Algebra

Matrices & Determinants

Medium

common

If $A = \begin{bmatrix} a & a & a \\ o & a & a \\ o & o & a \end{bmatrix}$ , then $|adj A|$ is equal to

$a^3$

$a^6$

$a^9$

$a^{27}$

✅ Correct Option: 2

Given matrix $A = \begin{bmatrix} a & a & a \\ 0 & a & a \\ 0 & 0 & a \end{bmatrix}$

For any $n \times n$ matrix $A$ , the relationship between the determinant of the adjoint and the determinant of $A$ is:

$|adj \, A| = |A|^{n-1}$

Since the given matrix is $3 \times 3$ , we have $n = 3$ :

$|adj \, A| = |A|^{3-1}$

$|adj \, A| = |A|^2$

The matrix $A$ is upper triangular (all entries below the main diagonal are zero).

For an upper triangular matrix, the determinant equals the product of the diagonal elements:

$|A| = a \times a \times a$

$|A| = a^3$

Using the formula:

$|adj \, A| = |A|^2$

$|adj \, A| = (a^3)^2$

$|adj \, A| = a^6$

Therefore, $|adj \, A| = a^6$

2026: 19 May Shift 1

Given that $Y_{3\times k}$ , $W_{n\times 3}$ and $P_{p\times k}$ are matrices of specified order, then the condition for n, p, k so that $3PY + 2WY$ is well defined, is:

2025: 30 May Shift 2

Let $A$ and $B$ be square matrices of order 3, then det $[(A - A^T) + (B - B^T)]$ is equal to

2023: 15 June Shift 2

If the points (2, 1), $(-1, 4)$ and (a, 3) are collinear then the value/(s) of a is/(are) :

2026: 29 May Shift 1

Let $A=\begin{bmatrix}a & 0 \\ b & 0\end{bmatrix}$ and $A^2=\begin{bmatrix}4 & c \\ 2 & 0\end{bmatrix}$ , $a,b,c \in \mathbb{Z}$ . Then the minimum value of $a+b+c$ is:

2025: 3 June Shift 2

The system of equations

$x + y + z = 4$

$x + 2y + 3z = 12$

$x + 3y + \lambda z = \mu$ has a unique solution if

2026: 29 May Shift 1

Match List-I with List-II

List-I (Singular matrix)	List-II (A value of $x$ )
(A) $A=\begin{bmatrix}x & 4 \\ 2 & 2x\end{bmatrix}$	(I) 3
(B) $A=\begin{bmatrix}3 & x \\ 3x & 9\end{bmatrix}$	(II) 4
(C) $A=\begin{bmatrix}x & 1 \\ 1 & x\end{bmatrix}$	(III) 2
(D) $A=\begin{bmatrix}x & 1 \\ 16 & x\end{bmatrix}$	(IV) 1

Choose the correct answer from the options given below:

If A=[aaaoaaooa]A = \begin{bmatrix} a & a & a \\ o & a & a \\ o & o & a \end{bmatrix}A=​aoo​aao​aaa​​, then ∣adjA∣|adj A|∣adjA∣ is equal to

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $A = \begin{bmatrix} a & a & a \\ o & a & a \\ o & o & a \end{bmatrix}$ , then $|adj A|$ is equal to