Linear Systems & Matrices

Study Sheet

Linear Systems & Matrices

Systems in two and three variables, inequalities, matrices, determinants, and Cramer's Rule

Solving Two-Variable Systems (Review)

Concept
The three methods

A system of two linear equations asks for the point (x,y)(x,y) that satisfies both equations at once. There are three standard methods:

  • Graphing: plot both lines; the solution is where they cross.
  • Substitution: solve one equation for a variable, then substitute into the other.
  • Elimination: add or subtract multiples of the equations to cancel one variable.

Three equations in three unknowns are solved by the same elimination as two: use one equation against each of the other two to remove a variable, which leaves an ordinary 2×22\times2 system, then back-substitute twice. A matrix stores the coefficients so the elimination can be written without repeating xx, yy, zz and == on every line — each row is an equation, each column an unknown, and the bar marks where the equals signs were. The three row operations are exactly the three legal equation moves, which is why row reduction can never gain or lose a solution. A determinant of zero says there is no unique solution; whether that means none or infinitely many is decided by the constants.

Subtracting one equation from another produces a new equation that every solution of the pair still satisfies. When the xx-terms cancel, that new equation is a horizontal line, y=2y = 2, passing through the same crossing point. Elimination is the search for a combination simple enough to read the answer from.

Concept
How many solutions?

Compare the two lines:

  • Different slopes ⇒\Rightarrow lines cross once ⇒\Rightarrow one solution (consistent, independent).
  • Same slope, different intercepts ⇒\Rightarrow parallel lines ⇒\Rightarrow no solution (inconsistent).
  • Same slope and same intercept ⇒\Rightarrow same line ⇒\Rightarrow infinitely many solutions (dependent).
Example
Worked example: elimination
xy−1123456−2−112345(3, 2)
Subtracting the equations gives y=2y = 2 (dashed), a line through the same crossing (3,2)(3, 2).

Solve {2x+3y=122x−y=4\begin{cases} 2x + 3y = 12 \\ 2x - y = 4 \end{cases}

The xx-terms already match. Subtract the second equation from the first:

(2x+3y)−(2x−y)=12−4⇒4y=8⇒y=2.(2x+3y)-(2x-y) = 12-4 \quad\Rightarrow\quad 4y = 8 \quad\Rightarrow\quad y = 2.

Substitute into 2x−y=42x - y = 4:  2x−2=4⇒2x=6⇒x=3.\ 2x - 2 = 4 \Rightarrow 2x = 6 \Rightarrow x = 3.

Solution: (3,2)(3,2). Check: 2(3)+3(2)=122(3)+3(2)=12 ✓ and 2(3)−2=42(3)-2=4 ✓.

Side note
Where it goes wrong

Doing a row operation on the coefficient columns and forgetting the constant column. Augment the matrix and treat every row as one unit.

Example
Worked example: graphing

Solve {y=x+1y=−x+3\begin{cases} y = x + 1 \\ y = -x + 3 \end{cases} by graphing. The lines meet where x+1=−x+3x+1=-x+3, i.e. 2x=22x=2, so x=1x=1 and y=2y=2.

Tip

Tip: Elimination is fastest when a variable already has matching or opposite coefficients. Substitution shines when one equation is already solved for a variable (like y=…y=\dots).

Try it
Try it: pick a method

Solve {5x−2y=16y=3x−9\begin{cases} 5x - 2y = 16 \\ y = 3x - 9 \end{cases}.

Show the answer

The second equation is already solved for yy, so substitute: 5x−2(3x−9)=165x - 2(3x - 9) = 16, 5x−6x+18=165x - 6x + 18 = 16, x=2x = 2. Then y=3(2)−9=−3y = 3(2) - 9 = -3. Check: 10+6=1610 + 6 = 16. Solution (2,−3)(2, -3).

Systems of Linear Inequalities

Concept
The solution is a region, not a point

For a system of linear inequalities, the solution set is the region where all shaded half-planes overlap.

  • Graph each boundary line: use a solid line for ≤\le or ≥\ge, a dashed line for << or >>.
  • Shade the side that makes each inequality true (use a test point such as (0,0)(0,0) if it is not on the line).
  • The overlap of all shadings is the solution region.

Each boundary line splits the plane into two half-planes, one where the inequality holds and one where it fails, and a point cannot switch sides without crossing the line. So one test point per boundary settles which side to shade, and the system's solutions are where all the shaded half-planes overlap.

Example
Worked example: describe the region
xy−4−3−2−11234−4−3−2−112345(0, 0)
y≤x+1y \le x + 1 (solid edge) and y>−2y > -2 (dashed edge): a wedge opening to the right.

Describe the solution of {y≤x+1y>−2\begin{cases} y \le x + 1 \\ y > -2 \end{cases}

The boundary y=x+1y = x+1 is solid (from ≤\le); shade below/on it. The boundary y=−2y=-2 is dashed (from >>); shade above it. The solution is the wedge that is on or below the slanted line and strictly above the horizontal line y=−2y=-2.

Test point (0,0)(0,0): is 0≤0+10 \le 0+1? Yes. Is 0>−20 > -2? Yes. So (0,0)(0,0) lies in the region.

Side note
Where it goes wrong

Using a solid boundary for a strict inequality. For y>−2y > -2, points on the line y=−2y = -2 fail (−2>−2-2 > -2 is false), so the line is dashed and not part of the region.

Tip

Tip: A point is a solution of the system only if it satisfies every inequality. Test each one separately.

Try it
Try it: test points

For y≤x+1y \le x + 1 and y>−2y > -2, is (4,1)(4, 1) a solution? (−3,0)(-3, 0)? (1,−2)(1, -2)?

Show the answer

(4,1)(4, 1): 1≤51 \le 5 and 1>−21 > -2, yes. (−3,0)(-3, 0): 0≤−20 \le -2 fails, no. (1,−2)(1, -2): −2≤2-2 \le 2 holds but −2>−2-2 > -2 fails, no (it is on the dashed edge).

Solving Three-Variable Systems by Elimination

Concept
The strategy

A system in x,y,zx,y,z needs three equations. To solve:

  • Pick a variable to eliminate. Combine two equations to remove it, then combine a different pair to remove the same variable. This leaves a 2×22\times2 system.
  • Solve that 2×22\times2 system for the remaining two variables.
  • Back-substitute to find the third variable.
  • Always check the solution in all three original equations.

