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

Examining consistency and solvability, solve the following equations by matrix method.

$x - y + z = 4$

$2 x + y - 3 z = 0$

$x + y + z = 2$

Accepted Answer

The given system of equations is given in the form $x - y + z = 4$ , where $A = {[a_{i j}]}_{3 \times 3} = [\begin{matrix} 1 & - 1 & 1 \\ 2 & 1 & - 3 \\ 1 & 1 & 1 \end{matrix}]$ , $X = [\begin{matrix} x \\ y \\ z \end{matrix}]$ and $B = [\begin{matrix} 4 \\ 0 \\ 2 \end{matrix}]$

Now, $|A| = 1 (1 + 3) + 1 (2 + 3) + 1 (2 - 1) = 10$

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

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

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

$A_{12} = (- 1)^{1 + 2} | \begin{array}{cc} 2 & - 3 \\ 1 & 1 \end{array} | = - 5$

$A_{13} = (- 1)^{1 + 3} | \begin{array}{ll} 2 & 1 \\ 1 & 1 \end{array} | = 1$

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

$A_{22} = (- 1)^{2 + 2} | \begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array} | = 0$

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

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

$A_{32} = (- 1)^{3 + 2} | \begin{array}{cc} 1 & 1 \\ 2 & - 3 \end{array} | = 5$

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

$∴ A d j A = [\begin{array}{ccc} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{array}]$

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

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

$X = \frac{1}{10} [\begin{array}{ccc} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{array}] [\begin{matrix} 4 \\ 0 \\ 2 \end{matrix}]$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{10} [\begin{matrix} 4 \times 4 + 2 \times 0 + 2 \times 2 \\ (- 5) \times 4 + 0 \times 0 + 5 \times 2 \\ 1 \times 4 + (- 2) \times 0 + 3 \times 2 \end{matrix}]$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{10} [\begin{matrix} 20 \\ - 10 \\ 10 \end{matrix}]$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 2 \\ - 1 \\ 1 \end{matrix}]$

$\Rightarrow x = 2, y = - 1, z = 1$

Examining consistency and solvability, solve the following equations by matrix method.

$x - y + z = 4$

$2 x + y - 3 z = 0$

$x + y + z = 2$

Important Questions on Determinants

Learn and Prepare for any exam you want !

Examining consistency and solvability, solve the following equations by matrix method. x−y+z=4 2x+y−3z=0 x+y+z=2

Important Questions on Determinants

Learn and Prepare for any exam you want !

Examining consistency and solvability, solve the following equations by matrix method.

$x - y + z = 4$

$2 x + y - 3 z = 0$

$x + y + z = 2$