Q.Examining consistency and solvability, solve the following equations by matrix method: $x - 2y = 3$, $3x + 4y - z = -2$, $5x - 3z = -1$.

Write the system as $AX=B$; since $|A|=-20\ne0$, the system is consistent with a unique solution, found here by Cramer's Rule.

System: $x-2y=3,\quad 3x+4y-z=-2,\quad 5x-3z=-1$, i.e.

$x-2y+0z=3$

$3x+4y-z=-2$

$5x+0y-3z=-1$

$A=\begin{bmatrix}1&-2&0\\3&4&-1\\5&0&-3\end{bmatrix},\quad B=\begin{bmatrix}3\\-2\\-1\end{bmatrix}$

Determinant of $A$:

$|A| = 1[4(-3)-(-1)(0)] - (-2)[3(-3)-(-1)(5)] + 0 = 1(-12) + 2(-9+5) = -12-8=-20$

Since $|A|=-20\ne0$, $A$ is non-singular, so the system is consistent and has a unique solution (by Cramer's Rule / matrix method).

By Cramer's Rule, $x=\dfrac{|A_1|}{|A|}$, etc., where $A_i$ replaces column $i$ of $A$ with $B$.

$A_1=\begin{bmatrix}3&-2&0\\-2&4&-1\\-1&0&-3\end{bmatrix}$: $|A_1| = 3[4(-3)-(-1)(0)] -(-2)[(-2)(-3)-(-1)(-1)] + 0 = 3(-12)+2(6-1) = -36+10=-26$

$x = \dfrac{-26}{-20} = \dfrac{13}{10}$

$A_2=\begin{bmatrix}1&3&0\\3&-2&-1\\5&-1&-3\end{bmatrix}$: $|A_2| = 1[(-2)(-3)-(-1)(-1)] - 3[3(-3)-(-1)(5)] + 0 = 1(6-1)-3(-9+5) = 5+12=17$

$y = \dfrac{17}{-20} = -\dfrac{17}{20}$

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