Let A = [alpha 1 2; 2 3 0; 0 4 5] and B = [1 0 0; 0 -5alpha 0; 0 4alpha -2alpha] + adj(A). If det(B)=66, then det(adj(A)) equals :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q4 – JEE-MAIN 2026

← Back

Home Maths MatricesQuestion #293

Let

A = [\begin{matrix} α & 1 & 2 \\ 2 & 3 & 0 \\ 0 & 4 & 5 \end{matrix}]

and

B = [\begin{matrix} 1 & 0 & 0 \\ 0 & - 5 α & 0 \\ 0 & 4 α & - 2 α \end{matrix}] + adj (A)

. If

det (B) = 66

, then

det (adj (A))

equals :

Clue 1 of 3: Matrix Properties

Recall the relationship between the determinant of a matrix and the determinant of its adjoint. For an $n \times n$ matrix $A$ , $det (adj (A)) = (det (A))^{n - 1}$ .

🧠 Full Solution Path

Step 1: Calculate

det (A)

and

adj (A)

in terms of

α

✦ Active

First, calculate the determinant of matrix $A$ :

det (A) = α (3 \times 5 - 0 \times 4) - 1 (2 \times 5 - 0 \times 0) + 2 (2 \times 4 - 3 \times 0) = 15 α - 10 + 16 = 15 α + 6

Next, calculate the adjoint of $A$ , $adj (A)$ , which is the transpose of its cofactor matrix. The cofactors are:

C_{11} = 15, C_{12} = - 10, C_{13} = 8

C_{21} = 3, C_{22} = 5 α, C_{23} = - 4 α

C_{31} = - 6, C_{32} = 4, C_{33} = 3 α - 2

So, the adjoint matrix is:

adj (A) = [\begin{matrix} 15 & 3 & - 6 \\ - 10 & 5 α & 4 \\ 8 & - 4 α & 3 α - 2 \end{matrix}]

Step 2: Form matrix

B

and calculate its determinant○ Expand

Given $B = [\begin{matrix} 1 & 0 & 0 \\ 0 & - 5 α & 0 \\ 0 & 4 α & - 2 α \end{matrix}] + adj (A)$ . Add the two matrices:

B = [\begin{matrix} 1 + 15 & 0 + 3 & 0 - 6 \\ 0 - 10 & - 5 α + 5 α & 0 + 4 \\ 0 + 8 & 4 α - 4 α & - 2 α + 3 α - 2 \end{matrix}] = [\begin{matrix} 16 & 3 & - 6 \\ - 10 & 0 & 4 \\ 8 & 0 & α - 2 \end{matrix}]

Now, calculate $det (B)$ by expanding along the second column (due to the zeros):

det (B) = - 3 det [\begin{matrix} - 10 & 4 \\ 8 & α - 2 \end{matrix}] = - 3 ((- 10) (α - 2) - (4) (8))

det (B) = - 3 (- 10 α + 20 - 32) = - 3 (- 10 α - 12) = 30 α + 36

💡 Teacher's Secret Hint

Notice the cancellation of terms involving $α$ in the second column of $B$ , which simplifies the determinant calculation.

Step 3: Solve for

α

and find

det (adj (A))

○ Expand

We are given that $det (B) = 66$ . Equate the calculated determinant to 66 to find $α$ :

30 α + 36 = 66 ⟹ 30 α = 30 ⟹ α = 1

Substitute $α = 1$ back into the expression for $det (A)$ :

det (A) = 15 (1) + 6 = 21

Finally, use the property that for an $n \times n$ matrix $A$ , $det (adj (A)) = (det (A))^{n - 1}$ . Since $A$ is a $3 \times 3$ matrix ( $n = 3$ ):

det (adj (A)) = (det (A))^{3 - 1} = (det (A))^{2} = (21)^{2} = 441

💡 Teacher's Secret Hint

Remember the fundamental property relating the determinant of a matrix to the determinant of its adjoint.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Maths Question 4 – JEE-MAIN 2026

Clue 1 of 3: Matrix Properties

🧠 Full Solution Path