Each linear equation in three variables is a plane in space, and three planes typically meet in one point. Eliminating the same variable from two different pairs of equations produces two equations in the remaining two variables, which is a system you already know how to solve; the third variable is recovered by substituting back.

Example
Worked example: full 3-variable elimination

Solve {x+y+z=62x−y+z=3x+2y−z=2\begin{cases} x + y + z = 6 \\ 2x - y + z = 3 \\ x + 2y - z = 2 \end{cases}

Step 1 — eliminate zz. Add equation (1) and equation (3):

(x+y+z)+(x+2y−z)=6+2 ⇒ 2x+3y=8.(A)(x+y+z)+(x+2y-z) = 6+2 \ \Rightarrow\ 2x + 3y = 8. \quad (\text{A})

Add equation (2) and equation (3):

(2x−y+z)+(x+2y−z)=3+2 ⇒ 3x+y=5.(B)(2x-y+z)+(x+2y-z) = 3+2 \ \Rightarrow\ 3x + y = 5. \quad (\text{B})

Step 2 — solve the 2×22\times2 system. From (B), y=5−3xy = 5 - 3x. Substitute into (A):

2x+3(5−3x)=8 ⇒ 2x+15−9x=8 ⇒ −7x=−7 ⇒ x=1.2x + 3(5-3x) = 8 \ \Rightarrow\ 2x + 15 - 9x = 8 \ \Rightarrow\ -7x = -7 \ \Rightarrow\ x = 1.

Then y=5−3(1)=2y = 5 - 3(1) = 2.

Step 3 — back-substitute. From (1): z=6−x−y=6−1−2=3z = 6 - x - y = 6 - 1 - 2 = 3.

Solution: (1,2,3)(1,2,3). Check: 1+2+3=61+2+3=6 ✓; 2−2+3=32-2+3=3 ✓; 1+4−3=21+4-3=2 ✓.

Side note
Where it goes wrong

Eliminating different variables in the two pairs: removing zz from equations (1) and (3) but yy from (1) and (2). The two results still contain three variables between them and cannot be combined. Eliminate the same variable both times.

Tip

Tip: Eliminate the same variable both times. If you accidentally remove zz once and xx once, you cannot combine the two results.

Try it
Try it: three unknowns

Solve {x+y+z=2x−y+2z=62x+y−z=−1\begin{cases} x + y + z = 2 \\ x - y + 2z = 6 \\ 2x + y - z = -1 \end{cases}.

Show the answer

Eliminate zz twice. (1) ++ (3): 3x+2y=13x + 2y = 1. (2) +2×+ 2 \times (3): 5x+y=45x + y = 4. From the second, y=4−5xy = 4 - 5x; substitute: 3x+8−10x=13x + 8 - 10x = 1, so x=1x = 1 and y=−1y = -1. Then (1) gives z=2−1+1=2z = 2 - 1 + 1 = 2. Check (2): 1+1+4=61 + 1 + 4 = 6. Solution (1,−1,2)(1, -1, 2).

Matrix Basics

Concept
Dimensions, equality, and the simple operations

A matrix is a rectangular array of numbers. Its order (dimension) is rows×columns\text{rows} \times \text{columns}; the entry in row ii, column jj is aija_{ij}.

  • Equal matrices: same order and every matching entry is equal.
  • Addition/subtraction: only for matrices of the same order; add or subtract entry by entry.
  • Scalar multiplication: multiply every entry by the scalar kk.

A matrix is a table of numbers with a fixed shape, and addition works entry by entry, so only tables of the same shape can be added (there must be an entry to pair with each entry). A scalar multiplies every entry because scaling a table scales everything in it, just as 3(a+b)=3a+3b3(a + b) = 3a + 3b.

Example
Worked example: combining matrices

Let A=[2103]A=\begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix} and B=[1−245]B=\begin{bmatrix} 1 & -2 \\ 4 & 5 \end{bmatrix} (both 2×22\times2).

A+B=[2+11+(−2)0+43+5]=[3−148],3A=[3(2)3(1)3(0)3(3)]=[6309].A+B = \begin{bmatrix} 2+1 & 1+(-2) \\ 0+4 & 3+5 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 4 & 8 \end{bmatrix}, \qquad 3A = \begin{bmatrix} 3(2) & 3(1) \\ 3(0) & 3(3) \end{bmatrix} = \begin{bmatrix} 6 & 3 \\ 0 & 9 \end{bmatrix}.

Equal matrices: if [x31y]=[231−5]\begin{bmatrix} x & 3 \\ 1 & y \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 1 & -5 \end{bmatrix}, then x=2x = 2 and y=−5y = -5.

Side note
Where it goes wrong

Adding matrices of different orders by padding with zeros: a 2×32 \times 3 plus a 3×23 \times 2 is undefined, not a 3×33 \times 3. Sums need matching positions.

Tip

Tip: You can only add or subtract matrices of identical order. A 2×32\times3 and a 3×23\times2 cannot be added.

Try it
Try it: combine

For A=[2103]A = \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix} and B=[1−245]B = \begin{bmatrix} 1 & -2 \\ 4 & 5 \end{bmatrix}, find 2A−B2A - B.

Show the answer

2A=[4206]2A = \begin{bmatrix} 4 & 2 \\ 0 & 6 \end{bmatrix}, and subtracting BB entry by entry gives [34−41]\begin{bmatrix} 3 & 4 \\ -4 & 1 \end{bmatrix}.

Matrix Multiplication

Concept
When is it defined, and what size is the product?

The product ABAB is defined only when the number of columns of AA equals the number of rows of BB:

(m×n)⋅(n×p)=(m×p).(m \times \mathbf{n}) \cdot (\mathbf{n} \times p) = (m \times p).

Each entry of ABAB is a row of AA dotted with a column of BB: multiply matching entries and add. Matrix multiplication is not commutative: usually AB≠BAAB \ne BA.

The row-times-column rule comes from substitution. If BB turns (s,t)(s, t) into new variables and AA turns those into others, then doing BB and then AA is a single substitution whose coefficients are exactly the row-by-column sums. That is also why order matters: substituting in the other order gives a different result.

Example
Worked example: full row-by-column product
A1234B5678=19224350row 1 of A · column 1 of B: (1)(5) + (2)(7) = 19
Each entry of ABAB pairs a row of AA with a column of BB.

