha_n \). Convenient? Absolutely.
Inverse of a Matrix in Practice: Testing the Method on Problems
Now let’s move to the practical part. Here it’s important not only to get the final answer, but also to see how sequential bordering builds the inverse step by step. And there’s one more big advantage: you will quickly notice the exact moment when the inverse of a matrix simply does not exist.
Example 1. What steps do we follow to find the inverse matrix by the bordering method?
- Start from the smallest case. For \( A_1 = \begin{pmatrix}a_{11}\end{pmatrix} \) we compute \( A_1^{-1}=\begin{pmatrix}a_{11}^{-1}\end{pmatrix} \).
- Build the sequence \( A_2, A_3, \dots, A_n \), where each new matrix is obtained by bordering the previous one. At the \( k \)-th step we write
\[ A_k= \begin{pmatrix} A_{k-1} & u_k\\ v_k & a_{kk} \end{pmatrix}. \]
- If \( A_{k-1}^{-1} \) is already known, compute the key scalar \( \alpha_k = a_{kk} – v_k \cdot A_{k-1}^{-1} \cdot u_k \).
- Check the applicability condition: we must have \( \alpha_k\neq 0 \). If \( \alpha_k=0 \), the bordering formula at this step does not directly give an inverse.
- Compute \( \beta_k=\frac{1}{\alpha_k} \), and then the vectors
\[ r_k = -A_{k-1}^{-1} \cdot u_k \cdot \beta_k, \qquad q_k = -v_k \cdot A_{k-1}^{-1} \cdot \beta_k. \]
- Update the main block \( P_{k-1} = A_{k-1}^{-1} + A_{k-1}^{-1} \cdot u_k \cdot v_k \cdot A_{k-1}^{-1} \cdot \beta_k \) and assemble the result:
\[ A_k^{-1}= \begin{pmatrix} P_{k-1} & r_k\\ q_k & \beta_k \end{pmatrix}. \]
Example 2. Find the inverse matrix
\[ A= \begin{pmatrix} 2 & -1 & 4 \\ 4 & -2 & 8 \\ 6 & -3 & 12 \end{pmatrix}. \]
Start with the first step:
\[ A_1=\begin{pmatrix}2\end{pmatrix}, \qquad A_1^{-1}=\begin{pmatrix}0.5\end{pmatrix}. \]
Move on to
\[ A_2= \begin{pmatrix} 2 & -1\\ 4 & -2 \end{pmatrix} = \begin{pmatrix} A_1 & u_2\\ v_2 & a_{22} \end{pmatrix}, \qquad u_2=\begin{pmatrix}-1\end{pmatrix}, \quad v_2=\begin{pmatrix}4\end{pmatrix}, \quad a_{22}=-2. \]
Now compute \( \alpha_2 = a_{22} – v_2 \cdot A_1^{-1} \cdot u_2 \). We have
\[ v_2 \cdot A_1^{-1} \cdot u_2= \begin{pmatrix}4\end{pmatrix} \cdot \begin{pmatrix}0.5\end{pmatrix} \cdot \begin{pmatrix}-1\end{pmatrix} =-2, \]
so \( \alpha_2=-2-(-2)=0 \).
Since \( \alpha_2=0 \), the number \( \beta_2 \) is not defined, which means \( A_2^{-1} \) does not exist via the bordering formula. This is not a coincidence: in the original matrix, the second row is double the first, and the third is triple. So the rows are linearly dependent, and the inverse of a matrix for \( A \) does not exist. If in a similar problem the dependence is not so obvious, you usually try a different bordering order—for example, by swapping rows or columns.
