Review : Matrices & Vectors — Question 7

Question 7 – Solution

Strategy. Scalar differentiation still applies to each entry, but matrix factors must retain their multiplication order.

Step 1: Derive and apply the product rule. For component $i$, $y_i=\sum_jM_{ij}v_j$. Differentiating the finite sum gives $y_i'=\sum_j(M_{ij}'v_j+M_{ij}v_j')$, hence $y'=M'v+Mv'$. Direct multiplication and differentiation give $$y=\begin{pmatrix}t^2+te^t\\2e^t\end{pmatrix},\qquad \boxed{y'=\begin{pmatrix}2t+(1+t)e^t\\2e^t\end{pmatrix}.}$$ The product-rule route gives $M'v=(e^t,0)^T$ and $Mv'=(2t+te^t,2e^t)^T$, confirming the same derivative.

Step 2: Differentiate the inverse with the correct order. The inverse is $N=\begin{pmatrix}1&-t/2\\0&1/2\end{pmatrix}$. Differentiating $MN=I_2$ gives $M'N+MN'=0$; multiplying on the left by $N$ yields $$\boxed{N'=-NM'N=\begin{pmatrix}0&-1/2\\0&0\end{pmatrix}.}$$ This agrees with direct entrywise differentiation of $N$.

Step 3: Test the square rule explicitly. Since $M^2=\begin{pmatrix}1&3t\\0&4\end{pmatrix}$, $$(M^2)'=\begin{pmatrix}0&3\\0&0\end{pmatrix} =\underbrace{\begin{pmatrix}0&2\\0&0\end{pmatrix}}_{M'M} +\underbrace{\begin{pmatrix}0&1\\0&0\end{pmatrix}}_{MM'}.$$ But $2MM'=\begin{pmatrix}0&2\\0&0\end{pmatrix}$, which is different. The shortcut works when $M$ and $M'$ commute; they do not here.

Step 4: Interpret the constant determinant. The determinant is always $2$, whereas $M(t)(0,1)^T=(t,2)^T$ has length $\sqrt{t^2+4}$. Thus signed area scaling remains fixed while shear and some vector lengths change. Determinant information alone cannot determine every entry or every geometric effect of a matrix.