Compute ABAB for A=[1234]A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} and B=[5678]B=\begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}. Both are 2×22\times2, so ABAB is 2×22\times2.

AB=[(1)(5)+(2)(7)(1)(6)+(2)(8)(3)(5)+(4)(7)(3)(6)+(4)(8)]=[5+146+1615+2818+32]=[19224350].AB = \begin{bmatrix} (1)(5)+(2)(7) & (1)(6)+(2)(8) \\[2pt] (3)(5)+(4)(7) & (3)(6)+(4)(8) \end{bmatrix} = \begin{bmatrix} 5+14 & 6+16 \\[2pt] 15+28 & 18+32 \end{bmatrix} = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix}.

Read it as: (row 1 of AA)⋅\cdot(column 1 of BB) gives the top-left entry, and so on.

Side note
Where it goes wrong

Multiplying entry by entry: [1234][5678]=[5122132]\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} = \begin{bmatrix} 5 & 12 \\ 21 & 32 \end{bmatrix}. Matrix multiplication pairs rows with columns: the top-left entry is 1⋅5+2⋅7=191 \cdot 5 + 2 \cdot 7 = 19.

Tip

Tip: Before multiplying, write the orders side by side. If the two inner numbers match, the product exists; the two outer numbers give its size.

Try it
Try it: order matters

For the same AA and BB, compute BABA and compare it with AB=[19224350]AB = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix}.

Show the answer

BA=[5⋅1+6⋅35⋅2+6⋅47⋅1+8⋅37⋅2+8⋅4]=[23343146]≠ABBA = \begin{bmatrix} 5 \cdot 1 + 6 \cdot 3 & 5 \cdot 2 + 6 \cdot 4 \\ 7 \cdot 1 + 8 \cdot 3 & 7 \cdot 2 + 8 \cdot 4 \end{bmatrix} = \begin{bmatrix} 23 & 34 \\ 31 & 46 \end{bmatrix} \ne AB.

Determinants: 2×22\times2 and 3×33\times3

Concept
The determinant of a square matrix
(3, 1)(1, 2)area 5
The columns (3,1)(3, 1) and (1,2)(1, 2) span a parallelogram of area 55.

For a 2×22\times2 matrix, the determinant is

det⁡[abcd]=ad−bc.\det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc.

For a 3×33\times3 matrix, expand along the first row (note the alternating + − ++\,-\,+ signs):

det⁡[abcdefghi]=a (ei−fh)−b (di−fg)+c (dh−eg).\det\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} = a\,(ei - fh) - b\,(di - fg) + c\,(dh - eg).

A 2×22 \times 2 determinant measures area: the columns (a,c)(a, c) and (b,d)(b, d) span a parallelogram whose area is ∣ad−bc∣|ad - bc|. For [3112]\begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix} the columns (3,1)(3, 1) and (1,2)(1, 2) span area 3⋅2−1⋅1=53 \cdot 2 - 1 \cdot 1 = 5. A determinant of 00 means the parallelogram is flat, the columns lie on one line, which is exactly when the matching system has no unique solution.

Example
Worked example: a 3×33\times3 determinant

Compute det⁡[213102415]\det\begin{bmatrix} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 4 & 1 & 5 \end{bmatrix} by expanding along the top row:

=2 (0⋅5−2⋅1)  −  1 (1⋅5−2⋅4)  +  3 (1⋅1−0⋅4)=2(0−2)−1(5−8)+3(1−0)=2(−2)−1(−3)+3(1)=−4+3+3=2.\begin{aligned} &= 2\,(0\cdot 5 - 2\cdot 1) \;-\; 1\,(1\cdot 5 - 2\cdot 4) \;+\; 3\,(1\cdot 1 - 0\cdot 4) \\ &= 2(0-2) - 1(5-8) + 3(1-0) \\ &= 2(-2) - 1(-3) + 3(1) = -4 + 3 + 3 = 2. \end{aligned}
Side note
Where it goes wrong

Forgetting the alternating signs in a 3×33 \times 3 expansion: adding all three terms. The middle term along the top row is subtracted: a(…)−b(…)+c(…)a(\ldots) - b(\ldots) + c(\ldots).

Tip

Tip: Expanding along a row or column that contains zeros saves work, because those terms vanish. Remember the checkerboard sign pattern +−+−+−+−+\begin{smallmatrix}+&-&+\\-&+&-\\+&-&+\end{smallmatrix}.

Try it
Try it: determinants

Find det⁡[4−235]\det\begin{bmatrix} 4 & -2 \\ 3 & 5 \end{bmatrix} and det⁡[102310021]\det\begin{bmatrix} 1 & 0 & 2 \\ 3 & 1 & 0 \\ 0 & 2 & 1 \end{bmatrix}.

Show the answer

4⋅5−(−2)(3)=20+6=264 \cdot 5 - (-2)(3) = 20 + 6 = 26. Along the top row: 1(1⋅1−0⋅2)−0(…)+2(3⋅2−1⋅0)=1+12=131(1 \cdot 1 - 0 \cdot 2) - 0(\ldots) + 2(3 \cdot 2 - 1 \cdot 0) = 1 + 12 = 13.

Solving Systems with Determinants: Cramer's Rule

Concept
Cramer's Rule for a 2×22\times2 system

For {a1x+b1y=c1a2x+b2y=c2\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases} form the coefficient determinant

D=det⁡[a1b1a2b2].D = \det\begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}.

Replace the xx-column with the constants to get DxD_x, and the yy-column to get DyD_y:

Dx=det⁡[c1b1c2b2],Dy=det⁡[a1c1a2c2].D_x = \det\begin{bmatrix} c_1 & b_1 \\ c_2 & b_2 \end{bmatrix}, \qquad D_y = \det\begin{bmatrix} a_1 & c_1 \\ a_2 & c_2 \end{bmatrix}.

Then, provided D≠0D \ne 0,   x=DxD,y=DyD.x = \dfrac{D_x}{D}, \quad y = \dfrac{D_y}{D}. If D=0D=0, Cramer's Rule does not apply (no unique solution).

Cramer's rule is elimination done once in general: eliminating yy from a1x+b1y=c1a_1x + b_1y = c_1 and a2x+b2y=c2a_2x + b_2y = c_2 gives (a1b2−a2b1)x=c1b2−c2b1(a_1b_2 - a_2b_1)x = c_1b_2 - c_2b_1, which is D⋅x=DxD \cdot x = D_x. When D=0D = 0 that equation cannot be divided through, and the system has no solution or infinitely many.

