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If [−110ab1121]\begin{bmatrix} -1 & 1 & 0 \\ a & b & 1 \\ 1 & 2 & 1 \end{bmatrix} is a singular matrix, then the relation between aa and bb is:

Solution

✅ Correct Option: 3

A matrix is singular when its determinant equals zero.

For the matrix [−110ab1121]\begin{bmatrix} -1 & 1 & 0 \\ a & b & 1 \\ 1 & 2 & 1 \end{bmatrix}, the determinant must equal zero.


Expanding along Row 1 (which contains a zero):

det⁡(A)=(−1)×∣b121∣−(1)×∣a111∣+(0)×∣ab12∣\det(A) = (-1) \times \begin{vmatrix} b & 1 \\ 2 & 1 \end{vmatrix} - (1) \times \begin{vmatrix} a & 1 \\ 1 & 1 \end{vmatrix} + (0) \times \begin{vmatrix} a & b \\ 1 & 2 \end{vmatrix}


The 2×2 determinants:

∣b121∣=(b)(1)−(1)(2)\begin{vmatrix} b & 1 \\ 2 & 1 \end{vmatrix} = (b)(1) - (1)(2)

=b−2= b - 2

∣a111∣=(a)(1)−(1)(1)\begin{vmatrix} a & 1 \\ 1 & 1 \end{vmatrix} = (a)(1) - (1)(1)

=a−1= a - 1


det⁡(A)=(−1)(b−2)−(1)(a−1)+0\det(A) = (-1)(b - 2) - (1)(a - 1) + 0

=−b+2−a+1= -b + 2 - a + 1

=−a−b+3= -a - b + 3


Since the matrix is singular:

det⁡(A)=0\det(A) = 0

−a−b+3=0-a - b + 3 = 0

−a−b=−3-a - b = -3

a+b=3a + b = 3

Therefore, the relation between aa and bb is a+b=3a + b = 3.

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