Exercises — Systems of linear equations — matrix form Ax = b

What we do: compare the two lines' slopes via the determinant of the coefficient matrix. $A=\left[\begin{array}{cc}1& 2\\ 3& -1\end{array}\right],\det A=(1)(-1)-(2)(3)=-1-6=-7.$ Why the determinant? For a square $2\times 2$ system, $\det A\ne 0$ means the columns point in genuinely different directions, so the transformation $x\mapsto Ax$ hits every target exactly once — a unique solution (parent note: invertible-matrix shortcut). Answer: $\det A=-7\ne 0\Rightarrow$ unique solution. Left picture of the figure above.

What we do: check if one equation is a scalar multiple of the other. Row 1 is exactly $2\times$ Row 2 (LHS: $2x+4y=2(x+2y)$; RHS: $6=2\cdot 3$). So $\det A=(2)(2)-(4)(1)=0$, and the augmented right-hand side agrees too. Why this matters: $\det A=0$ kills the invertible shortcut, so we drop to rank. Here $\mathrm{rank}(A)=1=\mathrm{rank}([A|b])$, but $n=2>1$. Answer: infinitely many solutions — the same line drawn twice (middle picture).

What we do: same left side, different right side. The left sides give $\mathrm{rank}(A)=1$; but the augmented matrix $[A|b]=\left[\begin{array}{cccc}1& -1& |& 1\\ 1& -1& |& 4\end{array}\right]$ has two independent rows ($1\ne 4$ forces a contradiction), so $\mathrm{rank}([A|b])=2$. Why this is "no solution": rank of the augmented matrix jumped above rank of $A$. The right-hand side $b$ demands a direction the columns cannot supply — $b\notin \mathrm{Col}(A)$. Answer: no solution — parallel lines (right picture).

From L1.1, $\det A=-7\ne 0$, so the inverse exists. Step 1 — build $A^{-1}$. For $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ the inverse is $\frac{1}{\det }\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ (swap diagonal, negate off-diagonal, divide by determinant). $A^{-1}=\frac{1}{-7}\left[\begin{array}{cc}-1& -2\\ -3& 1\end{array}\right].$ Step 2 — multiply by $b$. Why left-multiply? Because $A^{-1}Ax=Ix=x$ isolates the unknown. $x=\frac{1}{-7}\left[\begin{array}{cc}-1& -2\\ -3& 1\end{array}\right]\left[\begin{array}{c}4\\ 5\end{array}\right]=\frac{1}{-7}\left[\begin{array}{c}-4-10\\ -12+5\end{array}\right]=\frac{1}{-7}\left[\begin{array}{c}-14\\ -7\end{array}\right]=\left[\begin{array}{c}2\\ 1\end{array}\right].$ Answer: $x=2,y=1$. Check (column view): $2\left[\begin{array}{c}1\\ 3\end{array}\right]+1\left[\begin{array}{c}2\\ -1\end{array}\right]=\left[\begin{array}{c}4\\ 5\end{array}\right]=b$. ✓

Step 1 — eliminate $x$ from row 3. Row3 $\leftarrow$ Row3 $-2\times$Row1: $(2x+5y-z)-2(x+y+z)=27-12\Rightarrow 3y-3z=15.$ System now: $x+y+z=6,2y+5z=-4,3y-3z=15$. Step 2 — eliminate $y$ from row 3. Row3 $\leftarrow$ Row3 $-\frac{3}{2}\times$Row2: $3y-3z-\frac{3}{2}(2y+5z)=15-\frac{3}{2}(-4)\Rightarrow -3z-\frac{15}{2}z=15+6\Rightarrow -\frac{21}{2}z=21.$ So $z=-2$. Step 3 — back-substitute. From row 2: $2y+5(-2)=-4\Rightarrow 2y=6\Rightarrow y=3$. From row 1: $x+3+(-2)=6\Rightarrow x=5$. Answer: $x=5,y=3,z=-2$. Check: $2(5)+5(3)-(-2)=10+15+2=27$. ✓

What we notice: Row 2 $=2\times$ Row 1, so $\mathrm{rank}(A)=1$, $n=3$. Free parameters $=n-r=2$. Why homogeneous is always solvable: $x=0$ always works, so we never ask existence — only how big is the null space. Solve the one real equation $x=-2y+z$. Let $y=s,z=t$: $x=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=s\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right]+t\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right].$ Answer: the null space is $\mathrm{span}\left\{\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]\right\}$ (a plane through the origin).

$A=\left[\begin{array}{cc}1& 1\\ 1& k\end{array}\right]$, $\det A=k-1$. (a) Unique when $\det A\ne 0$, i.e. $k\ne 1$. (For every such $k$ the lines cross once.) (b)/(c) The knife-edge $k=1$. Then the system is $x+y=2$ and $x+y=4$ — same left side, different right sides → contradiction → no solution. Is there ANY $k$ giving infinitely many? We'd need $\det A=0$ and consistent RHS, i.e. $k=1$ and the second equation $=$ multiple of the first. But at $k=1$ the RHS is $4\ne 2$, so consistency fails. Hence no value of $k$ gives infinitely many here. Answer: unique for $k\ne 1$; no solution for $k=1$; never infinitely many.

