Verify that $A^2 = I$ when $A = \begin{bmatrix}0 & 1 & -1\\ 4 & -3 & 4\\ 3 & -3 & 4\end{bmatrix}$.

Verify that A² = I when A = 0 1 -1\4 -3 4\3 -3 4 .

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A^2 = I

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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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Multiplying $A$ by itself entry-by-entry gives the identity matrix, so $A^{2}=I$ .

"Verify" means: don't assume it, compute it. We form the product $A\cdot A$ and show every entry lands exactly on the identity. Recall each entry of the product is (a row of the first matrix) $\cdot$ (a column of the second).

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

The columns of $A$ are $\begin{bmatrix}0\\4\\3\end{bmatrix},\ \begin{bmatrix}1\\-3\\-3\end{bmatrix},\ \begin{bmatrix}-1\\4\\4\end{bmatrix}$ .

Row 1 of $A^{2}$ (row $[0,\;1,\;-1]$ against each column)

$[0,1,-1]\cdot[0,4,3]=0+4-3=1$
$[0,1,-1]\cdot[1,-3,-3]=0-3+3=0$
$[0,1,-1]\cdot[-1,4,4]=0+4-4=0$

Row 2 (row $[4,\;-3,\;4]$ )

$[4,-3,4]\cdot[0,4,3]=0-12+12=0$
$[4,-3,4]\cdot[1,-3,-3]=4+9-12=1$
$[4,-3,4]\cdot[-1,4,4]=-4-12+16=0$

Row 3 (row $[3,\;-3,\;4]$ )

$[3,-3,4]\cdot[0,4,3]=0-12+12=0$ …

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

TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
$P(x)$
›Reveal solutionSolution
The key idea is to notice that the given values follow a simple quadratic pattern, so the quartic polynomial can be expressed as a known quadratic plus a term that vanishes at those four points, leading to $P(5)=76$ .

We are given a quartic polynomial $P(x)=2x^4+ax^3+bx^2+cx+d$ and four values: $P(1)=4$ , $P(2)=7$ , $P(3)=12$ , $P(4)=19$ .

We need $P(5)$ .

Concept and intuition

The values $4,7,12,19$ are not random — look at their differences:

$7-4=3$ , $12-7=5$ , $19-12=7$ . The second differences are constant: $5-3=2$ , $7-5=2$ .

That means these four points lie on a quadratic function. But $P(x)$ is a quartic (degree 4). So $P(x)$ must be that quadratic plus something that is zero at $x=1,2,3,4$ — a multiple of $(x-1)(x-2)(x-3)(x-4)$ . This lets us find $P(5)$ without solving for all coefficients.

Step-by-step solution

Find the quadratic that fits the given points

Let $Q(x)$ be a quadratic such that $Q(1)=4$ , $Q(2)=7$ , $Q(3)=12$ , $Q(4)=19$ .

Since second differences are constant ( $2$ ), the leading coefficient of $Q$ is $\frac{2}{2!}=1$ .

So $Q(x)=x^2+px+q$ .

Using $Q(1)=4$ : $1+p+q=4 \Rightarrow p+q=3$ .

Using $Q(2)=7$ : $4+2p+q=7 \Rightarrow 2p+q=3$ .

Subtract: $(2p+q)-(p+q)=3-3 \Rightarrow p=0$ , then $q=3$ .

Thus $Q(x)=x^2+3$ .

Relate $P(x)$ to $Q(x)$

Since $P$ and $Q$ agree at four points, their difference is a polynomial of degree at most 4 that has roots at $x=1,2,3,4$ .

Hence

$P(x) - Q(x) = k(x-1)(x-2)(x-3)(x-4)$
for some constant $k$ .

Determine $k$ using the leading coefficient $P(x)=2x^4+ax^3+bx^2+cx+d$ has leading term $2x^4$ . $Q(x)=x^2+3$ has leading term $0x^4$ . The product $(x-1)(x-2)(x-3)(x-4)$ expands to $x^4 -10x^3 + \dots$ , so its leading term is $x^4$ . …
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Q.Verify that $A^2 = I$ when $A = \begin{bmatrix}0 & 1 & -1\\ 4 & -3 & 4\\ 3 & -3 & 4\end{bmatrix}$ .

Concept understanding — Matrix Polynomial Evaluation

Matrix Polynomial Evaluation

The Definition

A Worked Example

A Shortcut for Diagonal Matrices

Row 1 of $A^{2}$ (row $[0,\;1,\;-1]$ against each column)

Row 2 (row $[4,\;-3,\;4]$ )

Row 3 (row $[3,\;-3,\;4]$ )

Method: Verifying a matrix satisfies a given equation (involution check)

Steps

Common Mistakes

Q.Verify that A2=IA^2 = IA2=I when A=[01−14−343−34]A = \begin{bmatrix}0 & 1 & -1\\ 4 & -3 & 4\\ 3 & -3 & 4\end{bmatrix}A=​043​1−3−3​−144​​.

Row 1 of A2A^{2}A2 (row [0, 1, −1][0,\;1,\;-1][0,1,−1] against each column)

Row 2 (row [4, −3, 4][4,\;-3,\;4][4,−3,4])

Row 3 (row [3, −3, 4][3,\;-3,\;4][3,−3,4])

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Q.Verify that $A^2 = I$ when $A = \begin{bmatrix}0 & 1 & -1\\ 4 & -3 & 4\\ 3 & -3 & 4\end{bmatrix}$ .

Row 1 of $A^{2}$ (row $[0,\;1,\;-1]$ against each column)

Row 2 (row $[4,\;-3,\;4]$ )

Row 3 (row $[3,\;-3,\;4]$ )

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