Name: Solving the Systems Using Cramer's Rule
Uploaded: 2019-07-19
Duration: 15 min 2 s
Description: In this video, you will learn how to solve the system of linear equations with the help of cramer's rule? So let's watch this video to know more about this concept.

Question

Given the matrices

A = [\begin{array}{rrr} 1 & 2 & 3 \\ 3 & - 2 & 1 \\ 4 & 2 & 1 \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}] and C = [\begin{array}{l} 1 \\ 2 \\ 3 \end{array}]

, write down the linear equations given by

A X = C

and solve it for

x, y, z

. by matrix method.

Accepted Answer

Given the matrices $A = [\begin{array}{rrr} 1 & 2 & 3 \\ 3 & - 2 & 1 \\ 4 & 2 & 1 \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}] and C = [\begin{array}{l} 1 \\ 2 \\ 3 \end{array}]$

$A X = C \Rightarrow [\begin{array}{rrr} 1 & 2 & 3 \\ 3 & - 2 & 1 \\ 4 & 2 & 1 \end{array}] [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{array}{l} 1 \\ 2 \\ 3 \end{array}]$

$\Rightarrow [\begin{array}{l} x + 2 y + 3 z \\ 3 x - 2 y + z \\ 4 x + 2 y + z \end{array}] = [\begin{array}{l} 1 \\ 2 \\ 3 \end{array}]$

$\Rightarrow x + 2 y + 3 z = 1, 3 x - 2 y + z = 2, 4 x + 2 y + z = 3$ which are the required linear equations.

The given system of equations is in the form $A X = B$ , where $A = {[a_{i j}]}_{3 \times 3} = [\begin{matrix} 1 & 2 & 3 \\ 3 & - 2 & 1 \\ 4 & 2 & 1 \end{matrix}]$ , $X = [\begin{matrix} x \\ y \\ z \end{matrix}]$ and $B = [\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}]$

Now, $|A| = 1 (- 2 - 2) - 2 (3 - 4) + 3 (6 + 8) = 40$

So, the given system is consistent and has a unique solution.

$A_{i j}$ is the cofactor of the $a_{i j}$

$A_{11} = (- 1)^{1 + 1} |\begin{matrix} - 2 & 1 \\ 2 & 1 \end{matrix}| = - 4$

$A_{12} = (- 1)^{1 + 2} |\begin{array}{l} 3 & 1 \\ 4 & 1 \end{array}| = 1$

$A_{13} = (- 1)^{1 + 3} |\begin{matrix} 3 & - 2 \\ 4 & 2 \end{matrix}| = 14$

$A_{21} = (- 1)^{2 + 1} | \begin{array}{ll} 2 & 3 \\ 2 & 1 \end{array} | = 4$

$A_{22} = (- 1)^{2 + 2} | \begin{array}{ll} 1 & 3 \\ 4 & 1 \end{array} | = - 11$

$A_{23} = (- 1)^{2 + 3} | \begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array} | = 6$

$A_{31} = (- 1)^{3 + 1} | \begin{array}{cc} 2 & 3 \\ - 2 & 1 \end{array} | = 8$

$A_{32} = (- 1)^{3 + 1} | \begin{array}{cc} 2 & 3 \\ - 2 & 1 \end{array} | = 8$

$A_{33} = (- 1)^{3 + 3} | \begin{array}{cc} 1 & 2 \\ 3 & - 2 \end{array} | = - 8$

$∴ A d j A = [\begin{array}{ccc} - 4 & 4 & 8 \\ 1 & - 11 & 8 \\ 14 & 6 & - 8 \end{array}]$

The solution of $A X = C$ is $X = A^{- 1} C$ .

$X = \frac{A d j A}{|A|} C$

$X = \frac{1}{40} [\begin{array}{ccc} - 4 & 4 & 8 \\ 1 & - 11 & 8 \\ 14 & 6 & - 8 \end{array}] [\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}]$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{40} [\begin{matrix} (- 4) \times 1 + 4 \times 2 + 8 \times 3 \\ 1 \times 1 + (- 11) \times 2 + 8 \times 3 \\ 14 \times 1 + 6 \times 2 + (- 8) \times 3 \end{matrix}]$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 7 / 10 \\ 3 / 40 \\ 1 / 20 \end{matrix}]$

$\Rightarrow x = \frac{7}{10}, y = \frac{3}{40}, z = \frac{1}{20}$

Given the matrices $A = [\begin{array}{rrr} 1 & 2 & 3 \\ 3 & - 2 & 1 \\ 4 & 2 & 1 \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}] and C = [\begin{array}{l} 1 \\ 2 \\ 3 \end{array}]$ , write down the linear equations given by $A X = C$ and solve it for $x, y, z$ . by matrix method.

Important Questions on Determinants

Learn and Prepare for any exam you want !

Given the matrices A=[1233−21421],X=[xyz] and C=[123], write down the linear equations given by AX=C and solve it for x,y,z. by matrix method.

Important Questions on Determinants

Learn and Prepare for any exam you want !

Given the matrices $A = [\begin{array}{rrr} 1 & 2 & 3 \\ 3 & - 2 & 1 \\ 4 & 2 & 1 \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}] and C = [\begin{array}{l} 1 \\ 2 \\ 3 \end{array}]$ , write down the linear equations given by $A X = C$ and solve it for $x, y, z$ . by matrix method.