If A={bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 {bmatrix}, then what is 23A3 - 19A2 - 4A equal to ?

Correct Answer is : Null matrix of order 3

{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 {bmatrix}

{bmatrix} 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7 {bmatrix}

Correct Answer is : Null matrix of order 3

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UPSC NDA II 2023 Mathematics Solved Paper

Question : 18 of 120

Marks: +1, -0

A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

, then what is 23A3 - 19A2 - 4A equal to ?

Solution:

Concept:

A

is the identity matrix

I

of order 3.

For any integer

n

I^{n} = I

Explanation:

The given matrix is

A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I

Since

A = I

, we have

A^{2} = I

A^{3} = I

, and so on.

The expression is

23A^{3} - 19A^{2} - 4A

Substitute

A^{3} = I

A^{2} = I

, and

A = I

into the expression.

This gives

23I - 19I - 4I = (23 - 19 - 4)I = 0 \cdot I

The scalar

0

multiplied by matrix

I

results in the null matrix of order 3.

Therefore, the value is the zero matrix of size

3 \times 3

Answer:

Null matrix of order 3

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