Question

If matrix A =\(\begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}\) 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 1

The Correct Option is D

We have A = \(\begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}\) 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 = \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) 
Ax = I
\(\begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}\)\(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
\(\begin{bmatrix} a+2c & b+2d \\ 4a+3c & 4b+3d \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
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 = -\(\frac {3}{5}\), b = \(\frac {2}{5}\), c = \(\frac {4}{5}\) and d = -\(\frac {1}{5}\)
Substituting we get x = \(\begin{bmatrix} -\frac{3}{5} & \frac{2}{5} \\ \frac{4}{5} & -\frac{1}{5} \end{bmatrix}\)
 ∴x = \(\frac {1}{5}\)\(\begin{bmatrix} -3 & 2 \\ 4 & -1 \end{bmatrix}\)