Question

Solve the system of linear equations x – y = 3, 2x + 3y + 4z = 17 and y + 2z = 7 using cramer rule​

Answer 1

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Answer 2

Given:

Given equations are -

x - y = 3

2x + 3y + 4z = 17

y + 2z = 7

To find:

Values of x, y, z

Solution:

The given equation are -

x - y = 3        -(1)

a₁ = 1

b₁ = -1

m₁ = 0

c₁ = 3

2x + 3y + 4z = 17       -(2)

a₂ = 2

b₂ = 3

m₂ = 4

c₂ = 17

y + 2z = 7      -(3)

a₃ = 0

b₃ = 1

m₃ = 2

c₃ = 7

here a, b, m are coefficient of x, y and z respectively. c are constants.

Using Cramer's rule,

$D = \left[\begin{array}{ccc}a_1&b_1&m_1\\a_2&b_2&m_2\\a_3&b_3&m_3\end{array}\right]$

putting all values, we get

$D = \left[\begin{array}{ccc}1&-1&0\\2&3&4\\0&1&2\end{array}\right]$

D = 1(6 - 4) - (-1)(4-0) + 0(2-0)

D = 2 + 4

D = 6

now, solve

$D_x=\left[\begin{array}{ccc}c_1&b_1&m_1\\c_2&b_2&m_2\\c_3&b_3&m_3\end{array}\right]$

putting all values, we get

  $D_x = \left[\begin{array}{ccc}3&-1&0\\17&3&4\\7&1&2\end{array}\right]$

$D_x$ = 3(6-4) -(-1)(34-28) + 0(17-21)

    = 6 + 6

    = 12

$D_y = \left[\begin{array}{ccc}a_1&c_1&m_1\\a_2&c_2&m_2\\a_3&c_3&m_3\end{array}\right]$

putting all values, we get

$D_y = \left[\begin{array}{ccc}1&3&0\\2&17&4\\0&7&2\end{array}\right]$

$D_y$ = 1(34-28) -3(4-0) +0(14-0)

    = 6 -12

    = -6

and,

$D_z = \left[\begin{array}{ccc}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{array}\right]$

putting all values, we get

$D_z = \left[\begin{array}{ccc}1&-1&3\\2&3&17\\0&1&7\\\end{array}\right]$

$D_z$ = 1(21-17) -(-1)(14-0) +3(2-0)

    = 4 + 14 + 6

    = 24

so, by Cramer's rule

$x = \frac{D_x}{D} = \frac{12}{6}$

x = 2

$y = \frac{D_y}{D} = \frac{-6}{6}$

y = -1

$z = \frac{D_z}{D} = \frac{24}{6}$

z = 4

Therefore, value of x = 2, y = -1 and z = 4.