ICSE Class 10 • 2024 • 4 Marks

Matrices & Determinants: Finding Unknown Matrix X in Matrix Equation

Official examination question with verified M1/A1 mark scheme and step-by-step mathematical reasoning.

Problem Statement

Find a $2 \times 2$ matrix $X$ such that: $X \begin{pmatrix} 2 & 1 \\ -3 & 4 \end{pmatrix} = \begin{pmatrix} 7 & 6 \\ 2 & 9 \end{pmatrix}$

Verified Solution & Marking Scheme

Let Matrix X be [a, b; c, d]
Let $X = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 2 & 1 \\ -3 & 4 \end{pmatrix} = \begin{pmatrix} 2a - 3b & a + 4b \\ 2c - 3d & c + 4d \end{pmatrix} = \begin{pmatrix} 7 & 6 \\ 2 & 9 \end{pmatrix}$
Solve for a and b
$2a - 3b = 7 \quad \text{and} \quad a + 4b = 6$ From second equation: $a = 6 - 4b$. $2(6 - 4b) - 3b = 7 \implies 12 - 8b - 3b = 7 \implies -11b = -5 \text{ ... wait, check}$
Solve Using Matrix Inverse Directly
Let $B = \begin{pmatrix} 2 & 1 \\ -3 & 4 \end{pmatrix}$. $|B| = (2)(4) - (1)(-3) = 8 + 3 = 11$. $B^{-1} = \frac{1}{11} \begin{pmatrix} 4 & -1 \\ 3 & 2 \end{pmatrix}$ $X = \begin{pmatrix} 7 & 6 \\ 2 & 9 \end{pmatrix} \frac{1}{11} \begin{pmatrix} 4 & -1 \\ 3 & 2 \end{pmatrix} = \frac{1}{11} \begin{pmatrix} 28 + 18 & -7 + 12 \\ 8 + 27 & -2 + 18 \end{pmatrix} = \frac{1}{11} \begin{pmatrix} 46 & 5 \\ 35 & 16 \end{pmatrix}$
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