Examine linear dependence of five systems 1^° x, x+1; 2^°…
Examine linear dependence of five systems $1^\circ\; x,\ x+1;\qquad 2^\circ\; x^2,\ -2x^2;\qquad 3^\circ\; x,\ x^2,\ x^3;$ $4^\circ\; e^{x},\ e^{2x},\ e^{3x};\qquad 5^\circ\; \sin^2x,\ \cos^2x,\ 1$.
Explanation of Method
Test five short systems of functions for linear dependence, verifying each conclusion directly and by the Wronskian.
A system $f_1,\ldots,f_n$ is linearly dependent on an interval if constants $c_1,\ldots,c_n$, not all zero, exist with $c_1f_1+\cdots+c_nf_n\equiv0$ there; otherwise it is linearly independent. The Wronskian \[ W(x)=\begin{vmatrix} f_1&f_2&\cdots&f_n\\ f_1'&f_2'&\cdots&f_n'\\ \vdots& & &\vdots\\ f_1^{(n-1)}&f_2^{(n-1)}&\cdots&f_n^{(n-1)}\end{vmatrix}\] gives a one-directional test: $W\not\equiv0$ forces independence. $W\equiv0$ is only consistent with dependence – it does not, by itself, prove it (see Problem 66). Each system below is checked both directly and by Wronskian.
Step 1: System $1^\circ$: $x,\ x+1$
Direct check.
Suppose $c_1x+c_2(x+1)\equiv0$, i.e. $(c_1+c_2)x+c_2\equiv0$ for all $x$. Matching coefficients of $x^1$ and $x^0$: \[c_1+c_2=0,\qquad c_2=0\ \Longrightarrow\ c_1=c_2=0.\]
Only the trivial combination vanishes identically.
Wronskian check.
\[ W(x)=\begin{vmatrix}x&x+1\\1&1\end{vmatrix}=x-( x+1)=-1\neq0. \]
The system is linearly independent ($W\equiv-1\neq0$).
Step 2: System $2^\circ$: $x^2,\ -2x^2$
Direct check.
Take $c_1=2,\ c_2=1$ (both nonzero): $2\cdot x^2+1\cdot(-2x^2)=2x^2-2x^2=0.$
Indeed $-2x^2=(-2)\cdot x^2$: one function is a scalar multiple of the other.
Wronskian check.
\[ W(x)=\begin{vmatrix}x^2&-2x^2\\2x&-4x\end{vmatrix} =x^2(-4x)-(-2x^2)(2x)=-4x^3+4x^3=0. \]
The system is linearly dependent ($-2x^2=-2\cdot x^2$, $W\equiv0$).
Step 3: System $3^\circ$: $x,\ x^2,\ x^3$
Direct check.
Suppose $c_1x+c_2x^2+c_3x^3\equiv0$. Dividing by $x\neq0$ gives $c_1+c_2x+c_3x^2\equiv0$ for $x\neq0$, hence for all $x$ by continuity. A polynomial vanishing identically has all coefficients zero:
\[ c_1=c_2=c_3=0. \]
Wronskian check.
\[ W(x)=\begin{vmatrix}x&x^2&x^3\\1&2x&3x^2\\0&2&6x\end{vmatrix} =x\begin{vmatrix}2x&3x^2\\2&6x\end{vmatrix}-1\begin{vmatrix}x^2&x^3\\2&6x\end{vmatrix} \]
\[ =x(12x^2-6x^2)-(6x^3-2x^3)=6x^3-4x^3=2x^3\neq0. \]
The system is linearly independent ($W=2x^3\not\equiv0$).
Step 4: System $4^\circ$: $e^{x},\ e^{2x},\ e^{3x}$
Direct check.
Suppose $c_1e^x+c_2e^{2x}+c_3e^{3x}\equiv0$. Divide by $e^x>0$ and set $t=e^x>0$: the quadratic $c_1+c_2t+c_3t^2$ vanishes for every $t>0$ (infinitely many points), forcing every coefficient to zero:
\[ c_1=c_2=c_3=0. \]
Wronskian check.
Factoring $e^x,e^{2x},e^{3x}$ from the three columns leaves a Vandermonde determinant with nodes $1,2,3$: \[W(x)=e^{6x}\begin{vmatrix}1&1&1\\1&2&3\\1&4&9\end{vmatrix} =e^{6x}\big[1(18-12)-1(9-3)+1(4-2)\big]=e^{6x}(6-6+2)=2e^{6x}.\]
The system is linearly independent ($W=2e^{6x}\neq0$).
Step 5: System $5^\circ$: $\sin^2x,\ \cos^2x,\ 1$
Direct check.
The Pythagorean identity gives a non-trivial relation directly: $1\cdot\sin^2x+1\cdot\cos^2x+(-1)\cdot1=\sin^2x+\cos^2x-1=0.$
Wronskian check.
With $f_1'=\sin2x,\ f_2'=-\sin2x,\ f_1''=2\cos2x,\ f_2''=-2\cos2x$, expand along the third column (only the top entry is nonzero): \[W(x)=\begin{vmatrix}\sin^2x&\cos^2x&1\\ \sin2x&-\sin2x&0\\ 2\cos2x&-2\cos2x&0\end{vmatrix} =\begin{vmatrix}\sin2x&-\sin2x\\2\cos2x&-2\cos2x\end{vmatrix}\]
\[ =-2\sin2x\cos2x+2\sin2x\cos2x=0. \]
The system is linearly dependent ($\sin^2x+\cos^2x-1\equiv0$, $W\equiv0$).
Result
Final answer:
\[\boxed{\,\text{Linearly independent}\quad(W\equiv-1\neq0)\,}\]
Final answer:
\[\boxed{\,\text{Linearly dependent}\quad(-2x^2=-2\cdot x^2,\ \ W\equiv0)\,}\]
Final answer:
\[\boxed{\,\text{Linearly independent}\quad(W=2x^3\not\equiv0)\,}\]
Final answer:
\[\boxed{\,\text{Linearly independent}\quad(W=2e^{6x}\neq0)\,}\]
Final answer:
\[\boxed{\,\text{Linearly dependent}\quad(\sin^2x+\cos^2x-1\equiv0,\ \ W\equiv0)\,}\]
Remark
Summary.
$1^\circ$ independent ($W=-1$) · $2^\circ$ dependent ($-2x^2=-2\cdot x^2$) · $3^\circ$ independent ($W=2x^3$) · $4^\circ$ independent ($W=2e^{6x}$) · $5^\circ$ dependent ($\sin^2x+\cos^2x=1$).