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Q.If [1212313a1]\begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & a & 1 \end{bmatrix} is a non-singular matrix and a∈Aa \in A, then the set AA is:

(a) R\mathbb{R}
(b) {0}\{0\}
(c) {4}\{4\}
(d) R−{4}\mathbb{R} - \{4\}
CBSECBSE Class XII Board 2023MCQ· 1mImportance★★★★★
✓ Free question

A matrix is non-singular when its determinant is non-zero. For this 3×33 \times 3 matrix, the determinant simplifies to a−4a - 4, so the matrix is non-singular for all a≠4a \neq 4. Thus A=R−{4}A = \mathbb{R} - \{4\}.

The key idea here is simple: a square matrix is called non-singular (or invertible) exactly when its determinant is not zero. If the determinant is zero, the matrix is singular — it has no inverse, and its rows (or columns) are linearly dependent.

So the problem reduces to: find all real numbers aa for which the determinant of the given matrix is non-zero. Then the set AA is precisely that collection of aa values.

Let’s compute the determinant.

  1. Write the matrix:

M=[1212313a1]M = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & a & 1 \end{bmatrix}

  1. Compute det⁡(M)\det(M) using expansion along the first row (or any row/column). I’ll expand along the first row:

det⁡(M)=1⋅∣31a1∣−2⋅∣2131∣+1⋅∣233a∣\det(M) = 1 \cdot \begin{vmatrix} 3 & 1 \\ a & 1 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 3 \\ 3 & a \end{vmatrix}

  1. Evaluate each 2×22 \times 2 determinant:

    • ∣31a1∣=(3)(1)−(1)(a)=3−a\begin{vmatrix} 3 & 1 \\ a & 1 \end{vmatrix} = (3)(1) - (1)(a) = 3 - a
    • ∣2131∣=(2)(1)−(1)(3)=2−3=−1\begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} = (2)(1) - (1)(3) = 2 - 3 = -1
    • ∣233a∣=(2)(a)−(3)(3)=2a−9\begin{vmatrix} 2 & 3 \\ 3 & a \end{vmatrix} = (2)(a) - (3)(3) = 2a - 9
  2. Substitute back:

det⁡(M)=1⋅(3−a)−2⋅(−1)+1⋅(2a−9)\det(M) = 1 \cdot (3 - a) - 2 \cdot (-1) + 1 \cdot (2a - 9)

Simplify:

det⁡(M)=3−a+2+2a−9\det(M) = 3 - a + 2 + 2a - 9

Combine like terms:

det⁡(M)=(3+2−9)+(−a+2a)=(−4)+(a)=a−4\det(M) = (3 + 2 - 9) + (-a + 2a) = (-4) + (a) = a - 4

Tip

Notice how the first and third columns are identical except for the middle entry? That’s a quick check: if a=4a = 4, the second and third rows become proportional? Actually no — but the determinant simplifies so cleanly that you could also spot that the matrix has a pattern: subtracting row 1 from row 2 and row 3 gives a simpler form. Try it: R2→R2−2R1R_2 \to R_2 - 2R_1, R3→R3−3R1R_3 \to R_3 - 3R_1 yields [1210−1−10a−6−2]\begin{bmatrix}1&2&1\\0&-1&-1\\0&a-6&-2\end{bmatrix}, and the determinant is 1⋅∣−1−1a−6−2∣=(−1)(−2)−(−1)(a−6)=2+a−6=a−41 \cdot \begin{vmatrix}-1&-1\\a-6&-2\end{vmatrix} = (-1)(-2) - (-1)(a-6) = 2 + a - 6 = a - 4. Same result, faster.

  1. The matrix is non-singular when det⁡(M)≠0\det(M) \neq 0, i.e.:

a−4≠0⇒a≠4a - 4 \neq 0 \quad \Rightarrow \quad a \neq 4

  1. Since aa is a real number (the problem implies a∈Ra \in \mathbb{R} from the options), the set of all such aa is all real numbers except 44:

A=R−{4}A = \mathbb{R} - \{4\}

Watch out

A common mistake is to think "non-singular" means the determinant is positive, or that it must be something other than zero in some special sense. No — it simply means not zero. Also, don’t confuse this with the matrix being "non-invertible" — that’s the same as singular. So a=4a = 4 is the only value that makes the matrix singular.

✓Final answer

The correct option is (d) R−{4}\mathbb{R} - \{4\}.

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