Name: Properties of Determinants (Part 3)
Uploaded: 2019-07-19
Duration: 3 min 53 s
Description: Suppose we multiply a row of a determinant with a particular value, then what would be the change in the value of the determinant? Watch this video to learn...

Question

Using Cramer's rule, solve the following system of linear equations:

$x + y + z = 1$

$a x + b y + c z = k$

$a^{2} x + b^{2} y + c^{2} z = k^{2}$ where $a, b, c$ are all different.

Accepted Answer

The given system of equations be :

$x + y + z = 1$

$a x + b y + c z = k$

$a^{2} x + b^{2} y + c^{2} z = k^{2}$ where $a, b, c$ are all different.

The matrix form of the system of equations is as follows:

$|\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{matrix}| [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 1 \\ k \\ k^{2} \end{matrix}]$

$A X = B$

Let $D$ be the determinant of the coefficient of $x, y$ and $z$ , and $D_{1}, D_{2}$ and $D_{3}$ be the determinants obtained from $D$ by replacing the coefficients of $x, y$ and $z$ respectively by the constant terms i.e.,

$|D| = |\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{matrix}|$

Applying $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{1}$ [to create two zero in the first row]

$= |\begin{matrix} 1 & 0 & 0 \\ a & b - a & c - a \\ a^{2} & b^{2} - a^{2} & c^{2} - a^{2} \end{matrix}|$

Expanding the first row, we get

$= 1 |\begin{matrix} b - a & c - a \\ b^{2} - a^{2} & c^{2} - a^{2} \end{matrix}|$

$= |\begin{matrix} b - a & c - a \\ (b - a) (b + a) & (c - a) (c + a) \end{matrix}|$

$[∵ m^{2} - n^{2} = (m + n) (m - n)]$

Taking out $b - a$ common from first row and $c - a$ from second row, we get

$= (b - a) (c - a) |\begin{matrix} 1 & 1 \\ b + a & c + a \end{matrix}|$

Expanding determinant, we have

$= (b - a) (c - a) (c + a - a - b)$

$= (a - b) (b - c) (c - a) \neq 0$ since $a \neq b \neq c$ .

Thus $A$ is a nonsingular matrix, therefore the system of equations has a unique solution.

$|D_{1}| = |\begin{matrix} 1 & 1 & 1 \\ k & b & c \\ k^{2} & b^{2} & c^{2} \end{matrix}|$

If we put $k = a$ we get $|D_{1}| = |D|$ .

Therefore, putting $a = k$ in value of $|D|$ ,we get

$|D_{1}| = (k - b) (b - c) (c - k)$

Similarly

$|D_{2}| = |\begin{matrix} 1 & 1 & 1 \\ a & k & c \\ a^{2} & k^{2} & c^{2} \end{matrix}| = (a - k) (k - c) (c - a)$

and $|D_{3}| = |\begin{matrix} 1 & 1 & 1 \\ a & b & k \\ a^{2} & b^{2} & k^{2} \end{matrix}| = (a - b) (b - k) (k - a)$

Then, by Cramer's rule, we have

$x = \frac{|D_{1}|}{|D|} = \frac{(k - b) (b - c) (c - k)}{(a - b) (b - c) (c - a)} = \frac{(k - b) (c - k)}{(a - b) (c - a)}$

$y = \frac{|D_{2}|}{|D|} = \frac{(a - k) (k - c) (c - a)}{(a - b) (b - c) (c - a)} = \frac{(a - k) (k - c)}{(a - b) (b - c)}$ and

$z = \frac{|D_{3}|}{|D|} = \frac{(a - b) (b - k) (k - a)}{(a - b) (b - c) (c - a)} = \frac{(b - k) (k - a)}{(b - c) (c - a)}$ .

Using Cramer's rule, solve the following system of linear equations:

$x + y + z = 1$

$a x + b y + c z = k$

$a^{2} x + b^{2} y + c^{2} z = k^{2}$ where $a, b, c$ are all different.

Important Questions on Determinants

Learn and Prepare for any exam you want !

Using Cramer's rule, solve the following system of linear equations: x+y+z=1 ax+by+cz=k a2x+b2y+c2z=k2 where a, b, c are all different.

Important Questions on Determinants

Learn and Prepare for any exam you want !

Using Cramer's rule, solve the following system of linear equations:

$x + y + z = 1$

$a x + b y + c z = k$

$a^{2} x + b^{2} y + c^{2} z = k^{2}$ where $a, b, c$ are all different.