Matrices & Determinants
Matrix Multiplication
Grade 12
Question:
<p>Let \(A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\) and \(B = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}\), where \(a, b \in \mathbb{N}\). Then</p>
<p>(a) there cannot exist any B such that AB = BA</p>
<p>(b) there exist more than one but finite number of B's such that AB = BA</p>
<p>(c) there exists exactly one B such that AB = BA</p>
<p>(d) there exist infinitely many B's such that AB = BA</p>
Step-by-Step Solution
Key Concept: Two matrices commute if and only if specific equality conditions are satisfied for their entries.
<p><strong>Solution:</strong> We need AB = BA where $B = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$.<br>$AB = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} = \begin{pmatrix} a & 2b \\ 3a & 4b \end{pmatrix}$<br>$BA = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} a & 2a \\ 3b & 4b \end{pmatrix}$<br>For AB = BA: $2b = 2a$ and $3a = 3b$, which gives $a = b$.<br>Since $a, b \in \mathbb{N}$, there are infinitely many such pairs $(a, a)$. ∴ Answer is D.</p>
Correct Answer: D