If A is a square matrix such that $A^2 = A$, then $(I - A)^3 + A$ is equal to (A) $I$ (B) $0$ (C) $I - A$ (D) $I + A$

> [!TLDR] > The key idea is that $A^2 = A$ makes $A$ an **idempotent matrix**, which lets us simplify powers of $(I - A)$ using the binomial theorem. Expanding $(I - A)^3$ and adding $A$ gives $I$ as the final result. We are given that $A^2 = A$. This is the defining property of an **idempotent matrix** — a matrix that, when multiplied by itself, gives itself back. Such matrices behave like "projections" in linear algebra. The trick here is that $I - A$ also becomes idempotent, because: $$(I - A)^2 = I - 2A + A^2 = I - 2A + A = I - A$$ So $(I - A)^2 = I - A$ as well. This symmetry makes expansions clean. Now, let’s work through the problem step by step. 1. **Expand $(I - A)^3$ using the binomial theorem.** Since matrix multiplication is distributive and associative (and $I$ commutes with any matrix), we can expand just like numbers: $$(I - A)^3 = I^3 - 3I^2A + 3IA^2 - A^3$$ But $I^3 = I$, $I^2A = A$, $IA^2 = A^2$, and $A^3 = A \cdot A^2 = A \cdot A = A^2 = A$ (using $A^2 = A$ repeatedly). So: $$(I - A)^3 = I - 3A + 3A^2 - A^3$$ 2. **Replace $A^2$ and $A^3$ using $A^2 = A$.** Since $A^2 = A$, then $A^3 = A^2 \cdot A = A \cdot A = A$ as well. So: $$(I - A)^3 = I - 3A + 3A - A = I - A$$ > [!TIP] > Notice that $(I - A)^3 = I - A$ is actually a special case of a pattern: for idempotent $A$, $(I - A)^n = I - A$ for any $n \geq 1$. You can prove this by induction. 3. **Now add $A$ to this result.** The expression we need is $(I - A)^3 + A$. Substituting: $$(I - A)^3 + A = (I - A) + A = I$$ The $-A$ and $+A$ cancel perfectly, leaving just the identity matrix $I$. > [!WARNING] > A common mistake is to forget that $A^3 = A$ when $A^2 = A$. Students sometimes stop at $A^3 = A^2 \cdot A$ and leave it as $A^2$, then get a wrong cancellation. Always reduce powers fully using the given property. > [!ANSWER] > The value is $\boxed{I}$, which corresponds to option **(A)**.

If A is a square matrix such that A² = A, then (I - A)³ — CBSE Class 12 Mathematics 2020 PYQ

Question

A^2 = A

CBSECBSE Class XII Board 2020MCQ· 1mImportance★★★★★

✓ Free question

Concept understanding — Idempotent Matrix Property

Idempotent Matrix Property

Imagine a machine that cleans up a photo. Feed its own output back in, and nothing more changes — the whole job was done in one pass. An idempotent matrix behaves exactly like that.

The precise statement

A square matrix $A$ is idempotent if

A^2 = A,

that is, multiplying the matrix by itself gives the matrix back unchanged.

Note

The name is from Latin idem (same) + potens (power): raising to a higher power has no further effect.

What follows immediately

If $A^2 = A$ , then repeated multiplication never changes anything:

A^3 = A^2\cdot A = A\cdot A = A, \qquad \text{and in general } A^n = A \text{ for every } n \ge 1.

A simple example

A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \qquad A^2 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = A.

Geometrically this matrix flattens every vector onto the $x$ -axis; once flattened, doing it again changes nothing — which is the intuitive meaning of idempotent.

How it usually appears

In problems you are typically given that $A$ is idempotent and asked to simplify an expression. The identity $A^2 = A$ (and hence $A^n = A$ ) is the tool. For instance,

(I - A)^2 = I - 2A + A^2 = I - 2A + A = I - A,

so $I - A$ is idempotent too.

Watch out

Don't confuse idempotent with involutory. An involutory matrix satisfies $A^2 = I$ (it "resets" after two applications), whereas idempotent satisfies $A^2 = A$ (it "settles" after one). Different conditions, different behaviour.

Tip

When a question says " $A$ is idempotent," immediately replace any $A^2$ (or higher power) by $A$ — that single move solves most such problems.

Idempotent matrices, satisfying A² = A, appear in the miscellaneous exercises of the CBSE Class 12 Matrices chapter and are a favourite way to test whether students can manipulate matrix equations without over-computing. "Idempotent matrix questions and examples" is a commonly searched topic for board exam and JEE Main revision, especially when contrasted with involutory matrices satisfying A² = I.

