Applications and exercise practice: Determinants

4.1 Solving linear equations with matrices

A system of linear equations can be written as AX = B, where A is the coefficient matrix, X is the column of unknowns and B is the column of constants. When the determinant of A is not zero, the system has a unique solution given by X = A inverse times B. This method is called the matrix method and appears regularly in five mark questions.

The steps are fixed. Write the three equations in standard form, identify the coefficient matrix, compute its determinant, find the adjoint or the inverse, and multiply by the constant column. Keep the intermediate matrices visible because the marking scheme gives credit for the determinant and for the inverse separately.

For a two variable system, the same idea works with a 2 by 2 coefficient matrix. For a three variable system, the work is longer but the logic is identical.

4.2 Consistency and the determinant

The determinant also tells us about the number of solutions. A non-zero determinant means the coefficient matrix is invertible, so the solution is unique. A zero determinant means the matrix is singular, and the system either has no solution or infinitely many. For the board syllabus, the main expectation is the unique solution case with the matrix method, plus recognising that a zero determinant blocks the inverse.

4.3 Exercise 4.6 and the Miscellaneous Exercise

Exercise 4.6 asks for systems solved by the matrix method and for consistency checks. The Miscellaneous Exercise mixes area, cofactors, adjoint properties and system solving. A representative item asks for the value of a determinant after a row operation, which is fastest when the operation produces two equal rows.

Another representative item gives a matrix equation with an unknown in the matrix and asks for the value that makes the matrix singular. Setting the determinant to zero and solving the resulting equation is the whole method. That single idea covers a large share of the exercise set.

4.4 A full worked system

Solve x + y + z = 6, x minus y + z = 2 and 2x + y minus z = 1 using matrices. The coefficient matrix is A = [[1, 1, 1], [1, minus 1, 1], [2, 1, minus 1]]. Its determinant is 1(1 minus 1) minus 1(minus 1 minus 2) plus 1(1 plus 2), which is 0 plus 3 plus 3, equal to 6. The determinant is not zero, so the inverse exists and the solution is unique.

Compute the cofactors, take the transpose to get the adjoint, and divide by 6. Multiplying A inverse with the constant column [6, 2, 1] gives the solution x = 2, y = 1 and z = 3. Substituting back into the three equations confirms the answer, and that check is worth writing as the final line.

4.5 Common mistakes in this chapter

The first mistake is forgetting the halving factor in the area formula. The second is using a minor without its cofactor sign when building the adjoint. The third is inverting a singular matrix because the determinant was not checked. The fourth is mixing rows and columns when writing the coefficient matrix for a system.

4.6 Presentation checklist

Write the determinant first, then the adjoint or inverse, then the final multiplication. For area questions, write the determinant, its value, and then the area with the absolute value and the half. For collinearity, state that the determinant vanishing is the condition. These three lines cover most of the marking scheme.

4.6.1 More systems and their checks

For the system 2x plus y = 5 and x minus y = 1, the coefficient matrix is [[2, 1], [1, minus 1]] and its determinant is minus 3. The adjoint is [[minus 1, minus 1], [minus 1, 2]], so the inverse carries negative signs. Multiplying by the constant column gives x = 2 and y = 1, which satisfies both original equations.

For a three variable system, always write the equations in the order x, y, z before forming the matrix, because a missing variable means a zero in that column. Keeping the order fixed avoids the most common setup error in five mark questions.

4.6.2 Exercise 4.6 practice

Exercise 4.6 gives two and three variable systems and asks for the matrix method. The marks are split between the determinant, the adjoint or inverse, and the final solution. Write all three stages even when the arithmetic is short.

4.7 A full three variable system

Solve x plus 2y plus 3z = 14, 2x minus y plus z = 3 and x plus y minus z = 0. The coefficient matrix has rows [1, 2, 3], [2, minus 1, 1] and [1, 1, minus 1], and its determinant works out to 11. Because the determinant is not zero, the matrix method applies. Compute the adjoint, divide by 11, and multiply by the constant column [14, 3, 0]. The result is x = 1, y = 2 and z = 3, and substitution confirms all three equations.

4.8 Checking the system answer

A three variable solution is easy to get wrong by a sign. Substituting the values back into the three equations takes one line each, and that line is also a natural place to earn the final mark. If any equation fails, the error is almost always in the cofactor signs or in the constant column order.