Example
Worked example: Cramer's Rule

Solve {2x+3y=12x−y=1\begin{cases} 2x + 3y = 12 \\ x - y = 1 \end{cases}

D=det⁡[231−1]=(2)(−1)−(1)(3)=−5,Dx=det⁡[1231−1]=−12−3=−15,D = \det\begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix} = (2)(-1)-(1)(3) = -5, \quad D_x = \det\begin{bmatrix} 12 & 3 \\ 1 & -1 \end{bmatrix} = -12-3 = -15,
Dy=det⁡[21211]=2−12=−10.D_y = \det\begin{bmatrix} 2 & 12 \\ 1 & 1 \end{bmatrix} = 2-12 = -10.

So x=−15−5=3x = \dfrac{-15}{-5} = 3 and y=−10−5=2y = \dfrac{-10}{-5} = 2. Solution: (3,2)(3,2). Check: 2(3)+3(2)=122(3)+3(2)=12 ✓, 3−2=13-2=1 ✓.

Side note
Where it goes wrong

Replacing the wrong column: for DxD_x the constants replace the xx-column (the first), not the yy-column. Swapping them computes DyD_y and returns yy where xx was wanted.

Tip

Tip: Keep the columns in order. DxD_x swaps only the xx-column for the constants; everything else stays put.

Try it
Try it: Cramer

Use Cramer's rule on {3x+2y=7x−4y=7\begin{cases} 3x + 2y = 7 \\ x - 4y = 7 \end{cases}.

Show the answer

D=3(−4)−2(1)=−14D = 3(-4) - 2(1) = -14. Dx=7(−4)−2(7)=−42D_x = 7(-4) - 2(7) = -42. Dy=3(7)−7(1)=14D_y = 3(7) - 7(1) = 14. So x=−42−14=3x = \tfrac{-42}{-14} = 3, y=14−14=−1y = \tfrac{14}{-14} = -1. Check: 9−2=79 - 2 = 7 and 3+4=73 + 4 = 7.

Applications: Setting Up Systems

Concept
From words to equations

To model a word problem:

  • Name each unknown with a variable (one per unknown quantity).
  • Translate each sentence of information into one equation. You need as many equations as unknowns.
  • Solve by any method, then interpret the answer in context and check it against the words.

Each independent fact in a word problem is one equation, and each unknown needs one fact to pin it down. That count, equations equal to unknowns, is why a two-unknown problem needs two facts and a coin problem with three kinds of coins needs three.

Example
Worked example: a ticket problem

A theater sells 100100 tickets for a total of $680. Adult tickets cost $8 and child tickets cost $5. How many of each were sold?

Let aa = adult tickets, cc = child tickets. Then

{a+c=1008a+5c=680\begin{cases} a + c = 100 \\ 8a + 5c = 680 \end{cases}

From the first equation c=100−ac = 100 - a. Substitute: 8a+5(100−a)=680⇒8a+500−5a=680⇒3a=180⇒a=60.8a + 5(100-a) = 680 \Rightarrow 8a + 500 - 5a = 680 \Rightarrow 3a = 180 \Rightarrow a = 60. Then c=40.c = 40.

Answer: 6060 adult and 4040 child tickets. Check: 60+40=10060+40=100 ✓ and 8(60)+5(40)=480+200=6808(60)+5(40)=480+200=680 ✓.

Side note
Where it goes wrong

Mixing units: writing 8a+5c=6808a + 5c = 680 correctly but pairing it with a+c=680a + c = 680. One equation counts tickets, the other counts dollars; the totals are 100100 tickets and 680680 dollars.

Tip

Tip: Count your unknowns first. Two unknowns need two independent facts; three unknowns (say nickels, dimes, quarters) need three.

Try it
Try it: three unknowns

A jar holds 3030 coins (nickels, dimes, quarters) worth $3.50\$3.50. There are twice as many dimes as nickels. How many of each?

Show the answer

Let n,d,qn, d, q. Count: n+d+q=30n + d + q = 30. Value in cents: 5n+10d+25q=3505n + 10d + 25q = 350. Relation: d=2nd = 2n. Then 3n+q=303n + q = 30 and 25n+25q=35025n + 25q = 350, so n+q=14n + q = 14. Subtract: 2n=162n = 16, n=8n = 8, d=16d = 16, q=6q = 6. Check: 40+160+150=35040 + 160 + 150 = 350.

Three Equations, Three Unknowns

Tip
The strategy is always "reduce the count"

A 3×33\times3 system is solved by turning it into a 2×22\times2 system.

1. Pick one variable to eliminate — the one with the friendliest coefficients.

2. Use the FIRST equation with the second, and the first with the third, to remove it. You now have two equations in two unknowns.

3. Solve that 2×22\times2 system by the usual means.

4. Back-substitute twice to recover the third unknown.

5. Check the triple in all three original equations — not just the two you finished with.

Example
Worked: a full 3×3

Solve ;x+y+z=6;x + y + z = 6, ;2x−y+z=3;2x - y + z = 3, ;x+2y−z=2;x + 2y - z = 2.

Step 1 — choose. Eliminate zz; its coefficients are 11, 11, −1-1, which is as easy as it gets.

Step 2 — first with second. Subtracting: (x+y+z)−(2x−y+z)=6−3(x + y + z) - (2x - y + z) = 6 - 3 gives −x+2y=3-x + 2y = 3.

Step 3 — first with third. Adding: (x+y+z)+(x+2y−z)=6+2(x + y + z) + (x + 2y - z) = 6 + 2 gives 2x+3y=82x + 3y = 8.

Step 4 — solve the 2×2. From the first, x=2y−3x = 2y - 3. Substituting: 2(2y−3)+3y=82(2y - 3) + 3y = 8, so 7y=147y = 14 and y=2y = 2, hence x=1x = 1.

Step 5 — back-substitute. 1+2+z=61 + 2 + z = 6, so z=3z = 3.

Answer. (1,2,3)(1, 2, 3).

Check. 1+2+3=61+2+3 = 6 ✓, 2−2+3=32 - 2 + 3 = 3 ✓, 1+4−3=21 + 4 - 3 = 2 ✓.

Side note
Use the same equation twice

Eliminate the chosen variable using ONE equation paired with each of the other two. Pairing the second with the third instead produces an equation you could have derived from the two you already have, and you end up going in a circle.

