Systems of Three Linear Equations with Three Varaibles. Third-Order Determinants

Third-order determinant is a number that is defined by equality `|[a,b,c],[d,e,f],[g,h,i]|=aei+bfg+cdh-ceg-afh-bdi`.

This formula is a bit hard, but luckily you don't need to remember it. It is sufficiently to understand how this formula is derived.

So, there are three summands with "+" sign and three summands with "-" sign. With "+" sign there is product of elments onlaw of triangles for calculating third-order determinant the main diagonal and two products of elements that form a triangle, whose base is parallel to the main diagonal. With "-" sign there is product of elments off the main diagonal and two products of elements that form a triangle, whose base is perpendicular to the main diagonal. This rule is called law of triangles and is shown on figure to the right.

calculating third-order determinantAnother good method to remember formula for determinant is to write first two columns to the right of determinant and then calculate products along "diagonals". For better understanding see figure to the left (green lines are products with "+" and red lines are products with "-").

For example,

`|[2,4,1],[-1,3,5],[4,-1,3]|=2*3*3+4*5*4+1*(-1)*(-1)-1*3*4-2*5*(-1)-4*(-1)*3=109`.

Cofactor of the element `a_(ik)` of third-order is called number `A_(ik)`, that is second-order deterimnant, obtained by crossing out i-th row and k-th column of given determinant and take with "+" if `i+k` is even number, and with "-" if `i+k` is odd number.

For example, for determinant `|[a,b,c],[d,e,f],[g,h,i]|` `A_(11)=|[e,f],[h,i]|` (we crossed out first row and first column and since 1+1=2 is even number, then we take determinant with "+" sign) and `A_(32)=-|[a,c],[d,f]|` (we crossed out third row and second column and since 3+2=5 is odd number, then we take determinant with "-" sign).

Fact (Cofactor Expansion). Determinant equals sum of products of elements of any row (column) and corresponding cofactors.

For example, let's calculate determinant `Delta=|[2,4,1],[-1,3,5],[4,-1,3]|` in two ways:

  1. Expand along first row: `Delta=2*A_(11)+4*A_(12)+1*A_(13)=2|[3,5],[-1,3]|-4|[-1,5],[4,3]|+1*|[-1,3],[4,-1]|=`

    `=2(3*3-5*(-1))-4((-1)*3-5*4)+((-1)*(-1)-3*4)=109`.

  2. Expand along second column: `Delta=4*A_(21)+3*A_(22)-1*A_(23)=-4|[-1,5],[4,3]|+3|[2,1],[4,3]|+1*|[2,1],[-1,5]|=`

    `=-4((-1)*3-5*4)+3(2*3-1*4)+(2*5-1*(-1))=92+6+11=109`.

As can be seen we obtained same result. Once again, we can expand along any row or column.

For systems of three linear equations with three variables `{(a_(11)x+a_(12)y+a_(13)z=b_1),(a_(21)x+a_(22)y+a_(23)z=b_2),(a_(31)x+a_(32)y+a_(33)z=b_3):}` Cramer's Rule is also applicable: let `Delta =|[a_(11),a_(12),a_(13)],[a_(21),a_(22),a_(23)],[a_(31),a_(32),a_(33)]|`

`Delta_x =|[b_(1),a_(12),a_(13)],[b_(2),a_(22),a_(23)],[b_(3),a_(32),a_(33)]|,\ Delta_y=|[a_(11),b_(1),a_(13)],[a_(21),b_(2),a_(23)],[a_(31),b_(3),a_(33)]|,\ Delta_z=|[a_(11),a_(12),b_(1)],[a_(21),a_(22),b_(2)],[a_(31),a_(32),b_(3)]|` (where `Delta_x,Delta_y,Delta_z` are obtained from `Delta` by replacing column of coefficients near corresponding variable with column of free members), then:

  1. if `Delta!=0` then system has unique solution `x=(Delta x)/x,y=(Delta y)/y,z=(Delta z)/z`.
  2. if `Delta=Delta_x=Delta_y=Delta_z=0` then system has infinitely many solutions.
  3. if `Delta=0` and at least one of other determinants doesn't equal zero, then system is inconsistent, i.e. there are no solutions.

Example. Solve system of equations `{(2x+4y+z=-3),(-x+3y+5z=7),(4x-y+3z=-5):}`.

Here `Delta=|[2,4,1],[-1,3,5],[4,-1,3]|=109`.

`Delta_x=|[-3,4,1],[7,3,5],[-5,-1,3]|=-218`.

`Delta_y=|[2,-3,1],[-1,7,5],[4,-5,3]|=0`.

`Delta_z=|[2,4,-3],[-1,3,7],[4,-1,-5]|=109`.

Since `Delta!=0`, system has unique solution: `x=(Delta x)/(Delta)=(-218)/(109)=-2,\ y=(Delta y)/(Delta)=0/(109)=0,\ z=(Delta z)/(Delta)=(109)/(109)=1`.

So, x=-2, y=0, z=1 is solution of the initial system.