Question

What is the determinant of the coefficient matrix of the system

Answer 1

$\left[\begin{array}{ccc}-3&0&-2\\9&0&5\\6&0&-12\end{array}\right] \\D(system)= |-3|\left[\begin{array}{ccc}0&5\\0&-12\\\end{array}\right]-0 \left[\begin{array}{ccc}9&5\\6&-12\\\end{array}\right] +|-2| \left[\begin{array}{ccc}9&0\\6&0\\\end{array}\right] \\=3*0-0+0=0\\ determinant of system is 0$

Answer 2

ANSWER

The determinant is 0

EXPLANATION (METHOD 1)
This method involves expanding along any column.


For an n×n matrix A, the determinant of A, det(A), can be obtained by expanding along the kth column:

   $\det(A) = a_{1k} C_{1k} + a_{2k} C_{2k} + \ldots + a_{nk} C_{nk}$

where $a_{k1}$ is the entry of A in the kth row, 1st column, $a_{k2}$ is the entry of A in the kth row, 2nd column, etc., and $C_{kn}$ is the kn cofactor of A, defined as $C_{kn} = (-1)^{k+n} M_{kn}$. 

But we do not need to care about the cofactors as all the 2nd column entries are 

   a₁₂ = a₂₂ = a₃₂ = 0

We would end up with

   $\begin{aligned} \det\left(\begin{bmatrix} \bf -3 & \bf 0 & \bf -2\\ 9 & 0 & 5 \\ 6 & 0 & -12 \end{bmatrix}\right) &= (0) C_{12} + (0)C_{22} + (0)C_{32} \\ &= 0 \end{aligned}$

EXPLANATION (METHOD 2)|
This method involves expanding along a row

For an n×n matrix A, the determinant of A, det(A), can be obtained by expanding along the kth row:

$\det(A) = a_{k1} C_{k1} + a_{k2} C_{k2} + \ldots + a_{kn} C_{kn}$

where $a_{k1}$ is the entry of A in the kth row, 1st column, $a_{k2}$ is the entry of A in the kth row, 2nd column, etc., and $C_{kn}$ is the kn cofactor of A, defined as $C_{kn} = (-1)^{k+n} M_{kn}$.

$M_{kn}$ is the kn minor, obtained by getting the determinant of the matrix which is the matrix A with row k and column n deleted.

Applying this here, we can expand along the 1st row.
For convenience, let G be the coefficient matrix of this question, which is

$G=\begin{bmatrix} \bf -3 & \bf 0 & \bf -2\\ 9 & 0 & 5 \\ 6 & 0 & -12 \end{bmatrix}$

with the first row bolded.

The determinant is therefore

$\begin{aligned} \text{det}(G) &= g_{11}C_{11} + g_{12}C_{12} + g_{13}C_{13} \end{aligned}$

Note that g₁₁ is the matrix element of G that is in the 1st row, 1st column, g₁₂ is the matrix element of G that is in the 1st row, 2nd column, etc. Then we have

$\begin{aligned} \text{det}(G) &= g_{11}(-1)^{1+1}M_{11} + g_{12}(-1)^{1+2}M_{12} + g_{13}(-1)^{1+3}M_{13} \\ &= g_{11} M_{11} - g_{12}M_{12} + g_{13}M_{13} \end{aligned}$

M₁₁ is the determinant of the matrix that is matrix G with row 1 and column 1 removed. The bold entires are the row and the column we delete.

$\begin{aligned} G=\begin{bmatrix} \bf -3 & \bf 0 & \bf -2\\ \bf 9 & 0 & 5 \\ \bf 6 & 0 & -12 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&5 \\ 0&-12 \end{bmatrix} \right) \end{aligned}$

The determinant of a 2×2 matrix is

   $\det\left( \begin{bmatrix} a & b \\ c& d \end{bmatrix} \right) = ad-bc$

so it follows that

$\begin{aligned} G=\begin{bmatrix} \bf -3 & \bf 0 & \bf -2\\ \bf 9 & 0 & 5 \\ \bf 6 & 0 & -12 \end{bmatrix} \implies M_{11} &= \det\left(\begin{bmatrix} 0&5 \\ 0&-12 \end{bmatrix} \right) \\ &= (0)(-12) - (5)(0) \\ &= 0 \end{aligned}$

Applying the same for M₁₂ and M₁₃, we have

$\begin{aligned} G=\begin{bmatrix} \bf -3 & \bf 0 & \bf -2\\ 9 & \bf 0 & 5 \\ 6 & \bf 0 & -12 \end{bmatrix} \implies M_{12} &= \det\left(\begin{bmatrix} 9&5 \\ 6&-12 \end{bmatrix} \right) \\ &= (9)(-12) - (5)(6) \\ &= -138 \end{aligned}$

and

$\begin{aligned} G=\begin{bmatrix} \bf -3 & \bf 0 & \bf -2\\ 9 & 0 & \bf 5 \\ 6 & 0 & \bf -12 \end{bmatrix} \implies M_{13} &= \det\left(\begin{bmatrix} 9&0\\ 6&0 \end{bmatrix} \right) \\ &= (9)(0) - (0)(6) \\ &= 0 \end{aligned}$

so therefore

$\begin{aligned} \text{det}(G) &= g_{11} M_{11} - g_{12}M_{12} + g_{13}M_{13} \\ &= (-3)(0) - (0)(-138) + (-2)(0) \\ &= 0 \end{aligned}$