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.