Example 3. Find the inverse matrix
\[ A= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 1 & 2 & 3 \\ 3 & 2 & -1 & 2 \\ 4 & 3 & 2 & 1 \end{pmatrix}. \]
Start with the first step:
\[ A_1=\begin{pmatrix}1\end{pmatrix}, \qquad A_1^{-1}=\begin{pmatrix}1\end{pmatrix}. \]
Next, build
\[ A_2= \begin{pmatrix} 1 & 2\\ 2 & 1 \end{pmatrix} = \begin{pmatrix} A_1 & u_2\\ v_2 & a_{22} \end{pmatrix}, \qquad u_2=\begin{pmatrix}2\end{pmatrix}, \quad v_2=\begin{pmatrix}2\end{pmatrix}, \quad a_{22}=1. \]
Compute
\[ \alpha_2 = a_{22} – v_2 \cdot A_1^{-1} \cdot u_2 = 1- \begin{pmatrix}2\end{pmatrix} \cdot \begin{pmatrix}1\end{pmatrix} \cdot \begin{pmatrix}2\end{pmatrix} = -3, \qquad \beta_2=-0.333. \]
Now
\[ \begin{gathered} r_2 = -A_1^{-1}\cdot u_2\cdot \beta_2 = -\begin{pmatrix}1\end{pmatrix}\cdot\begin{pmatrix}2\end{pmatrix}\cdot(-0.333) = \begin{pmatrix}0.667\end{pmatrix}, \\[4pt] q_2 = -v_2\cdot A_1^{-1}\cdot \beta_2 = \begin{pmatrix}0.667\end{pmatrix}. \end{gathered} \]
Update the main block \( P_1 = A_1^{-1} + A_1^{-1} \cdot u_2 \cdot v_2 \cdot A_1^{-1} \cdot \beta_2 \). We get
\[ A_1^{-1} \cdot u_2 \cdot v_2 \cdot A_1^{-1} \cdot \beta_2 = \begin{pmatrix}1\end{pmatrix} \cdot \begin{pmatrix}2\end{pmatrix} \cdot \begin{pmatrix}2\end{pmatrix} \cdot \begin{pmatrix}1\end{pmatrix} \cdot (-0.333) = \begin{pmatrix}-1.333\end{pmatrix}, \]
so \( P_1 = \begin{pmatrix}1\end{pmatrix} + \begin{pmatrix}-1.333\end{pmatrix}=\begin{pmatrix}-0.333\end{pmatrix} \). Therefore,
\[ A_2^{-1}= \begin{pmatrix} -0.333 & 0.667\\ 0.667 & -0.333 \end{pmatrix}. \]
Move on to
\[ A_3= \begin{pmatrix} 1 & 2 & 3\\ 2 & 1 & 2\\ 3 & 2 & -1 \end{pmatrix} = \begin{pmatrix} A_2 & u_3\\ v_3 & a_{33} \end{pmatrix}, \qquad u_3=\begin{pmatrix}3 \\ 2\end{pmatrix},\quad v_3=\begin{pmatrix}3 & 2\end{pmatrix},\quad a_{33}=-1. \]
First compute the vectors needed for the formula, namely \( A_2^{-1} \cdot u_3 \) and \( v_3 \cdot A_2^{-1} \):
\[ \begin{gathered} A_2^{-1} \cdot u_3= \begin{pmatrix} -0.333 & 0.667\\ 0.667 & -0.333 \end{pmatrix} \cdot \begin{pmatrix} 3\\ 2 \end{pmatrix} = \begin{pmatrix} 0.333\\ 1.333 \end{pmatrix}, \\[4pt] v_3 \cdot A_2^{-1}= \begin{pmatrix} 3 & 2 \end{pmatrix} \cdot \begin{pmatrix} -0.333 & 0.667\\ 0.667 & -0.333 \end{pmatrix} = \begin{pmatrix} 0.333 & 1.333 \end{pmatrix}. \end{gathered} \]
Now
\[ \alpha_3 = a_{33} – v_3 \cdot A_2^{-1} \cdot u_3 = -1-\begin{pmatrix}3 & 2\end{pmatrix} \cdot \begin{pmatrix}0.333 \\ 1.333\end{pmatrix} =-4.667, \qquad \beta_3=-0.214. \]
Then
