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 .

NCERT Exemplar · Q35

A^2 = I

Jammu Kashmir JkboseShort· 3mImportance★★★★★

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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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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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