Find the matrix A satisfying the matrix equation

Engineering

Mathematics

Algebra of Matrices

Inverse of a Matrix

Question

Find the matrix A satisfying the matrix equation $[\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array}] A [\begin{array}{cc} - 3 & 2 \\ 5 & - 3 \end{array}] = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]$

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Let $A = [\begin{array}{cc} a & c \\ b & d \end{array}]$

so,

[\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array}] A [\begin{array}{cc} - 3 & 2 \\ 5 & - 3 \end{array}] = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]

\Rightarrow [\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array}] [\begin{array}{cc} a & c \\ b & d \end{array}] [\begin{array}{cc} - 3 & 2 \\ 5 & - 3 \end{array}] = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]

\Rightarrow [\begin{array}{cc} 2 a + b & 2 c + d \\ 3 a + 2 b & 3 c + 2 d \end{array}] [\begin{array}{cc} - 3 & 2 \\ 5 & - 3 \end{array}] [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]

\Rightarrow [\begin{array}{cc} - 6 a - 3 b + 10 c + 5 d & 4 a + 3 b - 6 c - 3 d \\ - 9 a - 6 b + 15 c + 10 d & 6 a + 4 b - 9 c - 6 d \end{array}] = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]

so, – 6a – 3b + 10c + 5d = 1

4a + 3b – 6c – 3d = 0

– 9a – 6b + 15c + 10d = 0

6a + 4b – 9c – 6d = 1

on squaring, we get

\Rightarrow a = \frac{1}{3}, b = 1, c = \frac{1}{3}, d = \frac{8}{15}

so,

A = [\begin{array}{cc} \frac{1}{3} & \frac{1}{3} \\ 0 & \frac{8}{15} \end{array}] = \frac{1}{15} [\begin{array}{cc} 5 & 5 \\ 15 & 8 \end{array}] .

Hence, the answer is

\frac{1}{15} [\begin{array}{cc} 5 & 5 \\ 15 & 8 \end{array}] .