Answer to Question #110824 in Linear Algebra for Hetisani Sewela

Question

Suppose
A =


4 3 5
1 3 −5
2 1 5

 .
6.1 Evaluate det (A) by expanding along the 2nd row. NO marks will be awarded if you use any other
method.
(5)
6.2 Can Cramer’s rule be used to solve the system
A


x
y
z

 =


0
0
a

 where A is given and a 6= 0?
Give reasons for your answer. If the answer is Yes, use Cramer’s rule to solve the system. If not, use
any other method.

Answer 1

We need to evaluate the determinant of matrix A using second row and also solve the system of equations by Cramer's rule .

The given matrix is

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

(6.1) Evaluation of the determinant of the given matrix A.

Determinant of A = det A = $\begin{vmatrix} 4 & 3 & 5\\ 1 & 3 & -5 \\ 2 & 1 & 5 \end{vmatrix}$

$= -1 \begin{vmatrix} 3 & 5 \\ 1 & 5 \end{vmatrix} + 3 \begin{vmatrix} 4 & 5 \\ 2 & 5 \end{vmatrix} - (-5) \begin{vmatrix} 4 & 3 \\ 2 & 1 \end{vmatrix}$

$= -1 ( 3\times5 - 1 \times 5) + 3 ( 4 \times 5 - 2 \times 5) +5 ( 4 \times 1 - 2 \times 3)$

$= -1( 15 - 5) + 3( 20 - 10) + 5(4 - 6)$

$= -10 + 30 -10 = 10$

$det \space A = 10$

(6.2). Solving of system of equation by using Cramer's rule

$A \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ a \end{bmatrix} and \space a\ne 0$

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

$D=|A | = 10 \ne 0$

Yes, we can use the cramer's rule to solve the system of equations

Answer column is = $\begin{bmatrix} 0 \\ 0 \\ a \end{bmatrix}$

$D_x =$ coefficient determinant with answer-column values in x-column

$= \begin{vmatrix} 0 & 3 & 5\\ 0 & 3 & -5\\ a & 1 & 5 \end{vmatrix}$

$= - 0\begin{vmatrix} 3 & 5 \\ 1 & 5 \end{vmatrix} + 3 \begin{vmatrix} 0 & 5 \\ a & 5 \end{vmatrix} - (-5) \begin{vmatrix} 0 & 3 \\ a & 1 \end{vmatrix}$

$= 0 + 3 (0 - 5a) +5 (0-3a) = - 15a -15a =-30a$

$D_y =$ coefficient determinant with answer-column values in y-column

$= \begin{vmatrix} 4 & 0 & 5\\ 1 & 0 & -5 \\ 2 & a & 5 \end{vmatrix}$

$= -1 \begin{vmatrix} 0 & 5 \\ a & 5 \end{vmatrix} + 0 \begin{vmatrix} 4 & 5 \\ 2 & 5 \end{vmatrix} - (-5) \begin{vmatrix} 4 & 0 \\ 2 & a \end{vmatrix}$

$D_y = -1 (0-5a) + 0 +5 (4a -0) = 5a +20a = 25a$

$D_z =$ coefficient determinant with answer-column values in z-column

$=\begin{vmatrix} 4 & 3 & 0\\ 1 & 3 & 0 \\ 2 & 1 & a \end{vmatrix}$

$= -1 \begin{vmatrix} 3 & 0 \\ 1 & a \end{vmatrix} + 3 \begin{vmatrix} 4 & 0 \\ 2 & a \end{vmatrix} -0 \begin{vmatrix} 4 & 3 \\ 2 & 1 \end{vmatrix}$

$= -1(3a -0) +3(4a -0) +0 = - 3a +12a = 9a$

Using Cramer's rule,

$x = \frac {D_x} {D} = \frac {-30a} {10} = -3a$

$y = \frac {D_y}{D} = \frac {25a}{10}= \frac {5a}{2}$

$z = \frac {D_z}{D} = \frac {9a}{10}$