\[ \begin{gathered} r_3 = -(A_2^{-1}\cdot u_3)\cdot \beta_3 = -\begin{pmatrix}0.333\\1.333\end{pmatrix}\cdot(-0.214) = \begin{pmatrix}0.071\\0.286\end{pmatrix}, \\[4pt] q_3 = -(v_3\cdot A_2^{-1})\cdot \beta_3 = -\begin{pmatrix}0.333 & 1.333\end{pmatrix}\cdot(-0.214) = \begin{pmatrix}0.071 & 0.286\end{pmatrix}. \end{gathered} \]
Now update the main block \( P_2 = A_2^{-1} + (A_2^{-1} \cdot u_3) \cdot (v_3 \cdot A_2^{-1}) \cdot \beta_3 \). First form the outer product:
\[ (A_2^{-1} \cdot u_3) \cdot (v_3 \cdot A_2^{-1}) = \begin{pmatrix}0.333 \ 1.333\end{pmatrix} \cdot \begin{pmatrix}0.333 & 1.333\end{pmatrix} = \begin{pmatrix} 0.111 & 0.444\\ 0.444 & 1.778 \end{pmatrix}, \]
multiply by \( \beta_3 = -0.214 \):
\[ \begin{pmatrix} 0.111 & 0.444\\ 0.444 & 1.778 \end{pmatrix} \cdot (-0.214) = \begin{pmatrix} -0.024 & -0.095\\ -0.095 & -0.381 \end{pmatrix}, \]
and add to \( A_2^{-1} \):
\[ P_2= \begin{pmatrix} -0.333 & 0.667\\ 0.667 & -0.333 \end{pmatrix} + \begin{pmatrix} -0.024 & -0.095\\ -0.095 & -0.381 \end{pmatrix} = \begin{pmatrix} -0.357 & 0.571\\ 0.571 & -0.714 \end{pmatrix}. \]
So
\[ A_3^{-1}= \begin{pmatrix} -0.357 & 0.571 & 0.071 \\ 0.571 & -0.714 & 0.286 \\ 0.071 & 0.286 & -0.214 \end{pmatrix}. \]
Now the final step: build the full matrix \( A_4 = A \) as a bordered matrix:
\[ A_4= \begin{pmatrix} A_3 & u_4\\ v_4 & a_{44} \end{pmatrix}, \qquad u_4=\begin{pmatrix}4 \\ 3 \\ 2\end{pmatrix},\quad v_4=\begin{pmatrix}4 & 3 & 2\end{pmatrix},\quad a_{44}=1. \]
Again, first compute the vectors \( A_3^{-1} \cdot u_4 \) and \( v_4 \cdot A_3^{-1} \):
\[ \begin{gathered} A_3^{-1} \cdot u_4= \begin{pmatrix} -0.357 & 0.571 & 0.071 \\ 0.571 & -0.714 & 0.286 \\ 0.071 & 0.286 & -0.214 \end{pmatrix} \cdot \begin{pmatrix} 4\\ 3\\ 2 \end{pmatrix} = \begin{pmatrix} 0.429\\ 0.714\\ 0.714 \end{pmatrix}, \\[6pt] v_4 \cdot A_3^{-1}= \begin{pmatrix} 4 & 3 & 2 \end{pmatrix} \cdot \begin{pmatrix} -0.357 & 0.571 & 0.071 \\ 0.571 & -0.714 & 0.286 \\ 0.071 & 0.286 & -0.214 \end{pmatrix} = \begin{pmatrix} 0.429 & 0.714 & 0.714 \end{pmatrix}. \end{gathered} \]
Now
\[ \alpha_4 = a_{44} – v_4 \cdot A_3^{-1} \cdot u_4 = 1-\begin{pmatrix}4 & 3 & 2\end{pmatrix} \cdot \begin{pmatrix}0.429 \\ 0.714 \\ 0.714\end{pmatrix} = 1-(1.716+2.142+1.428) =-4.286, \qquad \beta_4=-0.233. \]
From this,
\[ \begin{gathered} r_4 = -(A_3^{-1}\cdot u_4)\cdot \beta_4 = -\begin{pmatrix}0.429\\0.714\\0.714\end{pmatrix}\cdot(-0.233) = \begin{pmatrix}0.1\\0.167\\0.167\end{pmatrix}, \\[4pt] q_4 = -(v_4\cdot A_3^{-1})\cdot \beta_4 = -\begin{pmatrix}0.429 & 0.714 & 0.714\end{pmatrix}\cdot(-0.233) = \begin{pmatrix}0.1 & 0.167 & 0.167\end{pmatrix}. \end{gathered} \]
Now update the main block \( P_3 = A_3^{-1} + (A_3^{-1} \cdot u_4) \cdot (v_4 \cdot A_3^{-1}) \cdot \beta_4 \). First form the outer product:
