Applications to linear systems

A linear system $AX = B$ with $A$ invertible has the unique solution $X = A^{-1} B$. Two techniques compute this solution: the matrix method, which inverts $A$ explicitly (using the adjoint), and Cramer's rule, which gives each component of $X$ directly as a ratio of determinants. Both are theoretically clean. For practical hand computation, Cramer's rule beats the matrix method when the determinants are small.

The matrix method

For $AX = B$ with $\det A \neq 0$:

$X = A^{-1} B = \frac{1}{\det A}\,\text{adj}(A) \cdot B.$

This was illustrated in the previous subtopic.

Cramer's rule

Let $A_i$ denote the matrix obtained by replacing the $i$-th column of $A$ with $B$. Then, provided $\det A \neq 0$, the unique solution of $AX = B$ has

$x_i = \frac{\det A_i}{\det A}, \quad i = 1, 2, \ldots, n.$

This is Cramer's rule. The derivation uses cofactor orthogonality: multiply both sides of $AX = B$ on the left by $\text{adj}(A)$.

Consistency of a system

Given $AX = B$:

• If $\det A \neq 0$, the system has a unique solution.
• If $\det A = 0$ and $\text{adj}(A) \cdot B = 0$ (i.e., each $\det A_i = 0$), the system has either infinitely many solutions or no solution, depending on rank.
• If $\det A = 0$ but some $\det A_i \neq 0$, the system is inconsistent.

For a homogeneous system $AX = O$:

• $\det A \neq 0 \Rightarrow$ only the trivial solution $X = 0$.
• $\det A = 0 \Rightarrow$ infinitely many non-trivial solutions.

Worked example: Cramer's rule on a $3 \times 3$ system

Solve $\begin{cases} x + y + z = 6 \\ x - y + z = 2 \\ 2x + y - z = 1 \end{cases}$.

$A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{pmatrix}$, $B = \begin{pmatrix} 6 \\ 2 \\ 1 \end{pmatrix}$.

$\det A = 1(1 - 1) - 1(-1 - 2) + 1(1 + 2) = 0 + 3 + 3 = 6$.

$A_1$: replace column $1$ with $B$: $\det\begin{pmatrix} 6 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 1 & -1 \end{pmatrix} = 6(1 - 1) - 1(-2 - 1) + 1(2 + 1) = 0 + 3 + 3 = 6$.

$A_2$: $\det\begin{pmatrix} 1 & 6 & 1 \\ 1 & 2 & 1 \\ 2 & 1 & -1 \end{pmatrix} = 1(-2 - 1) - 6(-1 - 2) + 1(1 - 4) = -3 + 18 - 3 = 12$.

$A_3$: $\det\begin{pmatrix} 1 & 1 & 6 \\ 1 & -1 & 2 \\ 2 & 1 & 1 \end{pmatrix} = 1(-1 - 2) - 1(1 - 4) + 6(1 + 2) = -3 + 3 + 18 = 18$.

$x = 6/6 = 1, y = 12/6 = 2, z = 18/6 = 3$.

Verify: $1 + 2 + 3 = 6$, $1 - 2 + 3 = 2$, $2 + 2 - 3 = 1$. $\checkmark$

Geometric interpretation

A $3 \times 3$ linear system $AX = B$ represents three planes in $\mathbb{R}^3$. Their intersection geometry:

• $\det A \neq 0$: planes meet at a single point.
• $\det A = 0$: planes are mutually parallel, or two coincide, or all three intersect in a line , depending on the specific $A$ and $B$.

Worked examples

Example 1. Solve $\begin{cases} 2x + 3y = 5 \\ x + 2y = 3 \end{cases}$ by Cramer.

$\det A = 4 - 3 = 1$. $\det A_1 = \det\begin{pmatrix} 5 & 3 \\ 3 & 2 \end{pmatrix} = 1$. $\det A_2 = \det\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} = 1$. So $x = 1, y = 1$.

Example 2. Show that the system $\begin{cases} x + y = 1 \\ 2x + 2y = 3 \end{cases}$ has no solution.

$\det A = 2 - 2 = 0$. $\det A_1 = \det\begin{pmatrix} 1 & 1 \\ 3 & 2 \end{pmatrix} = -1 \neq 0$. So inconsistent, no solution.

Example 3. Determine $k$ for which $\begin{cases} kx + y = 0 \\ x + ky = 0 \end{cases}$ has a non-trivial solution.

Non-trivial iff $\det A = 0$: $k^2 - 1 = 0$, so $k = \pm 1$.

Example 4. Solve $\begin{cases} x + 2y - z = 1 \\ 3x - y + 2z = 4 \\ 2x + y + z = 5 \end{cases}$ by Cramer.

$\det A = 1(-1 - 2) - 2(3 - 4) + (-1)(3 + 2) = -3 + 2 - 5 = -6$.

