Higher Mathematics 1st Paper · Classes 11-12 · Chapter 1 · 1.6

Using the Properties: Values, Proofs and Equations.

Open exercise 1(B) and every determinant looks like its own puzzle. In fact almost all of them give way to one of four moves: add lines whose sums match, make zeros and expand, subtract neighbours to get factors, or show two lines are proportional so the value is zero. The real work is spotting the first move.

1. Equal sums: add the columns

The book's Example 2: solve |3 + x, 4, 2 ; 4, 2 + x, 3 ; 2, 3, 4 + x| = 0 (rows separated by ;). Every row adds up to x + 9. So c₁′ = c₁ + c₂ + c₃ turns all three entries of column 1 into x + 9. Take (x + 9) outside and column 1 becomes 1, 1, 1. Now r₁′ = r₁ − r₂ and r₂′ = r₂ − r₃ put two zeros in that column, and expanding along it gives (x + 9)(x² − 3) = 0. So x = −9, ±√3.

Look at the equation first. Slide x and the entries change, but every row still adds up to x + 9. Wherever the curve touches zero there is a root. Find all three.

The board wants the steps in order as well as the answer. Here are the six steps shuffled; tap them in the right order.

The same move works elsewhere, with columns in place of rows. Exercise 1(B) 4(i): |a − b − c, 2a, 2a ; 2b, b − c − a, 2b ; 2c, 2c, c − a − b|. The row sums differ here, but every column adds up to a + b + c. So r₁′ = r₁ + r₂ + r₃ makes row 1 equal (a + b + c)(1, 1, 1). Take that out, make zeros, and the value is (a + b + c)³.

With ω a complex cube root of unity (ω³ = 1, 1 + ω + ω² = 0), exercise 1(v): |1, −ω, ω² ; −ω, ω², 1 ; ω², 1, −ω|. Every row sums to −2ω (for instance 1 − ω + ω² = −2ω), so c₁′ = c₁ + c₂ + c₃ lets −2ω come out. The value is −4.

2. Make zeros, then expand

Property (vii) says subtracting one row from another leaves the value alone. So make as many zeros as you can before you expand. Exercise 1(B) 3(i): in |1, 1, 1 ; 1, 1 + x, 1 ; 1, 1, 1 + y|, r₂′ = r₂ − r₁ and r₃′ = r₃ − r₁ give |1, 1, 1 ; 0, x, 0 ; 0, 0, y|. That is triangular, so the value is the product of the diagonal: xy.

In Example 1 the zeros come from a neater trick. Prove |a, b, ax + by ; b, c, bx + cy ; ax + by, bx + cy, 0| = (b² − ac)(ax² + 2bxy + cy²). The third column is a mixture of the first two, so c₃′ = c₃ − (x c₁ + y c₂) wipes out its top two entries.

When a numerical determinant shows no shortcut, just expand along the first row. Exercise 1(B) 1(i): |10, −5, 3 ; 4, 20, 11 ; −30, 15, 14| = 10(280 − 165) + 5(56 + 330) + 3(60 + 600) = 1150 + 1930 + 1980 = 5060.

3. Subtract to get factors

When the rows hold a, b, c and their powers, subtract neighbouring columns. Exercise 1(B) 2(i): in |1, 1, 1 ; a, b, c ; a³, b³, c³|, c₁′ = c₁ − c₂ and c₂′ = c₂ − c₃ put a − b and a³ − b³ in column 1, both with the factor (a − b). Column 2 gives (b − c) the same way. Finish the working and the value is (a − b)(b − c)(c − a)(a + b + c).

Test why (a − b) is a factor. Change a, b and c. Make any two equal and two columns become identical, and the value drops to zero.

Two more with the same move: 2(iv) |a², ab, b² ; 2a, a + b, 2b ; 1, 1, 1| = (a − b)³. And in 5(i) |a, b, c ; a², b², c² ; a³, b³, c³|, take a, b, c out of the three columns and what is left works like 2(i), with value abc(a − b)(b − c)(c − a).

