What is det(adj(A)) for a 3×3 matrix?

For a 3×3 matrix, det(adj(A)) = (det A)². This is derived from A·adj(A) = det(A)·I.

How to find det(A) for the given matrix?

det(A) = α(15−0)−1(10−0)+2(8−0) = 15α−10+16 = 15α+6.

How to find α using det(B)=66?

With α=1: det(A)=21, adj(A) can be computed. B = [[1,0,0],[0,−5,0],[0,4,−2]] + adj(A). Computing det(B)=66 confirms α=1.

What is det(adj(A)) with α=1?

det(A)=21, so det(adj(A))=(det A)²=441.

What is the formula A·adj(A)?

A·adj(A) = adj(A)·A = det(A)·I for any square matrix A.

JEERANKERS

MathMatrices

Q MCQ Matrices & Determinants

Let $A=\begin{pmatrix}\alpha&1&2\\2&3&0\\0&4&5\end{pmatrix}$ and $B=\begin{pmatrix}1&0&0\\0&-5\alpha&0\\0&4\alpha&-2\alpha\end{pmatrix}+\text{adj}(A)$.

If $\det(B)=66$, then $\det(\text{adj}(A))$ equals:

A) $289$ B) $361$ C) $441$ D) $529$

✅ Correct Answer

C) 441

✓

Solution

Compute det(A) in terms of α

Expanding along Row 1:

$\det(A)=\alpha(3\times5-0\times4)-1(2\times5-0\times0)+2(2\times4-3\times0)$

$=\alpha(15)-1(10)+2(8)$

$=15\alpha-10+16=15\alpha+6$

Find α using det(B) = 66

Try $\alpha=1$: $\det(A)=15(1)+6=21$.

With $\alpha=1$, compute $\text{adj}(A)$ where $A=\begin{pmatrix}1&1&2\\2&3&0\\0&4&5\end{pmatrix}$:

$C_{11}=+(15-0)=15,\quad C_{12}=-(10-0)=-10,\quad C_{13}=+(8-0)=8$
$C_{21}=-(5-8)=3,\quad C_{22}=+(5-0)=5,\quad C_{23}=-(4-0)=-4$
$C_{31}=+(0-6)=-6,\quad C_{32}=-(0-4)=4,\quad C_{33}=+(3-2)=1$

$\text{adj}(A)=\begin{pmatrix}15&3&-6\\-10&5&4\\8&-4&1\end{pmatrix}$

Form B and verify det(B) = 66

$B=\begin{pmatrix}1&0&0\\0&-5&0\\0&4&-2\end{pmatrix}+\begin{pmatrix}15&3&-6\\-10&5&4\\8&-4&1\end{pmatrix}=\begin{pmatrix}16&3&-6\\-10&0&4\\8&0&-1\end{pmatrix}$

$\det(B)=16(0\cdot(-1)-4\cdot0)-3((-10)(-1)-4\cdot8)+(-6)(0-0)$

$=16(0)-3(10-32)+(-6)(0)$
$=0-3(-22)+0=66\ \checkmark$

Compute det(adj(A))

Using the standard formula for $3\times3$ matrices:

$$\det(\text{adj}(A))=(\det A)^{n-1}=(\det A)^2=21^2=\boxed{441}$$

📘 Key Formula

For an $n\times n$ matrix $A$:
• $A\cdot\text{adj}(A)=\det(A)\cdot I_n$
• $\det(\text{adj}(A))=(\det A)^{n-1}$
• For $3\times3$: $\det(\text{adj}(A))=(\det A)^2$
These are must-memorise formulas for JEE.

💡

Important Concepts

Adjoint of a Matrix $\text{adj}(A)$ is the transpose of the cofactor matrix. For $3\times3$: compute all 9 cofactors $C_{ij}=(-1)^{i+j}M_{ij}$, then transpose the resulting matrix.

det(adj A) Formula $\det(\text{adj}(A))=(\det A)^{n-1}$. For $n=3$: $(\det A)^2$. So if $\det A=21$, then $\det(\text{adj}(A))=441$.

Cofactor Sign Rule Cofactor $C_{ij}=(-1)^{i+j}\times$(minor $M_{ij}$). Signs follow the checkerboard pattern: $+,-,+\,/\,-,+,-\,/\,+,-,+$.

Strategy for These Problems Find $\alpha$ by trying integer values — JEE problems always have clean integer answers. With $\alpha=1$: $\det(A)=21$, $\det(\text{adj}(A))=441$ ✓. Always verify with the given condition ($\det(B)=66$).

FAQs

Where does the formula det(adj A) = (det A)² come from?

⌄

From $A\cdot\text{adj}(A)=\det(A)\cdot I$. Taking determinant of both sides: $\det(A)\cdot\det(\text{adj}(A))=(\det A)^n$. Dividing: $\det(\text{adj}(A))=(\det A)^{n-1}$.

How to compute cofactor C₁₁ for this matrix?

⌄

$C_{11}=(-1)^{1+1}\cdot M_{11}$ where $M_{11}$ is the minor obtained by deleting row 1, col 1: $\begin{vmatrix}3&0\\4&5\end{vmatrix}=15-0=15$. So $C_{11}=+15$.

How to find α quickly without computing all of adj(A)?

⌄

Since the answer choices for $\det(\text{adj}(A))$ are perfect squares ($289=17^2$, $361=19^2$, $441=21^2$, $529=23^2$), check which value of $\alpha$ gives $\det(A)\in\{17,19,21,23\}$. $15\alpha+6=21\Rightarrow\alpha=1$ is the cleanest.

What are the other key adjoint formulas?

⌄

$\text{adj}(\text{adj}(A))=(\det A)^{n-2}\cdot A$, $\det(\text{adj}(\text{adj}(A)))=(\det A)^{(n-1)^2}$, $\text{adj}(AB)=\text{adj}(B)\cdot\text{adj}(A)$.

Is this from JEE Main 2026?

⌄

Yes, this question appeared in JEE Main 2026.

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