IB Mathematics AA Systems of Liner Equations Study Notes | New Syllabus

From equation (1): \( z = 6 – x – y \)

Substitute into (2) and (3) and simplify:

(2): \( 2x – y + 3(6 – x – y) = 14 \)

→ \( 2x – y + 18 – 3x – 3y = 14 \)

→ \( -x – 4y = -4 \) → (4)

(3): \( -x + 4y + (6 – x – y) = -2 \)

→ \( -2x + 3y + 6 = -2 \)

→ \( -2x + 3y = -8 \) → (5)

Solve (4) and (5):

(4): \( x = -4y + 4 \)

Substitute into (5): \( -2(-4y + 4) + 3y = -8 \)

→ \( 8y – 8 + 3y = -8 \)

→ \( 11y = 0 \) → \( y = 0 \)

Then \( x = 4 \), \( z = 6 – 4 – 0 = 2 \)

Solution: \( x = 4, y = 0, z = 2 \)

Form matrix:

Step 1: Write the Augmented Matrix

$
\left[
\begin{array}{ccc|c}
1 & -1 & 1 & 8 \\
2 & 3 & -1 & -2 \\
3 & -2 & -9 & 9 \\
\end{array}
\right]
$

Step 2: Row Operations to Row-Echelon Form
First, eliminate below the pivot in column 1:

$
\begin{aligned}
R_2 &\leftarrow R_2 – 2R_1 \\
R_3 &\leftarrow R_3 – 3R_1
\end{aligned}
\Rightarrow
\left[
\begin{array}{ccc|c}
1 & -1 & 1 & 8 \\
0 & 5 & -3 & -18 \\
0 & 1 & -12 & -15 \\
\end{array}
\right]
$

Swap \( R_2 \leftrightarrow R_3 \) to get a leading 1 in the second row:

$
\left[
\begin{array}{ccc|c}
1 & -1 & 1 & 8 \\
0 & 1 & -12 & -15 \\
0 & 5 & -3 & -18 \\
\end{array}
\right]
$

Next, eliminate the 5 below the pivot in column 2:

$
R_3 \leftarrow R_3 – 5R_2
\Rightarrow
\left[
\begin{array}{ccc|c}
1 & -1 & 1 & 8 \\
0 & 1 & -12 & -15 \\
0 & 0 & 57 & 57 \\
\end{array}
\right]
$

Normalize the third row:

$
R_3 \leftarrow \frac{1}{57} R_3
\Rightarrow
\left[
\begin{array}{ccc|c}
1 & -1 & 1 & 8 \\
0 & 1 & -12 & -15 \\
0 & 0 & 1 & 1 \\
\end{array}
\right]
$

Step 3: Back Substitution

From row 3: \( z = 1 \) 
From row 2: \( y – 12z = -15 \Rightarrow y = -3 \) 
From row 1: \( x – y + z = 8 \Rightarrow x = 4 \)

$
\rm{(x, y, z) = (4, -3, 1)}
$

Press [2nd], then [x⁻¹] to open the MATRIX menu.

Go to EDIT and choose matrix [A].

Enter a 3×4 matrix:

[2 1 1 | 4]

[3 2 3 | 13]

[1 1 1 | 6]

Quit to the Home screen [2nd] → [MODE].

Go to MATRIX → MATH → rref( and apply it to matrix [A].

The reduced matrix will be:

[1 0 0 | 1]

[0 1 0 | 2]

[0 0 1 | 3]

Solution:

\( x = 1 \) ,\( y = 2 \) , \( z = 3 \)