If $A = \begin{bmatrix}1 & 0 & -1\\ 2 & 1 & 3\\ 0 & 1 & 1\end{bmatrix}$, then verify that $A^2 + A = A(A + I)$, where $I$ is the $3 \times 3$ unit matrix.

> [!TLDR] > Computing directly, $A^2+A$ and $A(A+I)$ both equal $\begin{bmatrix} 2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5 \end{bmatrix}$, so the identity is verified. ## Why the identity should hold Matrix multiplication distributes over addition and $AI=A$, so $A(A+I)=A\cdot A+A\cdot I=A^2+A$. The problem asks us to **verify** this by explicit computation for the given matrix. ## Step 1 — compute $A^2=A\cdot A$ Using rows of $A$ against columns of $A$: - Row 1 $[1,0,-1]$: $(1,1)=1(1)+0(2)+(-1)(0)=1$; $(1,2)=1(0)+0(1)+(-1)(1)=-1$; $(1,3)=1(-1)+0(3)+(-1)(1)=-2$. - Row 2 $[2,1,3]$: $(2,1)=2(1)+1(2)+3(0)=4$; $(2,2)=2(0)+1(1)+3(1)=4$; $(2,3)=2(-1)+1(3)+3(1)=4$. - Row 3 $[0,1,1]$: $(3,1)=0(1)+1(2)+1(0)=2$; $(3,2)=0(0)+1(1)+1(1)=2$; $(3,3)=0(-1)+1(3)+1(1)=4$. $$A^2=\begin{bmatrix} 1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4 \end{bmatrix}.$$ ## Step 2 — compute $A^2+A$ Add entrywise: $$A^2+A=\begin{bmatrix} 1+1 & -1+0 & -2-1 \\ 4+2 & 4+1 & 4+3 \\ 2+0 & 2+1 & 4+1 \end{bmatrix}=\begin{bmatrix} 2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5 \end{bmatrix}.$$ ## Step 3 — compute $A(A+I)$ First $A+I=\begin{bmatrix} 2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2 \end{bmatrix}$. Then …

If A = 1 0 -1\2 1 3\0 1 1 , then verify that A² + A = A(A + I), where I is the 3 × 3 unit matrix.

NCERT Exemplar · Q26

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

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Concept understanding — Matrix Polynomial Evaluation

Matrix Polynomial Evaluation

You know how to evaluate a polynomial like $p(x) = 2x^2 - 3x + 5$ at a number: plug in $x$ , get a number out. Now plug in a square matrix $A$ instead. The variable becomes $A$ , and — crucially — the constant term becomes a multiple of the identity matrix $I$ , because you cannot add a bare number to a matrix.

The Definition

p(x) = a_n x^n + \cdots + a_1 x + a_0

Here $A^k$ is $k$ -fold matrix multiplication, $a_k A^k$ is scalar multiplication, and $a_0 I$ replaces the constant. The result is a square matrix of the same size as $A$ .

Note

There is no ambiguity from non-commutativity here: a polynomial only ever multiplies $A$ by itself, and $A$ always commutes with $A$ .

A Worked Example

Let $p(x) = x^2 - 4x + 3$ and $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ .

A^2 = \begin{pmatrix} 4 & 5 \\ 0 & 9 \end{pmatrix}, \quad -4A = \begin{pmatrix} -8 & -4 \\ 0 & -12 \end{pmatrix}, \quad 3I = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}.

Adding term by term,

p(A) = \begin{pmatrix} -1 & 1 \\ 0 & 0 \end{pmatrix}.

A Shortcut for Diagonal Matrices

If $A = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$ , then $A^k = \begin{pmatrix} \lambda_1^k & 0 \\ 0 & \lambda_2^k \end{pmatrix}$ , so

p(A) = \begin{pmatrix} p(\lambda_1) & 0 \\ 0 & p(\lambda_2) \end{pmatrix}.

You simply evaluate $p$ at each diagonal entry. …

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Computing directly, $A^2+A$ and $A(A+I)$ both equal $\begin{bmatrix} 2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5 \end{bmatrix}$ , so the identity is verified.

Why the identity should hold

Matrix multiplication distributes over addition and $AI=A$ , so $A(A+I)=A\cdot A+A\cdot I=A^2+A$ . The problem asks us to verify this by explicit computation for the given matrix.

