Daily math problems for July 20, 2026 — Probability, Calculus, Trigonometry & more

$$\det A(3)=4\cdot(-4)-6\cdot 3=-34$$

$\bullet$ Evaluate $A(2)$ and its determinant:

$$A(2)=\begin{pmatrix}3&4\\3&-5\end{pmatrix}$$

$$\det A(2)=3\cdot(-5)-4\cdot 3=-27$$

$\bullet$ $A(3)$ has trace $0$, so by Cayley–Hamilton $A(3)^2=34I$. Inverting:

$$A^{-2}(3)=\tfrac{1}{34}I$$

$\bullet$ Invert $A(2)$:

$$A^{-1}(2)=\tfrac{1}{27}\begin{pmatrix}5&4\\3&-3\end{pmatrix}$$

$\bullet$ Square it:

$$A^{-2}(2)=\tfrac{1}{729}\begin{pmatrix}37&8\\6&21\end{pmatrix}$$

$\bullet$ One more product:

$$A^{-3}(2)=\tfrac{1}{19683}\begin{pmatrix}209&124\\93&-39\end{pmatrix}$$

$\bullet$ Add the matrices and apply $\det(sI+M)=s^2+s\,\operatorname{tr}M+\det M$ with $s=\tfrac{1}{34}$, $\operatorname{tr}M=\tfrac{170}{19683}$, $\det M=-\tfrac{1}{19683}$:

$$\det=\tfrac{1}{1156}+\tfrac{170}{34\cdot 19683}-\tfrac{1}{19683}$$

$$=\tfrac{1}{1156}+\tfrac{4}{19683}$$

$$=\tfrac{24307}{1156\cdot 19683}$$