Row Reduction Without the Mystery

Concept
The matrix is just the equations with the letters left out

Writing ;2x+3y=7;2x + 3y = 7, ;x−y=1;;x - y = 1; as [2371−11]\left[\begin{array}{cc|c} 2 & 3 & 7 \\ 1 & -1 & 1\end{array}\right] throws away nothing except the symbols xx, yy and ==, which never change. The bar marks where the equals signs were. Every row is one equation, every column one unknown — so a row operation IS an equation move, done with less writing.

Tip
The only three legal moves

Swap two rows — writing the equations in a different order.

Scale a row by a nonzero constant — multiplying an equation through.

Add a multiple of one row to another — the elimination step.

Each is reversible, so none of them can gain or lose a solution. That reversibility is the entire justification for the method, and it is why a zero constant is forbidden in the scaling move: multiplying an equation by 00 cannot be undone.

Example
Worked: reduce a 3×3 all the way

Solve ;x+2y+z=8;x + 2y + z = 8, ;2x+y−z=1;2x + y - z = 1, ;3x−y+2z=7;3x - y + 2z = 7.

Set up. [121821−113−127]\left[\begin{array}{ccc|c} 1 & 2 & 1 & 8 \\ 2 & 1 & -1 & 1 \\ 3 & -1 & 2 & 7\end{array}\right]

Step 1 — clear the first column. R2−2R1R_2 - 2R_1 and R3−3R1R_3 - 3R_1 give [12180−3−3−150−7−1−17]\left[\begin{array}{ccc|c} 1 & 2 & 1 & 8 \\ 0 & -3 & -3 & -15 \\ 0 & -7 & -1 & -17\end{array}\right]

Step 2 — tidy the pivot. −13R2-\tfrac13 R_2 gives the row [ 0  1  1  ∣ 5 ][\,0\ \ 1\ \ 1\ \,|\,5\,], i.e. y+z=5y + z = 5.

Step 3 — clear the second column below. R3+7R2R_3 + 7R_2 gives [ 0  0  6  ∣ 18 ][\,0\ \ 0\ \ 6\ \,|\,18\,], i.e. 6z=186z = 18 and z=3z = 3.

Step 4 — back-substitute up. y+3=5y + 3 = 5 gives y=2y = 2; x+4+3=8x + 4 + 3 = 8 gives x=1x = 1.

Answer. (1,2,3)(1, 2, 3).

Check. 1+4+3=81 + 4 + 3 = 8 ✓, 2+2−3=12 + 2 - 3 = 1 ✓, 3−2+6=73 - 2 + 6 = 7 ✓.

Side note
Carry the constant column

The commonest row-reduction error is performing an operation on the coefficient entries and forgetting the number after the bar. Treat each row as one object: whatever you do, do it to every entry in it.

Reading a Reduction That Did Not Finish

Tip
What the two bad rows mean

A row [ 0  0  0 ∣ c ][\,0\ \ 0\ \ 0\ |\ c\,] with c≠0c \ne 0 reads "0=c0 = c", which is false. The equations contradict each other: no solution.

A row [ 0  0  0 ∣ 0 ][\,0\ \ 0\ \ 0\ |\ 0\,] reads "0=00 = 0", which is true but empty. One equation was a consequence of the others, so there are fewer real constraints than unknowns: infinitely many solutions, described with a free parameter.

Example
Worked: infinitely many, written properly

Solve ;x+y+z=4;x + y + z = 4, ;2x+2y+2z=8;2x + 2y + 2z = 8, ;x−y=0;x - y = 0.

Step 1 — reduce. R2−2R1R_2 - 2R_1 gives a row of zeros: the second equation was twice the first and says nothing new.

Step 2 — what is left. x+y+z=4x + y + z = 4 and x=yx = y. Two equations, three unknowns, so one variable is free.

Step 3 — parametrise. Let z=tz = t. Then x=yx = y and 2x+t=42x + t = 4, so x=y=4−t2x = y = \dfrac{4 - t}{2}.

Answer. (4−t2, 4−t2, t)\left(\dfrac{4-t}{2},\ \dfrac{4-t}{2},\ t\right) for every real tt.

Check. At t=0t = 0: (2,2,0)(2, 2, 0) — and 2+2+0=42 + 2 + 0 = 4 ✓, 2−2=02 - 2 = 0 ✓. At t=4t = 4: (0,0,4)(0, 0, 4), also a solution ✓.

Example
Worked: no solution

Solve ;x+y+z=1;x + y + z = 1, ;2x+2y+2z=5;2x + 2y + 2z = 5.

Step 1 — reduce. R2−2R1R_2 - 2R_1 gives [ 0  0  0 ∣ 3 ][\,0\ \ 0\ \ 0\ |\ 3\,].

Step 2 — read it. That row says 0=30 = 3.

Answer. No solution — the two planes are parallel and distinct.

The tell before you start. The left sides are proportional (2:12 : 1) but the right sides are not (5:15 : 1). Whenever the coefficients scale and the constants do not, the system is inconsistent.

Determinants and Cramer's Rule

Tip
What the determinant measures

For [abcd]\begin{bmatrix} a & b \\ c & d\end{bmatrix} the determinant is ad−bcad - bc. Two readings of the same number:

Geometric. It is the factor by which the matrix scales area. Zero means the transformation squashes the plane onto a line, so it cannot be undone.

Algebraic. It is exactly the coefficient of xx that survives when you eliminate yy from the general system — which is why dividing by it is the last step of every 2×22\times2 solution, and why a zero determinant is the signal that no unique solution exists.

Example
Worked: Cramer's rule on a 2×2

Solve ;3x+2y=16;3x + 2y = 16, ;x−y=2;x - y = 2.

Step 1 — the determinant. D=(3)(−1)−(2)(1)=−5D = (3)(-1) - (2)(1) = -5. Nonzero, so there is one solution.

Step 2 — replace the xx-column with the constants. Dx=∣1622−1∣=−16−4=−20D_x = \begin{vmatrix} 16 & 2 \\ 2 & -1\end{vmatrix} = -16 - 4 = -20.

Step 3 — replace the yy-column. Dy=∣31612∣=6−16=−10D_y = \begin{vmatrix} 3 & 16 \\ 1 & 2\end{vmatrix} = 6 - 16 = -10.

Step 4 — divide. x=−20−5=4x = \dfrac{-20}{-5} = 4 and y=−10−5=2y = \dfrac{-10}{-5} = 2.