Step 1 — compute $A^2=A\cdot A$

Using rows of $A$ against columns of $A$ :

Row 1 $[1,0,-1]$ : $(1,1)=1(1)+0(2)+(-1)(0)=1$ ; $(1,2)=1(0)+0(1)+(-1)(1)=-1$ ; $(1,3)=1(-1)+0(3)+(-1)(1)=-2$ .
Row 2 $[2,1,3]$ : $(2,1)=2(1)+1(2)+3(0)=4$ ; $(2,2)=2(0)+1(1)+3(1)=4$ ; $(2,3)=2(-1)+1(3)+3(1)=4$ .
Row 3 $[0,1,1]$ : $(3,1)=0(1)+1(2)+1(0)=2$ ; $(3,2)=0(0)+1(1)+1(1)=2$ ; $(3,3)=0(-1)+1(3)+1(1)=4$ .

A^2=\begin{bmatrix} 1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4 \end{bmatrix}.

Step 2 — compute $A^2+A$

Add entrywise:

A^2+A=\begin{bmatrix} 1+1 & -1+0 & -2-1 \\ 4+2 & 4+1 & 4+3 \\ 2+0 & 2+1 & 4+1 \end{bmatrix}=\begin{bmatrix} 2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5 \end{bmatrix}.

Step 3 — compute $A(A+I)$

First $A+I=\begin{bmatrix} 2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2 \end{bmatrix}$ . Then …

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

AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
$A = \begin{bmatrix} 2 & -5 \\ 3 & 1 \end{bmatrix}$
›Reveal solutionSolution
Cayley–Hamilton reduces $f(A)$ to $-14A-22I$ , giving $\begin{bmatrix}-50 & 70\\ -42 & -36\end{bmatrix}$ .

Concept and Intuition

Rather than computing $A^3$ by brute-force matrix multiplication three times, use the Cayley–Hamilton theorem: every square matrix satisfies its own characteristic equation, letting us reduce any polynomial in $A$ to a linear expression $\alpha A+\beta I$ .

Step-by-Step Solution

$A=\begin{bmatrix}2&-5\\3&1\end{bmatrix}$ : trace $=3$ , $\det=2(1)-(-5)(3)=2+15=17$ .

Characteristic equation: $A^2-3A+17I=0 \Rightarrow A^2=3A-17I$ .

$A^3=A\cdot A^2=A(3A-17I)=3A^2-17A=3(3A-17I)-17A=9A-51I-17A=-8A-51I$ .

$f(A)=A^3-2A^2-5I=(-8A-51I)-2(3A-17I)-5I=-8A-51I-6A+34I-5I=-14A-22I$ . …
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AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
$3\times 3$
›Reveal solutionSolution
$R$ is just the constant-term matrix extracted from each quadratic-in- $x$ entry; expanding its determinant gives $136$ .

Concept and Intuition

Since $Px^2+Qx+R$ must equal the given matrix of quadratics for every $x$ (matching coefficients entry-by-entry), $R$ is simply the matrix formed by the $x^0$ (constant) coefficient of every entry — no matrix algebra with $P,Q$ is even needed to answer this.

Step-by-Step Solution

Read off constant terms entry-by-entry: Row 1: $3,\,4,\,5$ ; Row 2: $2,\,-1,\,8$ ; Row 3: $5,\,-2,\,7$ . So $R=\begin{pmatrix}3&4&5\\2&-1&8\\5&-2&7\end{pmatrix}$ .

Expand along row 1: $\det R = 3\big[(-1)(7)-(8)(-2)\big]-4\big[(2)(7)-(8)(5)\big]+5\big[(2)(-2)-(-1)(5)\big]$ . …
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AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
$A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix}$
›Reveal solutionSolution
Straightforward matrix arithmetic: compute $A^2$ , then combine $A^2-5A+6I$ entrywise.

Concept and Intuition

This is a direct test of matrix multiplication and linear combination of matrices — no shortcut (like Cayley–Hamilton) is needed since we're not told $A$ 's characteristic polynomial matches this exact expression; computing directly is safest and fastest here.

Step-by-Step Solution

$A=\begin{bmatrix}1&0&2\\2&1&3\\3&2&4\end{bmatrix}$ .

Compute $A^2$ row by row:

Row 1: $(1\cdot1+0\cdot2+2\cdot3,\ 1\cdot0+0\cdot1+2\cdot2,\ 1\cdot2+0\cdot3+2\cdot4) = (7,4,10)$

Row 2: $(2\cdot1+1\cdot2+3\cdot3,\ 2\cdot0+1\cdot1+3\cdot2,\ 2\cdot2+1\cdot3+3\cdot4) = (13,7,19)$

Row 3: $(3\cdot1+2\cdot2+4\cdot3,\ 3\cdot0+2\cdot1+4\cdot2,\ 3\cdot2+2\cdot3+4\cdot4) = (19,10,28)$ So $A^2 = \begin{bmatrix}7&4&10\\13&7&19\\19&10&28\end{bmatrix}$ .

