Matrix arithmetic and invariants

Problem

Let

A=[1235],z=[1−1]A=\begin{bmatrix}1&2\\3&5\end{bmatrix},\qquad z=\begin{bmatrix}1\\-1\end{bmatrix}

Compute AzAz, det⁡(A)\det(A), tr⁡(A)\operatorname{tr}(A), A−1A^{-1}, and the Frobenius norm ∥A∥F\|A\|_F.

Reveal answer or reference solution

Direct multiplication gives Az=(−1,−2)TAz=(-1,-2)^{\mathsf T}. Also,

det⁡(A)=1⋅5−2⋅3=−1,tr⁡(A)=6\det(A)=1\cdot5-2\cdot3=-1,\qquad \operatorname{tr}(A)=6

Since the determinant is nonzero, AA is invertible, and

A−1=1−1[5−2−31]=[−523−1]A^{-1}=\frac1{-1}\begin{bmatrix}5&-2\\-3&1\end{bmatrix} =\begin{bmatrix}-5&2\\3&-1\end{bmatrix}

Finally,

∥A∥F=12+22+32+52=39\|A\|_F=\sqrt{1^2+2^2+3^2+5^2}=\sqrt{39}
Outcome

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