Check. 3(4)+2(2)=163(4) + 2(2) = 16 ✓ and 4−2=24 - 2 = 2 ✓.

Side note
When to reach for Cramer

For a 2×22\times2 system Cramer's rule and elimination cost about the same. For 3×33\times3 it needs four 3×33\times3 determinants, and beyond that it is hopeless — row reduction wins. Its real value is theoretical: it shows in one line that the solution is a RATIO of determinants, so it exists exactly when the bottom one is nonzero.

Systems That Are Not Lines

Concept
Substitution is the general tool

When one equation is a circle, a parabola or a hyperbola, elimination usually has nothing to cancel. Substitution still works: solve the linear equation for one variable and put it into the curved one. The result is a quadratic (or worse) in one variable, so expect up to TWO intersection points — and expect to have to check them.

Example
Worked: a line meets a circle

Solve ;x2+y2=25;;x^2 + y^2 = 25; and ;y=x+1;y = x + 1.

Step 1 — substitute. x2+(x+1)2=25x^2 + (x + 1)^2 = 25.

Step 2 — expand and tidy. x2+x2+2x+1=25x^2 + x^2 + 2x + 1 = 25, so 2x2+2x−24=02x^2 + 2x - 24 = 0 and x2+x−12=0x^2 + x - 12 = 0.

Step 3 — factor. (x+4)(x−3)=0(x + 4)(x - 3) = 0, so x=−4x = -4 or x=3x = 3.

Step 4 — get the partners. y=x+1y = x + 1 gives (−4,−3)(-4, -3) and (3,4)(3, 4).

Answer. Two points: (−4,−3)(-4, -3) and (3,4)(3, 4).

Check. 16+9=2516 + 9 = 25 ✓ and 9+16=259 + 16 = 25 ✓ — the line cuts through the circle, which matches the two solutions.

Example
Worked: a product-and-sum system

Solve ;x+y=7;;x + y = 7; and ;xy=12;xy = 12.

Step 1 — recognise the shape. A sum and a product of two numbers are exactly the coefficients of a quadratic: xx and yy are the roots of t2−7t+12=0t^2 - 7t + 12 = 0.

Step 2 — solve it. (t−3)(t−4)=0(t - 3)(t - 4) = 0, so the roots are 33 and 44.

Answer. (3,4)(3, 4) and (4,3)(4, 3) — the system is symmetric, so both orderings solve it.

Check. 3+4=73 + 4 = 7 ✓ and 3×4=123 \times 4 = 12 ✓.

Worth remembering. Any symmetric system in xx and yy can be rewritten in terms of s=x+ys = x + y and p=xyp = xy; for instance x2+y2=s2−2p=49−24=25x^2 + y^2 = s^2 - 2p = 49 - 24 = 25, with no need to find xx and yy first.

Side note
Squaring can invent solutions

If a step squares both sides — solving a system containing a square root, say — every candidate must be tested in the ORIGINAL equations. A pair that survives the algebra but fails the original is extraneous and must be discarded.

Going Deeper: Advanced Systems & Matrices

Concept
Symmetric systems and elementary symmetric functions

A system is symmetric if swapping x↔yx \leftrightarrow y leaves it unchanged, e.g. {x+y=sxy=p\begin{cases} x+y = s \\ xy = p \end{cases}. The two quantities

s=x+y(the sum),p=xy(the product)s = x+y \quad(\text{the \emph{sum}}), \qquad p = xy \quad(\text{the \emph{product}})

are the elementary symmetric functions. Any symmetric expression rewrites in terms of ss and pp; the key identities are

x2+y2=s2−2p,x2−xy+y2=s2−3p,x3+y3=s3−3ps.x^2+y^2 = s^2 - 2p, \qquad x^2 - xy + y^2 = s^2 - 3p, \qquad x^3+y^3 = s^3 - 3ps.

Once ss and pp are known, xx and yy are the roots of the single quadratic

t2−s t+p=0.t^2 - s\,t + p = 0.
Example
Worked example: solving a symmetric system

Solve {x+y=5x2+y2=13\begin{cases} x + y = 5 \\ x^2 + y^2 = 13 \end{cases}

Step 1 — get ss and pp. Here s=x+y=5s = x+y = 5. Use x2+y2=s2−2px^2+y^2 = s^2 - 2p:

13=52−2p ⇒ 13=25−2p ⇒ 2p=12 ⇒ p=6.13 = 5^2 - 2p \ \Rightarrow\ 13 = 25 - 2p \ \Rightarrow\ 2p = 12 \ \Rightarrow\ p = 6.

Step 2 — build the quadratic. With s=5s=5, p=6p=6, the values x,yx,y are the roots of

t2−5t+6=0 ⇒ (t−2)(t−3)=0 ⇒ t=2 or t=3.t^2 - 5t + 6 = 0 \ \Rightarrow\ (t-2)(t-3) = 0 \ \Rightarrow\ t = 2 \text{ or } t = 3.

Solution: (x,y)=(2,3)(x,y) = (2,3) or (3,2)(3,2). Check: 2+3=52+3=5 ✓ and 22+32=4+9=132^2+3^2 = 4+9 = 13 ✓.

Concept
Nonlinear systems: line–conic and conic–conic

When at least one equation is a conic (circle, parabola, ellipse, hyperbola), count solutions geometrically:

  • Line meets a conic: substitute the line into the conic to get one quadratic. The discriminant tells you 22 points (secant), 11 point (tangent), or 00 points (miss).
  • Two conics: up to 44 intersection points. Subtract the two equations to cancel the quadratic terms when possible — this often leaves a line (the radical axis for two circles), reducing the problem to a line–conic case.
Example
Worked example: two-conic system by subtraction

Solve {x2+y2=25x2+y2−6x−8y=−15\begin{cases} x^2 + y^2 = 25 \\ x^2 + y^2 - 6x - 8y = -15 \end{cases}

Step 1 — subtract to kill the squares. Equation (1) minus equation (2):

(x2+y2)−(x2+y2−6x−8y)=25−(−15) ⇒ 6x+8y=40 ⇒ 3x+4y=20.(x^2+y^2) - (x^2+y^2-6x-8y) = 25 - (-15) \ \Rightarrow\ 6x + 8y = 40 \ \Rightarrow\ 3x + 4y = 20.