Step 1 — eliminate $x$. Row2$-$Row1: $y+2z=1$. Row3$-$Row1: $3y+(a-1)z=b-1$. Step 2 — eliminate $y$. Row3$-3\times$(new Row2): $(a-1)z-6z=(b-1)-3$, i.e. $(a-7)z=b-4.$ Reasoning about the last pivot:

• If $a\ne 7$: a genuine pivot on $z$ → unique solution.
• If $a=7$ and $b\ne 4$: $0=b-4\ne 0$, a contradiction → no solution (augmented rank jumps).
• If $a=7$ and $b=4$: $0=0$, one equation vanishes, $\mathrm{rank}=2<3=n$ → infinitely many (one free parameter). Answer: $(a,b)=(7,4)$.

Step 1 — spot dependence. Row 2 $=2\times$ Row 1 (RHS too: $6=2\cdot 3$). So $\mathrm{rank}(A)=1=\mathrm{rank}([A|b])$, and $n-r=3-1=2$ free parameters → infinitely many. Step 2 — one particular solution $x_p$. Set the free variables $y=0,z=0$ in $x+2y-z=3$: $x=3$. So $x_p=(3,0,0)$. Step 3 — the null space $x_h$. This is exactly L2.3's answer: $\mathrm{span}\{(-2,1,0),(1,0,1)\}$. Step 4 — assemble using the parent's rule $x=x_p+x_h$ (valid because $A(x_p+x_h)=b+0=b$): $x=\left[\begin{array}{c}3\\ 0\\ 0\end{array}\right]+s\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right]+t\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right],s,t\in \mathbb{R}.$ Check with $s=1,t=1$: $x=(3-2+1,1,1)=(2,1,1)$, and $2+2-1=3$. ✓

Why column space? From the parent's existence condition, $Ax=b$ is solvable iff $b\in \mathrm{Col}(A)$ — $b$ must be a combination $x_1{a}_{(1)}+x_2{a}_{(2)}$ of the two columns. Step 1 — set up the combination. We need $x_1\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]+x_2\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 4\end{array}\right].$ Rows 1 and 2 force $x_1=1,x_2=2$. Step 2 — test the last row. It demands $x_1+x_2=4$, but $1+2=3\ne 4$. Answer: the required row-3 condition fails, so $b\notin \mathrm{Col}(A)$ and $Ax=b$ has no solution. (Here rank$(A)=2$ but rank$([A|b])=3$.)

Design logic: "one-parameter family" means $n-r=1$, so with $n=3$ we need $\mathrm{rank}(A)=\mathrm{rank}([A|b])=2$. Recipe: pick two genuinely independent equations, then make the third a consistent combination of them (so it adds no new rank). Construction. Take independent rows $x+y+z=1$ and $y+z=1$. Force the third row to be Row1 $+$ Row2 so it is redundant and consistent: $\{\begin{array}{l}x+y+z=1\\ y+z=1\\ x+2y+2z=2\end{array}$ Verify rank. Row3 $=$ Row1 $+$ Row2 exactly (both LHS and RHS: $2=1+1$), so $\mathrm{rank}(A)=\mathrm{rank}([A|b])=2$, giving $3-2=1$ free parameter. ✓ Solve to display the line. From Row2, $y+z=1$; from Row1, $x=1-(y+z)=0$. Let $z=t$, then $y=1-t$: $x=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]+t\left[\begin{array}{c}0\\ -1\\ 1\end{array}\right].$ A genuine line of solutions — exactly one parameter, as required.

Suppose $Ax=b$ has two different solutions $u\ne v$, so $Au=b$ and $Av=b$. Key step — subtract. By linearity of matrix multiplication (distributivity): $A(u-v)=Au-Av=b-b=0.$ So $w:=u-v$ is a nonzero null-space vector. Manufacture infinitely many. For any scalar $\lambda$, consider $u+\lambda w$: $A(u+\lambda w)=Au+\lambda Aw=b+\lambda \cdot 0=b.$ Every $\lambda$ gives a solution, and since $w\ne 0$ these are all distinct. Conclusion: the moment you have two solutions you automatically have infinitely many. Therefore "exactly two" is impossible — a linear system has $0$, $1$, or $\infty$ solutions. $■$

Given ranks $r_A=\mathrm{rank}(A)$, $r_{aug}=\mathrm{rank}([A|b])$, $n$ unknowns — state the verdict. Verify: $r_A<r_{aug}$ → none; $r_A=r_{aug}=n$ → unique; $r_A=r_{aug}<n$ → infinite ($n-r_A$ free params).