SME
Grade 12 Math · Precalculus
Systems in three variables, matrices and determinants
Use this for “Solve x + y + z = 6, 2x − y + z = 3 and x + 2y − z = 2” with an augmented matrix.
Diagram
3 equations in 3 unknowns, the right-hand sides after the bar.
The solution: x = 1, y = 2, z = 3.
This is an example. Type your own values to start a new problem.
1
2
1
1
−1
2
1
1
−1
d₁
d₂
d₃
x
=
x
,
y
=
y
,
z
=
z
Formulas
−7
×
z
=
d₂
−
5
×
d₁
+
3
×
d₃
−7 × 3 = 3 − 5 × 6 + 3 × 2
y
−
2
×
z
=
d₃
−
d₁
2 − 2 × 3 = 2 − 6
x
+
y
+
z
=
d₁
1 + 2 + 3 = 6
Assumptions
•
The system x + y + z = d₁, 2x − y + z = d₂, x + 2y − z = d₃ as an augmented matrix; type the right sides.
•
Each row operation keeps the same solutions; the goal is zeros under the diagonal, then back-substitution.
•
A row that reads 0 = a number that is not 0 means the system has no solution.
Step-by-step
WE KNOW
d₁ = 6, d₂ = 3, d₃ = 2
FIND
z (z), y (y), x (x)
Step 1 · Find z (z)
Use −7 × z = d₂ − 5 × d₁ + 3 × d₃
Row
operations:
R₂
−
2R₁
and
R₃
−
R₁,
swap
those
two
rows,
then
R₃
+
3R₂;
the
last
row
is
−
7
z
=
d₂
−
5
d₁
+
3
d₃
,
so
divide
by
−7.
z
=
d₂
− 5 ×
d₁
+ 3 ×
d₃
−7
z
=
3 − 5 × 6 + 3 × 2
−7
z
=
3 − 30 + 6
−7
z
=
−
21
−7
z
=
3
Step 2 · Find y (y)
Use y − 2 × z = d₃ − d₁
Put
z
into
the
second
row
(after
the
swap)
and
add
2
z
to
both
sides.
y
=
d₃
−
d₁
+
2
×
z
y
=
2
−
6
+
2
×
3
y
=
2
−
6
+
6
y
=
2
Step 3 · Find x (x)
Use x + y + z = d₁
Put
y
and
z
into
the
first
row
and
take
them
from
both
sides.
x
=
d₁
−
y
−
z
x
=
6
−
2
−
3
x
=
1
Check
Put the numbers back in. Both sides should match.
−7 × 3 = 3 − 5 × 6 + 3 × 2 ✓
2 − 2 × 3 = 2 − 6 ✓
1 + 2 + 3 = 6 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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