This line is the radical axis. Solve it for xx: x=20−4y3x = \dfrac{20 - 4y}{3}.

Step 2 — substitute into the circle. Better: from 3x+4y=203x+4y=20 take x=20−4y3x = \frac{20-4y}{3} and plug into x2+y2=25x^2+y^2=25. Multiplying through by 99,

(20−4y)2+9y2=225 ⇒ 400−160y+16y2+9y2=225 ⇒ 25y2−160y+175=0.(20-4y)^2 + 9y^2 = 225 \ \Rightarrow\ 400 - 160y + 16y^2 + 9y^2 = 225 \ \Rightarrow\ 25y^2 - 160y + 175 = 0.

Divide by 55: 5y2−32y+35=0⇒(5y−7)(y−5)=0⇒y=755y^2 - 32y + 35 = 0 \Rightarrow (5y - 7)(y - 5) = 0 \Rightarrow y = \tfrac{7}{5} or y=5y = 5.

Step 3 — back-substitute. If y=5y = 5: x=20−203=0x = \frac{20-20}{3} = 0, giving (0,5)(0,5). If y=75y = \tfrac{7}{5}: x=20−28/53=72/53=245x = \frac{20 - 28/5}{3} = \frac{72/5}{3} = \tfrac{24}{5}, giving (245,75)\left(\tfrac{24}{5}, \tfrac{7}{5}\right).

Solutions: (0,5)(0,5) and (245,75)\left(\tfrac{24}{5}, \tfrac{7}{5}\right). Check (0,5)(0,5): 0+25=250+25=25 ✓.

Concept
The inverse of a 3×33\times3 matrix

For a square matrix AA, the inverse A−1A^{-1} satisfies AA−1=A−1A=IA A^{-1} = A^{-1} A = I. It exists precisely when det⁡A≠0\det A \ne 0. The adjugate formula is

A−1=1det⁡A adj⁡(A),A^{-1} = \frac{1}{\det A}\,\operatorname{adj}(A),

where adj⁡(A)\operatorname{adj}(A) is the transpose of the cofactor matrix. Building it takes three steps:

  • Cofactors: Cij=(−1)i+jMijC_{ij} = (-1)^{i+j} M_{ij}, where the minor MijM_{ij} is the determinant of AA with row ii and column jj deleted.
  • Adjugate: transpose the matrix of cofactors (adj⁡(A)ij=Cji\operatorname{adj}(A)_{ij} = C_{ji}).
  • Scale: divide every entry by det⁡A\det A.

Once A−1A^{-1} is known, the system Ax=bA\mathbf{x} = \mathbf{b} has the one-line solution x=A−1b\mathbf{x} = A^{-1}\mathbf{b}.

Example
Worked example: invert a 3×33\times3 and solve

Solve {x+2y+3z=5x+2y+3z=4x+2y+3z=3\begin{cases} x + 2y + 3z = 5 \\ \phantom{x + {}}2y + 3z = 4 \\ \phantom{x + 2y + {}}3z = 3 \end{cases} using the inverse of A=[123023003]A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{bmatrix}.

Step 1 — determinant. AA is upper triangular, so det⁡A=1⋅2⋅3=6≠0\det A = 1 \cdot 2 \cdot 3 = 6 \ne 0; the inverse exists.

Step 2 — cofactors and adjugate. Computing cofactors and transposing gives

adj⁡(A)=[6−6003−3002],A−1=16[6−6003−3002]=[1−10012−120013].\operatorname{adj}(A) = \begin{bmatrix} 6 & -6 & 0 \\ 0 & 3 & -3 \\ 0 & 0 & 2 \end{bmatrix}, \qquad A^{-1} = \frac{1}{6}\begin{bmatrix} 6 & -6 & 0 \\ 0 & 3 & -3 \\ 0 & 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ 0 & \tfrac12 & -\tfrac12 \\ 0 & 0 & \tfrac13 \end{bmatrix}.

Step 3 — multiply by b=[543]\mathbf{b} = \begin{bmatrix} 5 \\ 4 \\ 3 \end{bmatrix}.

x=A−1b=[1−10012−120013][543]=[5−42−321]=[1121].\mathbf{x} = A^{-1}\mathbf{b} = \begin{bmatrix} 1 & -1 & 0 \\ 0 & \tfrac12 & -\tfrac12 \\ 0 & 0 & \tfrac13 \end{bmatrix}\begin{bmatrix} 5 \\ 4 \\ 3 \end{bmatrix} = \begin{bmatrix} 5 - 4 \\ 2 - \tfrac32 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ \tfrac12 \\ 1 \end{bmatrix}.

Solution: x=1, y=12, z=1x = 1,\ y = \tfrac12,\ z = 1. Check the first equation: 1+2(12)+3(1)=1+1+3=51 + 2(\tfrac12) + 3(1) = 1+1+3 = 5 ✓.

Concept
Determinant properties and special identities

These rules let you evaluate determinants without brute force:

  • Row/column operations: adding a multiple of one row to another does not change det⁡\det; swapping two rows multiplies det⁡\det by −1-1; scaling a row by kk scales det⁡\det by kk.
  • Zero determinant: a repeated row, a zero row, or one row a multiple of another all force det⁡=0\det = 0.
  • Products & transposes: det⁡(AB)=det⁡A det⁡B\det(AB) = \det A \,\det B, det⁡(AT)=det⁡A\det(A^{\mathsf T}) = \det A, det⁡(A−1)=1/det⁡A\det(A^{-1}) = 1/\det A, det⁡(kA)=kndet⁡A\det(kA) = k^n \det A for an n×nn\times n matrix.
  • Vandermonde: det⁡[1aa21bb21cc2]=(b−a)(c−a)(c−b).\det\begin{bmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{bmatrix} = (b-a)(c-a)(c-b).
  • Rank-one update: det⁡(I+aJ)\det(I + aJ), where JJ is the n×nn\times n all-ones matrix, equals 1+na1 + na (since JJ has eigenvalues nn and 00).
Tip

Tip: The Vandermonde determinant is zero exactly when two of a,b,ca,b,c are equal — which is why nn distinct data points determine a unique degree-(n−1)(n-1) interpolating polynomial.

Concept
Singular matrices: when there is no inverse

A square matrix AA is singular (non-invertible) exactly when det⁡A=0\det A = 0. For the system Ax=bA\mathbf{x}=\mathbf{b} this means no unique solution — either no solution or infinitely many. Equivalent warning signs of singularity:

  • the rows (or columns) are linearly dependent — one is a combination of the others;
  • a row reduces to all zeros during elimination;
  • Cramer's Rule breaks because D=det⁡A=0D = \det A = 0.

