91Ó°ÊÓ

A square matrix \(A\) is invertible if there exists another square matrix \(B\) such that \(AB = BA = I\), where \(I\) is the identity matrix with \(1\)s on the diagonal and \(0\)s elsewhere. In this problem, we are dealing with an upper triangular matrix. An upper triangular matrix has the form: \[A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ 0 & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \end{pmatrix}\] Where each \(a_{ii}\) is an entry on the diagonal and other entries are either zero or nonzero as long as they are above the diagonal.

In order to prove that \(A\) has an inverse, we need to find another upper triangular matrix \(B\) whose product with \(A\) yields the identity matrix. Define matrix \(B\) in the form of: \[B = \begin{pmatrix} b_{11} & b_{12} & \cdots & b_{1n} \\ 0 & b_{22} & \cdots & b_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & b_{nn} \end{pmatrix}\] We want to find the entries of \(B\) that would make the product \(AB = I\). The product is: \[AB = \begin{pmatrix} a_{11}b_{11} & a_{11}b_{12} + a_{12}b_{22} & \cdots & a_{11}b_{1n} + a_{12}b_{2n} + \cdots + a_{1n}b_{nn}\\ 0 & a_{22}b_{22} & \cdots & a_{22}b_{2n} + a_{23}b_{3n} + \cdots + a_{2n}b_{nn} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn}b_{nn} \end{pmatrix}\] For this to be an identity matrix, we need to have: 1. Diagonal: \(a_{ii}b_{ii} = 1\) for all \(i\). 2. Above diagonal: \(a_{ij}b_{jk} = 0\) for all \(i < j\) and \(j \leq k\). Let's find a suitable \(B\) that would make the product \(AB\) equal to the identity matrix: 1. For the diagonal entries, we can choose \(b_{ii} = \frac{1}{a_{ii}}\) for all \(i\), since we know that all \(a_{ii} \neq 0\). Thus, \(a_{ii}b_{ii} = a_{ii}\left(\frac{1}{a_{ii}}\right) = 1\) for all \(i\). 2. For the entries above the diagonal, we can choose entries \(b_{ij}\) such that the sum \(a_{ij}b_{jk}\) evaluates to zero. We can set \(b_{ij}\) equal to the negative sum of the other product terms in the sum (for \(j < k\)), divided by \(a_{jj}\). It's guaranteed that we can find such \(b_{ij}\) values that make \(AB = I\) since all \(a_{ii} \neq 0\). Therefore, we have found a matrix \(B\) that, when multiplied with \(A\), yields the identity matrix, demonstrating that an upper triangular matrix with nonzero diagonal entries is invertible.

Suppose an upper triangular matrix \(A\) has an inverse \(B\). As demonstrated in Step 2, every diagonal entry of the product \(AB\) must be equal to 1. This means that for each \(i\): \[a_{ii}b_{ii} = 1\] If any diagonal entry \(a_{ii}\) were zero, the equation \(a_{ii}b_{ii} = 1\) could not hold. Hence, if an upper triangular matrix has an inverse, all its diagonal entries must be nonzero. Combining the results of Steps 2 and 3, we have shown that an upper triangular \(n \times n\) matrix is invertible if and only if all its diagonal entries are nonzero.