Given the matrix A = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}, if A^2 - 7A + kI = \mat… — JAMB UTME

Given the matrix $A = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}$ , if $A^2 - 7A + kI = \mathbf{0}$ , where $I$ is the $2 \times 2$ identity matrix and $\mathbf{0}$ is the $2 \times 2$ zero matrix, what is the value of $k$ ?

Answer: 10

Answer

The value of

k

is 10.

By matrix multiplication and algebraic evaluation,

A^2 - 7A = -10I

. Substituting into

A^2 - 7A + kI = \mathbf{0}

yields

-10I + kI = \mathbf{0}

, which gives

k = 10

. Alternatively, by the Cayley-Hamilton Theorem, any

2 \times 2

matrix

A

satisfies

A^2 - \text{tr}(A)A + \det(A)I = \mathbf{0}

. Here

\text{tr}(A) = 3 + 4 = 7

and

\det(A) = (3)(4) - (2)(1) = 10

, directly giving

k = \det(A) = 10

Step-by-Step Solution

Calculate the matrix product

A^2

A^2 = \begin{pmatrix} (3\cdot 3 + 2\cdot 1) & (3\cdot 2 + 2\cdot 4) \\ (1\cdot 3 + 4\cdot 1) & (1\cdot 2 + 4\cdot 4) \end{pmatrix} = \begin{pmatrix} 11 & 14 \\ 7 & 18 \end{pmatrix}

Squaring matrix

A

involves multiplying rows of

A

by columns of

A

Perform scalar multiplication for

7A

7A = \begin{pmatrix} 21 & 14 \\ 7 & 28 \end{pmatrix}

Each entry of matrix

A

is multiplied by the scalar 7.

Subtract

7A

from

A^2

A^2 - 7A = \begin{pmatrix} 11 - 21 & 14 - 14 \\ 7 - 7 & 18 - 28 \end{pmatrix} = \begin{pmatrix} -10 & 0 \\ 0 & -10 \end{pmatrix} = -10I

Subtracting corresponding entries yields a scalar multiple of the identity matrix.

Solve for the unknown scalar

k

-10I + kI = \mathbf{0} \implies k = 10

Setting

(k - 10)I = \mathbf{0}

implies

k - 10 = 0

, so

k = 10

Key Concept

Matrix Polynomial Equations and Cayley-Hamilton Theorem

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