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}

Jammu Kashmir JkboseShort· 3mImportance★★★★★

59% · 107/182 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ25 — If A = 2 1 , B = 5 3 4\8 7 6 and C = -1 2 1\1 0 2 , verify that A(B + …→Next questionQ27 — If A = 0 -1 2\4 3 -4 and B = 4 0\1 3\2 6 , then verify that: (i) (A')'…

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

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)

🎓Unlock everything free for 14 days

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 $A(A+I)$

Unlock everything free for 14 days