If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, show that $A^2 - 5A + 7I = 0$.

If A = 3 1 \-1 2 , show that A² - 5A + 7I = 0. – Class 12 Mathematics

Miscellaneous Exercise · Q5

A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \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 $A^2$ directly and substituting gives $A^2 - 5A = -7I$ , so $A^2 - 5A + 7I = O$ .

We must show $A^2 - 5A + 7I = O$ for $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ , where $I$ is the $2\times2$ identity and $O$ the zero matrix. The safest in-syllabus method is to compute $A^2$ by direct multiplication and substitute.

Step 1 — Compute $A^2$

A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 3(3)+1(-1) & 3(1)+1(2) \\ -1(3)+2(-1) & -1(1)+2(2) \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}.

Step 2 — Compute $5A$ and $7I$

5A = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix},\qquad 7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}.

Step 3 — Substitute …

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Q.If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ , show that $A^2 - 5A + 7I = 0$ .

Concept understanding — Matrix Polynomial Evaluation

Matrix Polynomial Evaluation

The Definition

A Worked Example

A Shortcut for Diagonal Matrices

Step 1 — Compute $A^2$

Step 2 — Compute $5A$ and $7I$

Step 3 — Substitute …

Method: Verifying a matrix satisfies a polynomial via its characteristic equation

Steps

Common Mistakes

Q.If A=[31−12]A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}A=[3−1​12​], show that A2−5A+7I=0A^2 - 5A + 7I = 0A2−5A+7I=0.

Step 1 — Compute A2A^2A2

Step 2 — Compute 5A5A5A and 7I7I7I

Step 3 — Substitute …

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Q.If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ , show that $A^2 - 5A + 7I = 0$ .

Step 1 — Compute $A^2$

Step 2 — Compute $5A$ and $7I$

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