\[ (A_3^{-1} \cdot u_4) \cdot (v_4 \cdot A_3^{-1}) = \begin{pmatrix}0.429 \ 0.714 \ 0.714\end{pmatrix} \cdot \begin{pmatrix}0.429 & 0.714 & 0.714\end{pmatrix} = \begin{pmatrix} 0.184 & 0.306 & 0.306\\ 0.306 & 0.51 & 0.51\\ 0.306 & 0.51 & 0.51 \end{pmatrix}, \]
multiply by \( \beta_4=-0.233 \):
\[ \begin{pmatrix} 0.184 & 0.306 & 0.306\\ 0.306 & 0.51 & 0.51\\ 0.306 & 0.51 & 0.51 \end{pmatrix} \cdot (-0.233) = \begin{pmatrix} -0.043 & -0.071 & -0.071\\ -0.071 & -0.119 & -0.119\\ -0.071 & -0.119 & -0.119 \end{pmatrix}, \]
and add to \( A_3^{-1} \):
\[ P_3= \begin{pmatrix} -0.357 & 0.571 & 0.071 \\ 0.571 & -0.714 & 0.286 \\ 0.071 & 0.286 & -0.214 \end{pmatrix} + \begin{pmatrix} -0.043 & -0.071 & -0.071\\ -0.071 & -0.119 & -0.119\\ -0.071 & -0.119 & -0.119 \end{pmatrix} = \begin{pmatrix} -0.4 & 0.5 & 0 \\ 0.5 & -0.833 & 0.167 \\ 0 & 0.167 & -0.333 \end{pmatrix}. \]
So the final result is:
\[ A^{-1}= \begin{pmatrix} -0.4 & 0.5 & 0 & 0.1 \\ 0.5 & -0.833 & 0.167 & 0.167 \\ 0 & 0.167 & -0.333 & 0.167 \\ 0.1 & 0.167 & 0.167 & -0.233 \end{pmatrix}. \]
Where to Go Next: Topics That Add New Tools
You’ve already seen how the bordering method makes computations more manageable—especially when the matrix gets large. So it’s natural to ask: what should you study next to feel more confident with different types of problems and with the inverse of a matrix in general? Here are three topics that continue this material in a very natural way:
- Inverse of a Matrix via Cofactors: The Logic of Minors and Determinants — We’ll go through the classic approach using minors and cofactors, so you can clearly see how the final answer is built step by step.
- Inverse of a Matrix Using the Coefficients of the Characteristic Polynomial: A Different Route to Inversion Without the Standard Schemes — We’ll explain how to reach an inverse through its connection to the characteristic polynomial, and what this method looks like in real calculations.
- Pseudoinverse of a Matrix: How to Work with Rectangular and Singular Cases — We’ll discuss what to do when a standard inverse is not defined, and how to find a pseudoinverse in typical examples.
Inverse of a Matrix in Code: Try Building Your Own Mini Calculator
Now imagine that the flowchart below is not just a picture, but a clear plan for a small programming project. Why not use it as a guide and write a compact program in your favorite language that computes the inverse of a matrix using the bordering method?
It’s a great way to see how the formulas turn into real computations. At the same time, you’ll get a practical tool you can run on your own matrices and quickly compare results.