The key idea is that $A^2 = A$ makes $A$ an idempotent matrix, which lets us simplify powers of $(I - A)$ using the binomial theorem. Expanding $(I - A)^3$ and adding $A$ gives $I$ as the final result.

We are given that $A^2 = A$ . This is the defining property of an idempotent matrix — a matrix that, when multiplied by itself, gives itself back. Such matrices behave like "projections" in linear algebra. The trick here is that $I - A$ also becomes idempotent, because:

(I - A)^2 = I - 2A + A^2 = I - 2A + A = I - A

So $(I - A)^2 = I - A$ as well. This symmetry makes expansions clean.

Now, let’s work through the problem step by step.

Expand $(I - A)^3$ using the binomial theorem. Since matrix multiplication is distributive and associative (and $I$ commutes with any matrix), we can expand just like numbers:

(I - A)^3 = I^3 - 3I^2A + 3IA^2 - A^3

But $I^3 = I$ , $I^2A = A$ , $IA^2 = A^2$ , and $A^3 = A \cdot A^2 = A \cdot A = A^2 = A$ (using $A^2 = A$ repeatedly). So:

(I - A)^3 = I - 3A + 3A^2 - A^3

Replace $A^2$ and $A^3$ using $A^2 = A$ . Since $A^2 = A$ , then $A^3 = A^2 \cdot A = A \cdot A = A$ as well. So:

(I - A)^3 = I - 3A + 3A - A = I - A

Tip

Notice that $(I - A)^3 = I - A$ is actually a special case of a pattern: for idempotent $A$ , $(I - A)^n = I - A$ for any $n \geq 1$ . You can prove this by induction.

Now add $A$ to this result. The expression we need is $(I - A)^3 + A$ . Substituting:

(I - A)^3 + A = (I - A) + A = I

The $-A$ and $+A$ cancel perfectly, leaving just the identity matrix $I$ .

Watch out

A common mistake is to forget that $A^3 = A$ when $A^2 = A$ . Students sometimes stop at $A^3 = A^2 \cdot A$ and leave it as $A^2$ , then get a wrong cancellation. Always reduce powers fully using the given property.

✓Final answer

\boxed{I}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set 65/3/11 markMCQ
$A$
›Reveal solutionSolution
The key idea is that $A^2 = A$ (idempotent property) lets us simplify powers of $(I - A)$ by expanding and using $A^n = A$ for $n \ge 1$ . Expanding $(I - A)^3$ gives $I - 3A + 3A^2 - A^3 = I - 3A + 3A - A = I - A$ . Comparing with $xA + I$ , we get $x = -1$ .

Why This Works

The condition $A^2 = A$ is called idempotence — it means $A$ is a projection matrix. Once you square it, you get the same matrix back. This has a powerful consequence: for any integer $n \ge 1$ , $A^n = A$ . That’s because $A^3 = A^2 \cdot A = A \cdot A = A$ , and by induction it holds for all higher powers.

So when we expand $(I - A)^3$ , every power of $A$ beyond the first collapses to just $A$ . That makes the algebra extremely clean — no infinite series, no diagonalization, just a simple substitution.

Step-by-Step

Write the expansion. $(I - A)^3 = (I - A)(I - A)(I - A)$ . Expand using the binomial theorem (since $I$ and $A$ commute — $I$ commutes with everything):

$(I - A)^3 = I^3 - 3I^2 A + 3I A^2 - A^3 = I - 3A + 3A^2 - A^3.$

Apply the idempotent property. Because $A^2 = A$ , we have $A^3 = A^2 \cdot A = A \cdot A = A$ . So $A^2 = A$ and $A^3 = A$ . Substitute:

$I - 3A + 3A^2 - A^3 = I - 3A + 3A - A.$

Simplify. The $-3A$ and $+3A$ cancel:

$I - 3A + 3A - A = I - A.$
So $(I - A)^3 = I - A$ .

Compare with the given form. The problem states $(I - A)^3 = xA + I$ . We just found $(I - A)^3 = I - A$ . Therefore:

$I - A = xA + I.$

Solve for $x$ . Subtract $I$ from both sides:

$-A = xA.$
This is a matrix equation. Factor $A$ (careful: we cannot divide by a matrix, but we can compare coefficients if $A \neq 0$ ).

Rewrite as:
$-A = xA \quad \Rightarrow \quad (x + 1)A = 0.$
For this to hold for a non-zero matrix $A$ (the problem doesn’t say $A$ is zero, and if $A=0$ then any $x$ works, but the options are fixed), we need $x + 1 = 0$ , so $x = -1$ .