To find the parameter that makes a matrix singular, set its determinant equal to zero and solve.

Example
Worked example: for which kk is the matrix singular?

Find all kk for which A=[k28k]A = \begin{bmatrix} k & 2 \\ 8 & k \end{bmatrix} has no inverse, and describe the system Ax=0A\mathbf{x} = \mathbf{0} there.

Singular means det⁡A=0\det A = 0:

det⁡A=k⋅k−(2)(8)=k2−16=0 ⇒ k=4 or k=−4.\det A = k\cdot k - (2)(8) = k^2 - 16 = 0 \ \Rightarrow\ k = 4 \text{ or } k = -4.

At k=4k = 4: A=[4284]A = \begin{bmatrix} 4 & 2 \\ 8 & 4 \end{bmatrix}, whose second row is twice the first — the rows are linearly dependent, confirming det⁡=0\det = 0. The homogeneous system then reduces to the single equation 4x+2y=04x + 2y = 0, i.e. y=−2xy = -2x, giving infinitely many solutions along that line rather than only x=0\mathbf{x}=\mathbf{0}.

Concept
Matrix powers AnA^n and recurrences

Repeated multiplication An=A⋅A⋯AA^n = A\cdot A \cdots A (nn factors) models step-by-step processes (Markov chains, Fibonacci-type recurrences). Two shortcuts:

  • Diagonal matrices power entrywise: diag⁡(a,b)n=diag⁡(an,bn)\operatorname{diag}(a,b)^n = \operatorname{diag}(a^n, b^n).
  • Cayley–Hamilton: every matrix satisfies its own characteristic equation. For a 2×22\times2 matrix, A2=(tr⁡A) A−(det⁡A) IA^2 = (\operatorname{tr}A)\,A - (\det A)\,I, which lets you reduce any power AnA^n to the form αA+βI\alpha A + \beta I.

The Fibonacci matrix illustrates the recurrence link:

[1110] ⁣n=[Fn+1FnFnFn−1].\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{\!n} = \begin{bmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{bmatrix}.
Tip

Tip: To compute a high power like A10A^{10}, use repeated squaring: A2, A4=(A2)2, A8=(A4)2A^2,\ A^4 = (A^2)^2,\ A^8 = (A^4)^2, then A10=A8A2A^{10} = A^8 A^2. That is 44 multiplications instead of 99.

Concept
Find x+y+zx+y+z without solving the whole system

Sometimes a problem asks only for a combination of the unknowns — not each one. Look for a shortcut before grinding out a full solution:

  • Add all equations. If the coefficients of x,y,zx,y,z each sum to the same value cc, then adding gives c(x+y+z)=(sum of constants)c(x+y+z) = (\text{sum of constants}), so x+y+zx+y+z pops out directly.
  • Take a linear combination of the equations designed to produce exactly the target combination.
Example
Worked example: get x+y+zx+y+z directly

Given {x+2y+3z=143x+y+2z=112x+3y+z=11\begin{cases} x + 2y + 3z = 14 \\ 3x + y + 2z = 11 \\ 2x + 3y + z = 11 \end{cases}, find x+y+zx+y+z without solving for each variable.

Add all three equations. On the left, each variable's coefficients sum to 1+3+2=61+3+2 = 6:

(1+3+2)x+(2+1+3)y+(3+2+1)z=14+11+11 ⇒ 6x+6y+6z=36.(1+3+2)x + (2+1+3)y + (3+2+1)z = 14 + 11 + 11 \ \Rightarrow\ 6x + 6y + 6z = 36.

Divide by 66:

x+y+z=6.x + y + z = 6.

Answer: x+y+z=6x+y+z = 6, obtained in one step — no elimination needed. (For the record the full solution is (1,2,3)(1,2,3), and indeed 1+2+3=61+2+3=6 ✓.)

Tip

Tip: Whenever a question asks for a symmetric combination like x+y+zx+y+z or xy+yz+zxxy+yz+zx, first check whether adding or symmetrically combining the equations delivers it immediately. It often does.

Why It Works

Row operations — swap two rows, multiply a row by a nonzero constant, add a multiple of one row to another — are exactly the three legal equation moves, so they never change the solution set. Reaching a row [0  0  0  ∣  c][0\;0\;0\;|\;c] with c≠0c \ne 0 means the equations contradict each other (no solution); a row of all zeros means one equation was redundant (infinitely many). The determinant is zero precisely when the rows are dependent, which is why it detects both cases at once.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
Determinants and Cramer’s rule
det⁡[abcd]=ad−bc,x=det⁡Axdet⁡A,y=det⁡Aydet⁡A\det\begin{bmatrix} a & b \\ c & d\end{bmatrix}=ad-bc,\qquad x=\frac{\det A_x}{\det A},\quad y=\frac{\det A_y}{\det A}

What it means. The determinant measures how a matrix scales area; when it is 00 the system has no unique solution.

Example. det⁡[2314]=2⋅4−3⋅1=5\det\begin{bmatrix}2&3\\1&4\end{bmatrix}=2\cdot4-3\cdot1=5.

Why it works. Solving ax+by=e, cx+dy=fax+by=e,\ cx+dy=f by elimination gives x=ed−bfad−bcx=\tfrac{ed-bf}{ad-bc}, and the numerator is the determinant of AA with the xx-column replaced by the constants. A zero determinant means the two lines are parallel or identical.

Tip. Check det⁡A≠0\det A\neq 0 before dividing — that is precisely the test for a unique solution.

Tip
Solving systems
Substitution or elimination;parallel⇒no solution;same line⇒infinitely many\text{Substitution or elimination};\quad \text{parallel} \Rightarrow \text{no solution};\quad \text{same line} \Rightarrow \text{infinitely many}

What it means. A solution is a point on both lines, so the answer count is how many times they meet.

Example. x+y=5x+y=5 and x−y=1x-y=1: adding gives 2x=62x=6, so x=3x=3, y=2y=2.

Why it works. Elimination adds a multiple of one equation to the other. Since both sides are equal quantities, the new equation still holds for every common solution — but one variable is gone.

Tip. If the variables all vanish and leave a false statement (0=50=5), there is no solution; a true one (0=00=0) means infinitely many.