$\det A_1 = \det\begin{pmatrix} 1 & 2 & -1 \\ 4 & -1 & 2 \\ 5 & 1 & 1 \end{pmatrix} = 1(-1 - 2) - 2(4 - 10) + (-1)(4 + 5) = -3 + 12 - 9 = 0$.

$\det A_2 = \det\begin{pmatrix} 1 & 1 & -1 \\ 3 & 4 & 2 \\ 2 & 5 & 1 \end{pmatrix} = 1(4 - 10) - 1(3 - 4) + (-1)(15 - 8) = -6 + 1 - 7 = -12$.

$\det A_3 = \det\begin{pmatrix} 1 & 2 & 1 \\ 3 & -1 & 4 \\ 2 & 1 & 5 \end{pmatrix} = 1(-5 - 4) - 2(15 - 8) + (15 + 2) = -9 - 14 + 17 = -6$.

$x = 0, y = 2, z = 1$.

Verify: $0 + 4 - 1 = 3$ , wait that doesn't match. Let me recompute. Equation 1: $x + 2y - z = 1$. With $x=0, y=2, z=1$: $0 + 4 - 1 = 3 \ne 1$.

There's an arithmetic error somewhere. Let me reread $\det A_1$. Expanding along row 1: $1 \cdot \det\begin{pmatrix} -1 & 2 \\ 1 & 1 \end{pmatrix} - 2 \cdot \det\begin{pmatrix} 4 & 2 \\ 5 & 1 \end{pmatrix} + (-1) \det\begin{pmatrix} 4 & -1 \\ 5 & 1 \end{pmatrix} = 1(-1 - 2) - 2(4 - 10) + (-1)(4 + 5) = -3 + 12 - 9 = 0$. OK.

$\det A_2 = 1(4 - 10) - 1(3 - 4) - 1(15 - 8) = -6 + 1 - 7 = -12$. OK.

$\det A_3 = 1(-5 - 4) - 2(15 - 8) + 1(15 + 2) = -9 - 14 + 17 = -6$.

$x = 0/(-6) = 0, y = -12/(-6) = 2, z = -6/(-6) = 1$.

Verify eq 1: $0 + 4 - 1 = 3$. But RHS is $1$. So the system has no solution with $(0, 2, 1)$. Let me reread the system. Actually if $\det A_1$ is really $0$ but $\det A \ne 0$, then $x = 0$ , but solution should be unique. Let me verify by substituting back into original.

Hmm , let me recompute the determinant of $A$. $\det\begin{pmatrix} 1 & 2 & -1 \\ 3 & -1 & 2 \\ 2 & 1 & 1 \end{pmatrix}$. Expand: $1(-1 \cdot 1 - 2 \cdot 1) - 2(3 \cdot 1 - 2 \cdot 2) + (-1)(3 \cdot 1 - (-1) \cdot 2) = 1(-3) - 2(-1) + (-1)(5) = -3 + 2 - 5 = -6$. OK.

Let me check $\det A_1$ again. $A_1$ replaces column 1 with $B = (1, 4, 5)$: $\det\begin{pmatrix} 1 & 2 & -1 \\ 4 & -1 & 2 \\ 5 & 1 & 1 \end{pmatrix} = 1(-1 - 2) - 2(4 - 10) + (-1)(4 + 5) = -3 + 12 - 9 = 0$. OK.

So $x = 0$. Let me actually solve by substitution to verify. From eq 1: $z = x + 2y - 1$. Sub into eq 2: $3x - y + 2(x + 2y - 1) = 4$, so $5x + 3y = 6$. Sub into eq 3: $2x + y + x + 2y - 1 = 5$, so $3x + 3y = 6$, i.e., $x + y = 2$.

From these: $5x + 3y = 6$ and $x + y = 2 \Rightarrow y = 2 - x$. Sub: $5x + 3(2 - x) = 6 \Rightarrow 2x = 0 \Rightarrow x = 0, y = 2, z = 0 + 4 - 1 = 3$.

So $z = 3$, not $1$. Let me recompute $\det A_3$. $A_3$ replaces column 3 with $B$: $\det\begin{pmatrix} 1 & 2 & 1 \\ 3 & -1 & 4 \\ 2 & 1 & 5 \end{pmatrix} = 1(-5 - 4) - 2(15 - 8) + 1(3 + 2) = -9 - 14 + 5 = -18$.

So $z = -18 / -6 = 3$. (I made an arithmetic error before with the last cofactor; it's $3 + 2 = 5$, not $15 + 2$.)

So solution: $x = 0, y = 2, z = 3$. Verify: $0 + 4 - 3 = 1$ $\checkmark$, $0 - 2 + 6 = 4$ $\checkmark$, $0 + 2 + 3 = 5$ $\checkmark$.

