For reasons that will become clear, the problem is made easier if each point is translated similarly so that one of them translates to the origin . Select the translation . Then the translations are:

Replace the variables in the matrix with these modified values:

The determinant can be found by choosing any row or column and adding the products of each value and its cofactor. Since the first row has three zeroes, thanks to our translations, choose this row:

,

where  is the minor - the determinant of the matrix formed when Row 1 and Column 4 are removed:

The determinant of a  matrix can be found by adding the products of the upper-left to lower-right diagonals and subtracting the products of the upper-right to lower-left diagonals:

,

and .

,

the volume of the tetrahedron.

The determinant of a  matrix can be calculated by subtracting the upper-right to lower-left product from the upper-left to lower-right product:

A square matrix  is nilpotent if, for some whole number ,  , the zero matrix of the same dimension as . 

The determinant of the product of matrices is equal to the product of their determinants. It follows that if , , so . Therefore, any nilpotent matrix must have determinant 0. 

However, not all matrices with determinant 0 are nilpotent, as is proved by counterexample. Let

This matrix is diagonal, as its only nonzero entry is alone its main diagonal. To raise this to a power , simply raise all of the diagonal elements to the power of , and preserve the off-diagonal zeroes. It follows that for all ,

Also, the determinant of  is . 

Since a non-nilpotent matrix with determinant zero exists, the biconditonal is false.

The determinant of the product of matrices is equal to the product of their determinants; thus,

.

A matrix has determinant 0 if and only if it is singular - that is, without an inverse. It therefore follows that  is singular.

If the three known points , , and , and an unknown point  are coplanar, then 

The variable equation can be formed from this determinant equation.  

Using the first row, this can be rewritten as

,

where , the minor, is the determinant of the matrix formed by striking out Row 1 and Column 4:

Adding the upper-left to lower-right products and subtracting the upper-right to lower left products:

Set this equal to 0 to get the equation of the plane:

The statement can be proved false through counterexample.

Let

Each of the matrices has only zero elements on its off-diagonal elements, so each is a diagonal matrix. Consequently, the determinant of each matrix is the product of its diagonal elements.

 is a zero matrix, so ;  is an identity matrix, so ; and

, , and  meet the conditions of the problem. Now, add the matrices by adding corresponding entries:

This matrix is also diagonal, so

A matrix has determinant 0 if and only if it is singular - that is, without an inverse. It therefore follows that  is singular. Thus, we have proved through counterexample that   need not be nonsingular.

The statement can be proved false through counterexample.

Let

Each of the matrices has only zero elements on its off-diagonal elements, so each is a diagonal matrix. Consequently, the determinant of each matrix is the product of its diagonal elements.

 is a zero matrix, so ;  is an identity matrix, so ; and 

, , and  meet the conditions of the problem. Now, add the matrices by adding corresponding entries:

This matrix is also diagonal, so

A matrix has a nonzero determinant if and only if it is nonsingular - that is, with an inverse. It therefore follows that  is nonsingular. Thus, we have proved through counterexample that   need not be singular.

The three points , , and , and a point of the form  

 for some real  are on the same plane. Therefore, it follows that we must find  so that

Using the first row, this can be rewritten as

,

where , the minor, is the determinant of the matrix formed by striking out Row 1 and Column 4:

Adding the upper-left to lower-right products and subtracting the upper-right to lower left products:

Thus, 

,

and

et this equal to 0 and solve for :

The desired point is .

The three points , , and , and a point of the form  for some real  are on the same plane. Therefore, it follows that we must find  so that

Using the first row, this can be rewritten as

,

where , the minor, is the determinant of the matrix formed by striking out Row 1 and Column 4:

Adding the upper-left to lower-right products and subtracting the upper-right to lower left products:

It follows that 

,

and

Set this equal to 0 to find :

The desired point is .

The statement can be proved false through counterexample.

Let

Each of the matrices has only zero elements on its off-diagonal elements, so each is a diagonal matrix. Consequently, the determinant of each matrix is the product of its diagonal elements.

 is a zero matrix, so ;  is an identity matrix, so ; and

, , and  meet the conditions of the problem. Now, add the matrices by adding corresponding entries:

This matrix is also diagonal, so

A matrix has determinant 0 if and only if it is singular - that is, without an inverse. It therefore follows that  is singular. Thus, we have proved through counterexample that   need not be nonsingular.