Name: Solving System of Linear Equations using Matrices
Uploaded: 2019-07-19
Duration: 11 min 30 s
Description: A system of linear equations can be represented in matrix form using its coefficient, variable and constant matrices. Learn about the inverse matrix method!

Question

Solve each of the following system of homogeneous linear equations.

\begin{array}{l} x + y - 2 z = 0 \\ 2 x + y - 3 z = 0 \\ 5 x + 4 y - 9 z = 0 \end{array}

Accepted Answer

Given:

$\begin{array}{l} x + y - 2 z = 0 \\ 2 x + y - 3 z = 0 \\ 5 x + 4 y - 9 z = 0 \end{array}$

For $a_{1} x + b_{1} y + c_{1} x = 0, a_{2} x + b_{2} y + c_{2} x = 0, a_{3} x + b_{3} y + c_{3} x = 0$ , $D = |\begin{array}{l} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}|$

Determinant for a matrix $X = [\begin{array}{lll} m & n & o \\ p & q & r \\ s & t & u \end{array}] \begin{array}{l} \end{array}$ is $|X| = |\begin{array}{l} m & n & o \\ p & q & r \\ s & t & u \end{array}| = m (q u - r t) - n (p u - r s) + o (p t - q s)$ .

Now, $D = |\begin{array}{l} 1 & 1 & - 2 \\ 2 & 1 & - 3 \\ 5 & 4 & - 9 \end{array}|$

$= 1 (- 9 + 12) - 1 (- 18 + 15) - 2 (8 - 5)$

$= 0$

So, the system has infinitely many solutions. Putting $z = k$ in the first two equations, we get

$x + y = 2 k$

$2 x + y = 3 k$

In Cramer's rule the solution to $a x + b y = e, c x + d y = f$ is $x = \frac{D_{1}}{D}, y = \frac{D_{2}}{D}$ , where $D = |\begin{array}{ll} a & b \\ c & d \end{array}|, D_{1} = |\begin{array}{ll} e & b \\ f & d \end{array}|, D_{2} = |\begin{array}{ll} a & e \\ c & f \end{array}|$ .

For a matrix $X = [\begin{matrix} m & n \\ o & p \end{matrix}]$ , determinant is $|X| = |\begin{matrix} m & n \\ o & p \end{matrix}| = m p - o n$ .

Using Cramer's rule, we get

$x = \frac{D_{1}}{D} = \frac{|\begin{array}{l} 2 k & 1 \\ 3 k & 1 \end{array}|}{|\begin{array}{l} 1 & 1 \\ 2 & 1 \end{array}|} = \frac{- k}{- 1} = k$

$y = \frac{D_{2}}{D} = \frac{|\begin{array}{l} 1 & 2 k \\ 2 & 3 k \end{array}|}{|\begin{array}{l} 1 & 1 \\ 2 & 1 \end{array}|} = \frac{- k}{- 1} = k$

$z = k$

Clearly, these values satisfy the third equation.

Thus, $x = y = z = k [k \in R]$ .

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