Example 5. Solve $\begin{cases} x + y + z = 9 \\ 2x + 5y + 7z = 52 \\ 2x + y - z = 0 \end{cases}$ using Cramer.

$\det A = 1(5 \cdot (-1) - 7 \cdot 1) - 1(2 \cdot (-1) - 7 \cdot 2) + 1(2 \cdot 1 - 5 \cdot 2) = -12 + 16 - 8 = -4$.

$\det A_1 = \det\begin{pmatrix} 9 & 1 & 1 \\ 52 & 5 & 7 \\ 0 & 1 & -1 \end{pmatrix} = 9(-5 - 7) - 1(-52 - 0) + 1(52 - 0) = -108 + 52 + 52 = -4$.

$\det A_2 = \det\begin{pmatrix} 1 & 9 & 1 \\ 2 & 52 & 7 \\ 2 & 0 & -1 \end{pmatrix} = 1(-52 - 0) - 9(-2 - 14) + 1(0 - 104) = -52 + 144 - 104 = -12$.

$\det A_3 = \det\begin{pmatrix} 1 & 1 & 9 \\ 2 & 5 & 52 \\ 2 & 1 & 0 \end{pmatrix} = 1(0 - 52) - 1(0 - 104) + 9(2 - 10) = -52 + 104 - 72 = -20$.

$x = 1, y = 3, z = 5$.

Example 6. Show that the homogeneous system $\begin{cases} 2x + y + z = 0 \\ x + 3y - z = 0 \\ x + 5y + 4z = 0 \end{cases}$ has only the trivial solution.

$\det A = 2(12 + 5) - 1(4 + 1) + 1(5 - 3) = 34 - 5 + 2 = 31 \neq 0$. So $X = O$ is the only solution.

Try it yourself

1. Solve $\begin{cases} 3x + 4y = 10 \\ 5x + 6y = 16 \end{cases}$ by Cramer.
2. Solve $\begin{cases} x + 2y + 3z = 14 \\ 2x + y + z = 7 \\ x + y + z = 6 \end{cases}$.
3. Determine $k$ such that $\begin{cases} 3x + y + z = 0 \\ x + ky + z = 0 \\ x + y + kz = 0 \end{cases}$ has a non-trivial solution.
4. Solve $\begin{cases} 2x + 3y = 8 \\ 3x + 2y = 7 \end{cases}$.
5. Show that $\begin{cases} x + y = 1 \\ 2x + 2y = 4 \end{cases}$ is inconsistent.
6. Find values of $\lambda$ for which $\begin{cases} (\lambda - 1) x + 3y + 4z = 0 \\ 3x + (\lambda - 1) y + 4z = 0 \\ 4x + 3y + (\lambda - 1) z = 0 \end{cases}$ has a non-trivial solution.
7. Solve $\begin{cases} 5x - 3y + 2z = 5 \\ 3x + 2y + z = 2 \\ x + y - z = 1 \end{cases}$.
8. Show that $\begin{cases} ax + by + c = 0 \\ a' x + b' y + c' = 0 \\ a'' x + b'' y + c'' = 0 \end{cases}$ has a solution iff $\det\begin{pmatrix} a & b & c \\ a' & b' & c' \\ a'' & b'' & c'' \end{pmatrix} = 0$.
9. Determine the number of solutions of $\begin{cases} x + y + z = 4 \\ 2x + 2y + 2z = 8 \\ 3x + 3y + 3z = 12 \end{cases}$.
10. Solve $\begin{cases} 3x + y - 2z = 3 \\ 2x + 3y + z = -3 \\ x - 2y + 2z = -2 \end{cases}$.
11. Find $\lambda$ if $\begin{cases} x - 2y = -2 \\ \lambda x + 4y = 4 \end{cases}$ has no solution.
12. Use Cramer to solve $\begin{cases} x + y + z = 1 \\ x + 2y + 3z = 4 \\ x + 3y + 6z = 10 \end{cases}$.
13. A boat goes $30$ km upstream in $5$ hours and $24$ km downstream in $3$ hours. Find the speed of the boat and the current. (Set up a linear system.)
14. Three persons $A, B, C$ together collected $\rupee\,1000$. $A$ and $B$ together collected $\rupee\,700$. $B$ and $C$ collected $\rupee\,600$. How much did each collect?

Pitfalls / Tricks

• Cramer's rule applies only when $\det A \neq 0$. For singular $A$, use row reduction to check consistency.
• Always verify your solution by substituting back into the original equations , Cramer's arithmetic is error-prone.
• Cramer is impractical for large systems; row reduction is the workhorse there.
• A homogeneous system always has the trivial solution; non-trivial solutions exist iff $\det A = 0$.

This concludes Chapter 4. The methods generalise enormously: every later chapter that uses linear systems (vectors, 3D geometry, linear programming) implicitly invokes these results.