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Mathematics Question on Matrices

If matrix A =[12 43]\begin{bmatrix} 1 & 2 \\\ 4 & 3 \end{bmatrix} is such that AX = I, where I is 2 x 2 unit matrix, then X =

A

\frac {1}{5}$$\begin{bmatrix} 3 & 2 \\\ 4 & 1 \end{bmatrix}

B

\frac {1}{5}$$\begin{bmatrix} 3 & -2 \\\ -4 & 1 \end{bmatrix}

C

\frac {1}{5}$$\begin{bmatrix} -3 & -2 \\\ 4 & 1 \end{bmatrix}

D

\frac {1}{5}$$\begin{bmatrix} -3 & 2 \\\ 4 & -1 \end{bmatrix}

Answer

\frac {1}{5}$$\begin{bmatrix} -3 & 2 \\\ 4 & -1 \end{bmatrix}

Explanation

Solution

We have A = [12 43]\begin{bmatrix} 1 & 2 \\\ 4 & 3 \end{bmatrix} such that Ax=I
Since, A and I are of order 2×2. So, x will be a matrix of order 2×2
let x = [ab cd]\begin{bmatrix} a & b \\\ c & d \end{bmatrix}
Ax = I
\begin{bmatrix} 1 & 2 \\\ 4 & 3 \end{bmatrix}$$\begin{bmatrix} a & b \\\ c & d \end{bmatrix} = [10 01]\begin{bmatrix} 1 & 0 \\\ 0 & 1 \end{bmatrix}
[a+2cb+2d 4a+3c4b+3d]\begin{bmatrix} a+2c & b+2d \\\ 4a+3c & 4b+3d \end{bmatrix} = [10 01]\begin{bmatrix} 1 & 0 \\\ 0 & 1 \end{bmatrix}
a + 2c = 1 ....……….(i)
4a + 3c = 0 ......……(ii)
b + 2d = 0 .....…….(iii)
4b + 3d = 1 .....…..(iv)
On solving eq (i),(ii) ,(iii) and (iv) we get
a = -35\frac {3}{5}, b = 25\frac {2}{5}, c = 45\frac {4}{5} and d = -15\frac {1}{5}
Substituting we get x = [−3525 45−15]\begin{bmatrix} -\frac{3}{5} & \frac{2}{5} \\\ \frac{4}{5} & -\frac{1}{5} \end{bmatrix}
∴x = \frac {1}{5}$$\begin{bmatrix} -3 & 2 \\\ 4 & -1 \end{bmatrix}