Question

we wish to solve the following system of linear equations by using cramers rule (the method of determinants).

\\\

$$\begin{cases} 3x - y = -4 \\\\ -7x + 2y + z = 0 \\\\ -7x + 3z = 1 \\end{cases}$$

\\

begin by calculating these four determinants:

(i) \\(\text{det} \

$$\begin{bmatrix} -4 & -1 & 0 \\\\ 0 & 2 & 1 \\\\ 1 & 0 & 3 \\end{bmatrix}$$

= \\)

(ii) \\(\text{det} \

$$\begin{bmatrix} 3 & -4 & 0 \\\\ -7 & 0 & 1 \\\\ -7 & 1 & 3 \\end{bmatrix}$$

= \\)

(iii) \\(\text{det} \

$$\begin{bmatrix} 3 & -1 & -4 \\\\ -7 & 2 & 0 \\\\ -7 & 0 & 1 \\end{bmatrix}$$

= \\)

(iv) \\(\text{det} \

$$\begin{bmatrix} 3 & -1 & 0 \\\\ -7 & 2 & 1 \\\\ -7 & 0 & 3 \\end{bmatrix}$$

= \\)

now divide these determinants appropriately to find \\(x\\), \\(y\\), and \\(z\\), rounding to two decimal places where necessary.

Explanation:

Calculate the determinant of the coefficient matrix (iv)

Using the Determinant and Cofactor Expansion knowledge points

$$ D = \det
LATEXBLOCK0
$$

Expand along the first row:

$$ D = 3
LATEXBLOCK1
- (-1)
LATEXBLOCK2
+ 0 $$

$$ D = 3(6 - 0) + 1(-21 - (-7)) = 18 - 14 = 4 $$

Calculate the determinant for x (i)

Using the Determinant and Cofactor Expansion knowledge points

$$ D_x = \det
LATEXBLOCK3
$$

Expand along the first row:

$$ D_x = -4
LATEXBLOCK4
- (-1)
LATEXBLOCK5
+ 0 $$

$$ D_x = -4(6 - 0) + 1(0 - 1) = -24 - 1 = -25 $$

Calculate the determinant for y (ii)

Using the Determinant and Cofactor Expansion knowledge points

$$ D_y = \det
LATEXBLOCK6
$$

Expand along the first row:

$$ D_y = 3
LATEXBLOCK7
- (-4)
LATEXBLOCK8
+ 0 $$

$$ D_y = 3(0 - 1) + 4(-21 - (-7)) = -3 + 4(-14) = -3 - 56 = -59 $$

Calculate the determinant for z (iii)

Using the Determinant and Cofactor Expansion knowledge points

$$ D_z = \det
LATEXBLOCK9
$$

Expand along the third row:

$$ D_z = -7
LATEXBLOCK10
- 0 + 1
LATEXBLOCK11
$$

$$ D_z = -7(0 - (-8)) + 1(6 - 7) = -7(8) - 1 = -56 - 1 = -57 $$

Solve for the variables using Cramer's Rule

We now introduce Cramer's Rule, which states that each variable is found by dividing its corresponding coordinate determinant by the main coefficient determinant \(D\):

$$ x = \frac{D_x}{D} = \frac{-25}{4} = -6.25 $$

$$ y = \frac{D_y}{D} = \frac{-59}{4} = -14.75 $$

$$ z = \frac{D_z}{D} = \frac{-57}{4} = -14.25 $$

Answer:

No.	Problem	Answer

| (i) | \(\det

$$\begin{bmatrix} -4 & -1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 3 \end{bmatrix}$$

\) | \(-25\) |
| (ii) | \(\det

$$\begin{bmatrix} 3 & -4 & 0 \\ -7 & 0 & 1 \\ -7 & 1 & 3 \end{bmatrix}$$

\) | \(-59\) |
| (iii) | \(\det

$$\begin{bmatrix} 3 & -1 & -4 \\ -7 & 2 & 0 \\ -7 & 0 & 1 \end{bmatrix}$$

\) | \(-57\) |
| (iv) | \(\det

$$\begin{bmatrix} 3 & -1 & 0 \\ -7 & 2 & 1 \\ -7 & 0 & 3 \end{bmatrix}$$

\) | \(4\) |

Solutions

Values of \(x, y, z\)

\(x = -6.25\), \(y = -14.75\), \(z = -14.25\)