1. **Problem:** Check if $\mathbf{v} = (3, 2, 1)$ can be written as a linear combination of $\mathbf{u}_1 = (1, 0, 1)$ and $\mathbf{u}_2 = (0, 1, 1)$. **Step:** We want to find scalars $x, y$ such that $$x(1,0,1) + y(0,1,1) = (3,2,1).$$ This gives the system: $$\begin{cases} x = 3 \\ y = 2 \\ x + y = 1 \end{cases}$$ Check if these are consistent: $3 + 2 = 5 \neq 1$, so no solution. **Answer:** $\mathbf{v}$ cannot be written as a linear combination of $\mathbf{u}_1$ and $\mathbf{u}_2$. 2. **Problem:** Check if $(2,1,3)$ can be expressed as a linear combination of $(1,0,1)$ and $(0,1,2)$. **Step:** Find $a,b$ such that $$a(1,0,1) + b(0,1,2) = (2,1,3).$$ This gives: $$\begin{cases} a = 2 \\ b = 1 \\ a + 2b = 3 \end{cases}$$ Check last: $2 + 2(1) = 4 \neq 3$, no solution. **Answer:** No, $(2,1,3)$ is not a linear combination of the given vectors. 3. **Problem:** Find constants $a,b$ such that $(1,2,3) = a(1,0,1) + b(0,1,2)$. **Step:** Set up system: $$\begin{cases} a = 1 \\ b = 2 \\ a + 2b = 3 \end{cases}$$ Check last: $1 + 2(2) = 5 \neq 3$, no solution. **Answer:** No such constants $a,b$ exist. 4. **Problem:** Verify if $(4,5,6)$ lies in the span of $(1,0,1)$, $(0,1,1)$, and $(1,1,0)$. **Step:** Find $x,y,z$ such that $$x(1,0,1) + y(0,1,1) + z(1,1,0) = (4,5,6).$$ This gives: $$\begin{cases} x + z = 4 \\ y + z = 5 \\ x + y = 6 \end{cases}$$ From third: $y = 6 - x$. Substitute into second: $6 - x + z = 5 \Rightarrow z = x - 1$. Substitute $z$ into first: $x + (x - 1) = 4 \Rightarrow 2x = 5 \Rightarrow x = \frac{5}{2}$. Then $z = \frac{5}{2} - 1 = \frac{3}{2}$ and $y = 6 - \frac{5}{2} = \frac{7}{2}$. **Answer:** Yes, $(4,5,6)$ lies in the span with $x=\frac{5}{2}, y=\frac{7}{2}, z=\frac{3}{2}$. 5. **Problem:** Express $(3,3,3)$ as a linear combination of $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. **Step:** Find $a,b,c$ such that $$a(1,0,0) + b(0,1,0) + c(0,0,1) = (3,3,3).$$ This gives: $$\begin{cases} a = 3 \\ b = 3 \\ c = 3 \end{cases}$$ **Answer:** $(3,3,3) = 3(1,0,0) + 3(0,1,0) + 3(0,0,1)$.