If $A = \begin{bmatrix}3 & -5\\ -4 & 2\end{bmatrix}$, then find $A^2 - 5A - 14I$. Hence, obtain $A^3$.

If A = 3 -5\-4 2 , then find A² - 5A - 14I. Hence, – Class 12 Mathematics

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A = \begin{bmatrix}3 & -5\\ -4 & 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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$A^{2}-5A-14I$ is the zero matrix, so $A^{2}=5A+14I$ ; using this, $A^{3}=39A+70I=\begin{bmatrix}187&-195\\-156&148\end{bmatrix}$ .

The smart move: instead of cubing $A$ by multiplying it three times, first find that $A$ satisfies a simple relation, then use that relation to bring every high power back down to a combination of $A$ and $I$ .

Step 1 — Square $A$

A^{2}=\begin{bmatrix}3&-5\\-4&2\end{bmatrix}\begin{bmatrix}3&-5\\-4&2\end{bmatrix}.

$(1,1):3\cdot3+(-5)(-4)=9+20=29$
$(1,2):3(-5)+(-5)(2)=-15-10=-25$
$(2,1):(-4)(3)+2(-4)=-12-8=-20$
$(2,2):(-4)(-5)+2\cdot2=20+4=24$

A^{2}=\begin{bmatrix}29&-25\\-20&24\end{bmatrix}.

Step 2 — Form $A^{2}-5A-14I$

5A=\begin{bmatrix}15&-25\\-20&10\end{bmatrix},\qquad 14I=\begin{bmatrix}14&0\\0&14\end{bmatrix}.

A^{2}-5A-14I=\begin{bmatrix}29-15-14&-25+25-0\\-20+20-0&24-10-14\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=O.

Watch out

$14I$ subtracts $14$ from the diagonal only, not from every entry — that is what a scalar times $I$ means.

So we have the key relation

A^{2}=5A+14I.

(This is just the Cayley–Hamilton relation in disguise: $A$ satisfies its own characteristic equation $\lambda^{2}-5\lambda-14=0$ .) …

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Q.If $A = \begin{bmatrix}3 & -5\\ -4 & 2\end{bmatrix}$ , then find $A^2 - 5A - 14I$ . Hence, obtain $A^3$ .

Concept understanding — Matrix Polynomial Evaluation

Matrix Polynomial Evaluation

The Definition

A Worked Example

A Shortcut for Diagonal Matrices

Step 1 — Square $A$

Step 2 — Form $A^{2}-5A-14I$

Method: Evaluating matrix polynomials and reducing higher powers with a satisfied relation

Steps

Common Mistakes

Q.If A=[3−5−42]A = \begin{bmatrix}3 & -5\\ -4 & 2\end{bmatrix}A=[3−4​−52​], then find A2−5A−14IA^2 - 5A - 14IA2−5A−14I. Hence, obtain A3A^3A3.

Step 1 — Square AAA

Step 2 — Form A2−5A−14IA^{2}-5A-14IA2−5A−14I

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Q.If $A = \begin{bmatrix}3 & -5\\ -4 & 2\end{bmatrix}$ , then find $A^2 - 5A - 14I$ . Hence, obtain $A^3$ .

Step 1 — Square $A$

Step 2 — Form $A^{2}-5A-14I$

Unlock everything free for 14 days