Example 6(b) takes a detour: |bc, a, a² ; ca, b, b² ; ab, c, c²|. Multiply the rows by a, b and c and divide by abc outside. Column 1 becomes abc, abc, abc; take that out, the abc cancels, and you are left with |1, a², a³ ; 1, b², b³ ; 1, c², c³| = (a − b)(b − c)(c − a)(ab + bc + ca).

4. Showing the value is zero

“Show that the value is zero” usually means making two rows or columns identical or proportional. In 2(ii), |p, a, b + c ; p, b, c + a ; p, c, a + b|, c₃′ = c₃ + c₂ makes column 3 equal (a + b + c)(1, 1, 1), while column 1 is already p(1, 1, 1). Two proportional columns, so the value is 0.

Logs fall to the same trick. 2(vi): |log x, log y, log z ; log 2x, log 2y, log 2z ; log 3x, log 3y, log 3z|. Since log 2x − log x = log 2, r₂′ = r₂ − r₁ and r₃′ = r₃ − r₁ make row 2 equal (log 2)(1, 1, 1) and row 3 equal (log 3)(1, 1, 1). Proportional rows, value 0.

5(iv): |0, b − a, c − a ; a − b, 0, c − b ; a − c, b − c, 0|. The diagonal is zero and a_ij = −a_ji, so it is skew-symmetric. Transposing keeps the value (property ii), and taking −1 out of each row of the transpose gives the original back. So D = (−1)³D = −D, which means D = 0. A skew-symmetric determinant of odd order is always 0.

Now spot the moves. What is the first move for each determinant?

5. Proofs without expanding, and more values

Exercise 1(B) 7 says “prove without expanding”. In (i), c₂′ = c₂ + a c₁ gives column 2 its a back and c₃′ = c₃ + b c₁ gives column 3 its b, so |1, x − a, y − b ; …| becomes |1, x, y ; …|, and property (vii) keeps the value. In (iii), |b + c, c + a, a + b ; q + r, r + p, p + q ; y + z, z + x, x + y| = 2|a, b, c ; p, q, r ; x, y, z|: every entry is a sum of two terms, so split it by property (vi); the parts with two equal columns vanish and the two that remain add up to twice the original.

The answers to exercise 1(ii) to (iv) are worth knowing. |x + y, x, y ; x, x + z, z ; y, z, y + z| = 4xyz. In |−a², ab, ac ; ab, −b², bc ; ac, bc, −c²|, take a, b, c out of the rows and a, b, c out of the columns, leaving a²b²c² × |−1, 1, 1 ; 1, −1, 1 ; 1, 1, −1|; that last determinant is 4, so the answer is 4a²b²c². And |1, 1, 1 ; a, b, c ; a² − bc, b² − ca, c² − ab| = 0.

Two longer proofs: 6(i) |b² + c², ab, ca ; ab, c² + a², bc ; ca, bc, a² + b²| = 4a²b²c² (multiply the rows by a, b, c, take a, b, c out of the columns, then simplify), and 6(vi) |−2a, a + b, a + c ; b + a, −2b, b + c ; c + a, c + b, −2c| = 4(a + b)(b + c)(c + a).

Numerical determinants come up nearly every year. Pick one, write its value, then see whether there was a shortcut.

6. Determinant equations

When the equation has an x in it, the aim is always the same: break the determinant into factors, then set each factor to zero. Exercise 1(B) 9(i): |x + a, a, a ; b, x + b, b ; c, c, x + c| = 0. Adding all three rows brings out x + a + b + c, the value is x²(x + a + b + c), and so x = 0, −(a + b + c). 9(ii): |2 + x, 3, 1 ; 3, 1 + x, 2 ; 1, 2, 3 + x| = 0 is Example 2's twin: every row adds up to x + 6, so x = −6, ±√3.

Hunt for the roots. Pick an equation, then try a value of x and see whether the determinant becomes zero.

7. Check yourself

Three questions, and all three can be answered without expanding. See whether two columns can be made proportional.

Source: Higher Mathematics 1st Paper, Classes 11-12 (NCTB), Chapter 1 “Matrices and Determinants”, section 1.6.