Answer to Question #112311 in Linear Algebra for Caylin

Question

Suppose

A= | 4 3 5 |
| 1 3 - 5 |
| 2 1 5 |.

1) Evaluate det(A) by expanding along the second row. No Other Method.

2) Can cramers rule be used to solve the system :

A | x |. | 0 |
| y | = | 0 |
| z | | a |

Where A is given and a can't equal 0?

Give reasons. IF YES, solve with cramers rule. IF NOT, solve with other method.

Please assist.

Answer 1

Given matrix A = $\left[ {\begin{array}{cc} 4 & 3 & 5 \\ 1 & 3 & -5 \\ 2 & 1 & 5 \\ \end{array} } \right]$

1) By expanding along the second row,

det(A) = (-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)(15-5) + 3(20-10) + 5(4-6) = -10 + 30 -10 = 10

2) Now since D = det(A) $\neq$ 0, so Cramers rule can be used to solve the given system of equation AX=b where X = [x y z]' , b = [0 0 a]' and a $\neq$ 0.

Now D_x = $\left| \begin{array}{cc} 0 & 3 & 5 \\ 0 & 3 & -5 \\ a & 1 & 5 \\ \end{array} \right|$ = a(-15-15) = -30a $\neq$ 0 (expanding along the first column)

D_y = $\left| \begin{array}{cc} 4 & 0 & 5 \\ 1 & 0 & -5 \\ 2 & a & 5 \\ \end{array} \right|$ = -a(-20-5) = 25a ​$\neq$ 0 (expanding along the second column)

D_z = $\left| \begin{array}{cc} 4 & 3 & 0 \\ 1 & 3 & 0 \\ 2 & 1 & a \\ \end{array} \right|$ = a(12-3) = 9a $\neq$ 0 (expanding along the third column)

So solution of the given system is

$x = \frac{D_x}{D} = \frac{-30a}{10} = -3a, y = \frac{D_y}{D} = \frac{25a}{10} = 2.5a, z = \frac{D_z}{D} = \frac{9a}{10} = 0.9a$