Question

Which one is correct, the following system of linear equations 2x – 3y + 4z = 7, 3x – 4y + 5z = 8, 4x – 5y + 6z = 9 has?

(a) No solutions

(b) Infinitely many solutions

(c) Unique Solution

(d) Can’t be predicted

The question was posed to me in an international level competition.

The doubt is from Application of Determinants topic in chapter Determinants of Mathematics – Class 12

Answer 1

Correct answer is (b) Infinitely many solutions

The explanation: Solving the given system of equation by Cramer’s rule, we get,

x = D1/D, y = D2/D, z = D3/D where,

D = \(\begin{vmatrix}2 & -3 & 4 \\3 & -4 & 5 \\4 & -5 & 6 \end {vmatrix}\)

D = –\(\begin{vmatrix}2 & 3 & 4 \\3 & 4 & 5 \\4 & 5 & 6 \end {vmatrix}\)

Now, performing, C3 = C3 – C2 and C2 = C2 – C1 we get,

D = –\(\begin{vmatrix}2 & 1 & 1 \\3 & 1 & 1 \\4 & 1 & 1 \end {vmatrix}\)

As two columns have identical values, so,

D = 0

Similarly,

D1 = \(\begin{vmatrix}7 & -3 & 4 \\8 & -4 & 5 \\9 & -5 & 6 \end {vmatrix}\)

Now, performing, C1 = C1 – C3

D1 = –\(\begin{vmatrix}3 & -3 & 4 \\3 & -4 & 5 \\3 & -5 & 6 \end {vmatrix}\)

Now, performing, C3 = C3 – C2

D1 = -3\(\begin{vmatrix}1 & -3 & 1 \\1 & -4 & 1 \\1 & -5 & 1 \end {vmatrix}\)

As two columns have identical values, so,

D1 = 0

D2 = \(\begin{vmatrix}2 & 7 & 4 \\3 & 9 & 5 \\4 & 8 & 6 \end {vmatrix}\)

Now, performing,  

D2 = –\(\begin{vmatrix}2 & 3 & 2 \\3 & 3 & 2 \\4 & 3 & 2 \end {vmatrix}\)

Now, performing, C2 = C2 – C3 and C3 = C3 – C1

D2 = 6\(\begin{vmatrix}2 & 1 & 1 \\3 & 1 & 1 \\4 & 1 & 1 \end {vmatrix}\)

As two columns have identical values, so,

D2 = 0

D3 = \(\begin{vmatrix}2 & -3 & 7 \\3 & -4 & 9 \\4 & -5 & 6 \end {vmatrix}\)

D3 = –\(\begin{vmatrix}2 & 3 & 4 \\3 & 4 & 4 \\4 & 5 & 4 \end {vmatrix}\)

Now, performing, C2 = C2 – C2 and C3 = C3 – C2

D3 = -4\(\begin{vmatrix}2 & 1 & 1 \\3 & 1 & 1 \\4 & 1 & 1 \end {vmatrix}\)

As two columns have identical values, so,

D3 = 0

Since, D = D1 = D2 = D3 = 0, thus, it has infinitely many solutions.