$5A = \begin{bmatrix}5&0&10\\10&5&15\\15&10&20\end{bmatrix}$ , and $6I=\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}$ .

Combine entrywise: $A^2-5A+6I$ : …
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AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
$A=\begin{pmatrix}i & 0\\0 & -i\end{pmatrix}$
›Reveal solutionSolution
Each of $A,B,C$ squares to $-I$ (they behave like quaternion-style imaginary units), and their product $ABC=I$ . Matching $A^2+B^2+C^2=-3I$ against the options shows it equals $3A^2B^2C^2$ .

Concept and Intuition

These three matrices are $2\times2$ representations reminiscent of Pauli-matrix-like objects, each satisfying $M^2=-I$ . Rather than manipulate symbols, direct matrix multiplication settles the identity quickly and unambiguously.

Step-by-Step Solution

$A^2 = \begin{pmatrix}i&0\\0&-i\end{pmatrix}^2=\begin{pmatrix}i^2&0\\0&(-i)^2\end{pmatrix}=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}=-I$ .

$B^2=\begin{pmatrix}0&-1\\1&0\end{pmatrix}^2=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}=-I$ (a $90^\circ$ rotation squared is a $180^\circ$ rotation, i.e. $-I$ ).

$C^2=\begin{pmatrix}0&i\\i&0\end{pmatrix}^2=\begin{pmatrix}i\cdot i&0\\0&i\cdot i\end{pmatrix}=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}=-I$ .

So $A^2+B^2+C^2=-I-I-I=-3I$ . …
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AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
$A = \begin{pmatrix}1 & 1 & 3\\1 & 7 & 9\\2 & 3 & 7\end{pmatrix}$
›Reveal solutionSolution
Using the linearity of trace, $Tr(A^2-A)=Tr(A^2)-Tr(A)$ , computed directly from A's diagonal entries and the diagonal entries of $A^2$ (each needing only one row-column dot product).

Concept and Intuition

Trace is linear: $Tr(M+N)=Tr(M)+Tr(N)$ for matrices of the same size, so $Tr(A^2-A)=Tr(A^2)-Tr(A)$ . This avoids computing the full matrix $A^2$ — we only need its three diagonal entries, each of which is just the dot product of the corresponding row of A with the corresponding column of A.

Step-by-Step Solution

$A=\begin{pmatrix}1&1&3\\1&7&9\\2&3&7\end{pmatrix}$ . $Tr(A)=1+7+7=15$ .

Columns of A: column 1 $=(1,1,2)$ , column 2 $=(1,7,3)$ , column 3 $=(3,9,7)$ .

$(A^2)_{11}=$ row 1 $\cdot$ column 1 $=1(1)+1(1)+3(2)=1+1+6=8$ .

$(A^2)_{22}=$ row 2 $\cdot$ column 2 $=1(1)+7(7)+9(3)=1+49+27=77$ .

$(A^2)_{33}=$ row 3 $\cdot$ column 3 $=2(3)+3(9)+7(7)=6+27+49=82$ .

$Tr(A^2)=8+77+82=167$ . …
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Q.If $A = \begin{bmatrix}1 & 0 & -1\\ 2 & 1 & 3\\ 0 & 1 & 1\end{bmatrix}$ , then verify that $A^2 + A = A(A + I)$ , where $I$ is the $3 \times 3$ unit matrix.

Concept understanding — Matrix Polynomial Evaluation

Matrix Polynomial Evaluation

The Definition

A Worked Example

A Shortcut for Diagonal Matrices

Why the identity should hold

Step 1 — compute $A^2=A\cdot A$

Step 2 — compute $A^2+A$

Step 3 — compute $A(A+I)$

Method: Verifying a matrix polynomial factorization

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.If A=[10−1213011]A = \begin{bmatrix}1 & 0 & -1\\ 2 & 1 & 3\\ 0 & 1 & 1\end{bmatrix}A=​120​011​−131​​, then verify that A2+A=A(A+I)A^2 + A = A(A + I)A2+A=A(A+I), where III is the 3×33 \times 33×3 unit matrix.

Why the identity should hold

Step 1 — compute A2=A⋅AA^2=A\cdot AA2=A⋅A

Step 2 — compute A2+AA^2+AA2+A

Step 3 — compute A(A+I)A(A+I)A(A+I)

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Q.If $A = \begin{bmatrix}1 & 0 & -1\\ 2 & 1 & 3\\ 0 & 1 & 1\end{bmatrix}$ , then verify that $A^2 + A = A(A + I)$ , where $I$ is the $3 \times 3$ unit matrix.

Step 1 — compute $A^2=A\cdot A$

Step 2 — compute $A^2+A$

Step 3 — compute $A(A+I)$

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