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GUJCET 2025 Math Paper

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Question : 19 of 40

Marks: +1, -0

For matrix

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

A^2 - 2I = KA

K = \dots

- $5$
$5$
- $7$
$7$

Solution:

Concept:

Use matrix multiplication and scalar multiplication to compare both sides of the given matrix equation.

Explanation:

Given

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

.

Compute

A^2=A\cdot A

:

A^2=\begin{bmatrix}2&3\\4&5\end{bmatrix}\begin{bmatrix}2&3\\4&5\end{bmatrix}=\begin{bmatrix}16&21\\28&37\end{bmatrix}

.

Also,

2I=\begin{bmatrix}2&0\\0&2\end{bmatrix}

.

So,

A^2-2I=\begin{bmatrix}16&21\\28&37\end{bmatrix}-\begin{bmatrix}2&0\\0&2\end{bmatrix}=\begin{bmatrix}14&21\\28&35\end{bmatrix}

.

Factor out

7

:

\begin{bmatrix}14&21\\28&35\end{bmatrix}=7\begin{bmatrix}2&3\\4&5\end{bmatrix}=7A

.

Given

A^2-2I=KA

, compare:

KA=7A\Rightarrow K=7

.

Answer:

K=7

, so the correct option is D.

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