Let A={bmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 {bmatrix}, then A is

Correct Answer is : None of these

Correct Answer is : None of these

Test Index

AP DSC School Assistant Math Model Test 2

Section: Subject Specific

Question : 100 of 160

Marks: +1, -0

Let

A=\begin{bmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{bmatrix}

, then A is

Solution:

Concept:

A square matrix is:

Nilpotent if $A^m = 0$ for some positive integer $m$ , and $A^{m-1} \neq 0$ .

Scalar if all diagonal entries are equal and all off‑diagonal entries are zero.

Explanation:

Given

A = \begin{bmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{bmatrix}

Check for nilpotent:

Compute

A^2

A^2 = A \times A = \begin{bmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{bmatrix} \begin{bmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{bmatrix}

= \begin{bmatrix} 1\cdot1+1\cdot5+3\cdot(-2) & 1\cdot1+1\cdot2+3\cdot(-1) & 1\cdot3+1\cdot6+3\cdot(-3) \\ 5\cdot1+2\cdot5+6\cdot(-2) & 5\cdot1+2\cdot2+6\cdot(-1) & 5\cdot3+2\cdot6+6\cdot(-3) \\ (-2)\cdot1+(-1)\cdot5+(-3)\cdot(-2) & (-2)\cdot1+(-1)\cdot2+(-3)\cdot(-1) & (-2)\cdot3+(-1)\cdot6+(-3)\cdot(-3) \end{bmatrix}

= \begin{bmatrix} 1+5-6 & 1+2-3 & 3+6-9 \\ 5+10-12 & 5+4-6 & 15+12-18 \\ -2-5+6 & -2-2+3 & -6-6+9 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 3 & 9 \\ -1 & -1 & -3 \end{bmatrix}

Since

A^2 \neq 0

A

is not nilpotent.

Check for idempotent:

A^2 = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 3 & 9 \\ -1 & -1 & -3 \end{bmatrix} \neq A

, so

A

is not idempotent.

Check for scalar:

The diagonal entries are

1, 2, -3

, which are not all equal, so

A

is not a scalar matrix.

Thus, none of the given options fit.

Answer:

D. None of these

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