Use A's QR Decomposition to determine if A is invertible and solvable

Question

If Q = \begin{bmatrix}1&1/√6&1/3√3\\1&-1/√6&-1/3√3\\0 & √2/√3 & -1/3√3\end{bmatrix}

and R = \begin{bmatrix}2√2&1/√2&1/√2\\0&4/√6&1/√6\\0&0&1/3√3\end{bmatrix}.

Is A invertible? (no computation required) Is the system Ax=b solvable for each b in $R^3$ (give the formula for its solutions).

Answer 1

Since $A=QR$ we have: $\det(A)=\det(QR)=\det(Q)\det(R)$

Since $Q$ is orthogonal (i.e. $Q^{-1}=Q^T$), we have: $1=\det(I)=\det(QQ^T)=\det(Q)\det(Q^T)=(\det(Q))^2$ so $\det(Q)=\pm 1\neq 0$.

Since $R$ is upper triangular, and has all nonzero elements on its diagonal, $\det(R)\neq 0$.

Thus $\det(A)\neq 0\implies A$ is invertible $\implies Ax=b$ has a unique solution for every $b\in\Bbb R^3$. I'm not sure what you mean by a formula for the solution to $Ax=b$. You could find $x$ as $x=A^{-1}b=(QR)^{-1}b=R^{-1}Q^Tb$ since $A$ is invertible, but I'm not sure if this is the "formula" you're looking for.

Answer 2

Hint. Consider the determinant of $A$. I assume you mean the $QR$-decomposition of $A$, i.e. $A = QR$.