Matrices · Form 5
Matrices: Worked Examples (Medium)
Puts techniques together: multiplying two 2×2 matrices, building the inverse of a 2×2 matrix, and solving a pair of simultaneous linear equations with the inverse matrix. For students who know the basics and want the two- and three-step methods.
Worked example 1
Given A = [[2, 1], [3, −1]] and B = [[4, 0], [−2, 5]], find AB.
- Multiply row by column: entry (1,1) = (2×4) + (1×(−2)) = 8 − 2 = 6.
- Entry (1,2) = (2×0) + (1×5) = 0 + 5 = 5.
- Entry (2,1) = (3×4) + ((−1)×(−2)) = 12 + 2 = 14.
- Entry (2,2) = (3×0) + ((−1)×5) = 0 − 5 = −5.
- Collect the entries: AB = [[6, 5], [14, −5]].
Worked example 2
Find the inverse of M = [[4, 2], [3, 5]].
- First find the determinant: det M = (4×5) − (2×3) = 20 − 6 = 14.
- Since det M = 14 ≠ 0, the inverse exists. Swap the main-diagonal entries and negate the other two: [[5, −2], [−3, 4]].
- Multiply by 1 ÷ det M: M⁻¹ = (1/14)[[5, −2], [−3, 4]].
Worked example 3
Solve the simultaneous equations 2x + y = 7 and 3x − 2y = 0 using the inverse matrix method.
- Write in matrix form: [[2, 1], [3, −2]] [[x], [y]] = [[7], [0]].
- Determinant of the coefficient matrix = (2×(−2)) − (1×3) = −4 − 3 = −7.
- Inverse = (1/(−7))[[−2, −1], [−3, 2]].
- Multiply the inverse by [[7], [0]]: [[x], [y]] = (1/(−7))[[(−2×7)+(−1×0)], [(−3×7)+(2×0)]] = (1/(−7))[[−14], [−21]].
- Divide each entry by −7: [[x], [y]] = [[2], [3]].
Worked example 4
Given A = [[1, 3], [2, −1]] and B = [[2, 4], [0, 5]], find AB.
- Each element of AB = (row of A) · (column of B).
- Row 1: [1×2 + 3×0, 1×4 + 3×5] = [2, 19].
- Row 2: [2×2 + (−1)×0, 2×4 + (−1)×5] = [4, 3].
Worked example 5
Find the inverse of the matrix [[3, 1], [5, 2]].
- Determinant = (3 × 2) − (1 × 5) = 6 − 5 = 1.
- Swap the leading diagonal and negate the other diagonal: [[2, −1], [−5, 3]].
- Multiply by 1/determinant = 1/1 = 1, so the matrix is unchanged.
Worked example 6
Using the inverse matrix method, solve the simultaneous equations x + 2y = 8 and 3x − y = 3.
- Write in matrix form: [[1, 2], [3, −1]] [[x], [y]] = [[8], [3]].
- Determinant = (1 × −1) − (2 × 3) = −1 − 6 = −7.
- Inverse = (1/−7) [[−1, −2], [−3, 1]].
- [[x], [y]] = (1/−7) [[−1, −2], [−3, 1]] [[8], [3]] = (1/−7) [[−14], [−21]] = [[2], [3]].
Worked example 7
Given A = [[5, −1], [2, 3]] and B = [[1, 4], [−2, 0]], find AB.
- Multiply row 1 of A by column 1 of B: (5×1) + (−1×−2) = 5 + 2 = 7.
- Multiply row 1 of A by column 2 of B: (5×4) + (−1×0) = 20 + 0 = 20.
- Multiply row 2 of A by column 1 of B: (2×1) + (3×−2) = 2 − 6 = −4.
- Multiply row 2 of A by column 2 of B: (2×4) + (3×0) = 8 + 0 = 8.
- AB = [[7, 20], [−4, 8]].
Worked example 8
Using the inverse matrix method, solve the simultaneous equations 3x + 2y = 12 and x − y = −1.
- Write in matrix form: [[3, 2], [1, −1]] [x, y] = [12, −1].
- Find the determinant: (3×−1) − (2×1) = −3 − 2 = −5.
- Find the inverse: (1/−5)[[−1, −2], [−1, 3]] = [[1/5, 2/5], [1/5, −3/5]].
- Multiply the inverse by [12, −1]: x = (1/5)(12) + (2/5)(−1) = 12/5 − 2/5 = 10/5 = 2.
- y = (1/5)(12) + (−3/5)(−1) = 12/5 + 3/5 = 15/5 = 3.
Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)
Frequently asked questions
What extra skills do medium-level matrix questions add?
Beyond basic operations, medium questions bring in the determinant and inverse of a 2×2 matrix, and using the inverse to solve a pair of simultaneous linear equations. You need to convert the equations into matrix form correctly, identify the coefficient matrix, and apply the inverse method carefully to reach both unknowns.
What mistake causes the inverse matrix method to fail?
A common error is forgetting to check that the determinant is not zero before finding the inverse, if it is zero, the inverse does not exist. Students also forget to multiply by 1/determinant, or swap the wrong diagonal entries when writing the adjoint matrix.
How should the final answer be presented when solving simultaneous equations by matrix method?
After finding the inverse and multiplying it by the constants matrix, write the resulting column matrix and clearly state the values of the two unknowns, matching them to the original variables. Leaving the answer as an unlabelled matrix instead of stating x = ...
and y = ... can cost marks.