Watch out

A common mistake is to forget that $A^3 = A$ follows from $A^2 = A$ — some students stop at $A^2$ and leave $A^3$ as is, getting a wrong coefficient. Always reduce every power $\ge 1$ to $A$ .

Tip

You can also think of this as: if $A$ is idempotent, then $I - A$ is also idempotent! Check: $(I - A)^2 = I - 2A + A^2 = I - 2A + A = I - A$ . So $(I - A)^3 = (I - A)^2 (I - A) = (I - A)(I - A) = I - A$ . That’s a one-liner.

✓Final answer
$x$
CBSE 2025Set E1 markMCQ
$A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \Rightarrow A^{100} =$
(a) $100A$
(b) $101A$
(c) $A$
(d) $99A$
›Reveal solutionSolution

$A$ is the identity matrix $I$ , and $I^{100} = I = A$ .

Here $A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$ . Any power of the identity is the identity:

$A^{100} = I^{100} = I = A.$

✓Final answer
$A$
CBSE 2023Set E1 markMCQ
$A=\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$
(a) $25A$
(b) $24A$
(c) $2A$
(d) $A$
›Reveal solutionSolution

$A^{25}=A$ because $A=I$ .

$A=\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$ is the $2\times 2$ identity matrix $I$ . Any power of $I$ is $I$ :

$A^{25}=I^{25}=I=A.$

✓Final answer
$A$
CBSE 2020Set 65/1/11 markMCQ
$A^2 = A$
›Reveal solutionSolution
The key idea is that $A^2 = A$ makes $A$ an idempotent matrix, which lets us simplify powers of $(I - A)$ using the binomial theorem. Expanding $(I - A)^3$ and adding $A$ gives $I$ as the final result.

We are given that $A^2 = A$ . This is the defining property of an idempotent matrix — a matrix that, when multiplied by itself, gives itself back. Such matrices behave like "projections" in linear algebra. The trick here is that $I - A$ also becomes idempotent, because:
$(I - A)^2 = I - 2A + A^2 = I - 2A + A = I - A$
So $(I - A)^2 = I - A$ as well. This symmetry makes expansions clean.

Now, let’s work through the problem step by step.

Expand $(I - A)^3$ using the binomial theorem. Since matrix multiplication is distributive and associative (and $I$ commutes with any matrix), we can expand just like numbers:

$(I - A)^3 = I^3 - 3I^2A + 3IA^2 - A^3$
But $I^3 = I$ , $I^2A = A$ , $IA^2 = A^2$ , and $A^3 = A \cdot A^2 = A \cdot A = A^2 = A$ (using $A^2 = A$ repeatedly). So:
$(I - A)^3 = I - 3A + 3A^2 - A^3$

Replace $A^2$ and $A^3$ using $A^2 = A$ . Since $A^2 = A$ , then $A^3 = A^2 \cdot A = A \cdot A = A$ as well. So:

$(I - A)^3 = I - 3A + 3A - A = I - A$
Tip

Notice that $(I - A)^3 = I - A$ is actually a special case of a pattern: for idempotent $A$ , $(I - A)^n = I - A$ for any $n \geq 1$ . You can prove this by induction.

Now add $A$ to this result. The expression we need is $(I - A)^3 + A$ . Substituting:

$(I - A)^3 + A = (I - A) + A = I$
The $-A$ and $+A$ cancel perfectly, leaving just the identity matrix $I$ .

Watch out

A common mistake is to forget that $A^3 = A$ when $A^2 = A$ . Students sometimes stop at $A^3 = A^2 \cdot A$ and leave it as $A^2$ , then get a wrong cancellation. Always reduce powers fully using the given property.

✓Final answer
$\boxed{I}$
CBSE 2018Set ANNUAL1 markMCQ
$A$
(a) $I_3$
(b) $A$
(c) $3A$
(d) $I_3 - A$
›Reveal solutionSolution

Idempotency $A^2=A$ collapses $(A+I)^3-7A$ to $I_3$ .

Since $A^2=A$ , also $A^3=A\cdot A^2=A\cdot A=A$ .

$(A+I)^3=A^3+3A^2+3A+I=A+3A+3A+I=7A+I$ .

Hence $(A+I)^3-7A=7A+I-7A=I_3$ .

✓Final answer
$I_3$

Q.If A is a square matrix such that A2=AA^2 = AA2=A, then (I−A)3+A(I - A)^3 + A(I−A)3+A is equal to (A) III (B) 000 (C) I−AI - AI−A (D) I+AI + AI+A

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Q.If A is a square matrix such that $A^2 = A$ , then $(I - A)^3 + A$ is equal to
(A) $I$
(B) $0$
(C) $I - A$
(D) $I + A$

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