Find the matrix A
A \(\begin {bmatrix} 0 & 1\\2 & -1 \end {bmatrix}\) = \(\begin {bmatrix} 2 & -1\\1 & 0 \end {bmatrix}\)
- \(\begin {bmatrix} 2 & 1\\-^1/_2 & -^1/_2 \end {bmatrix}\)
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\(\begin {bmatrix} 0 & 1\\^1/_2 & ^1/_2 \end {bmatrix}\)
- \(\begin {bmatrix} 2 & 1\\0 & -1 \end {bmatrix}\)
- \(\begin {bmatrix} 2 & 1\\^1/_2 & -2 \end {bmatrix}\)
The correct answer is: B
Explanation
Let A = \(\begin {bmatrix} a & b\\ c & d \end {bmatrix}\)
i.e \(\begin{bmatrix} a & b\\c & d \end{bmatrix}\) \(\begin{bmatrix} 0 & 1\\2 & -1 \end{bmatrix}\) = \(\begin{bmatrix} 2 & -1\\1 & 0 \end{bmatrix}\)
\(\implies \begin {bmatrix} a(0) + b(2) & a(1) + b(-1)\\ c(0) + d(2) & c(1) + d(-1)\end {bmatrix}\) = \(\begin {bmatrix} 2 & -1\\ 1 & 0 \end {bmatrix}\)
\(\implies \begin {bmatrix} 2b & a - b\\2d & c - d\end {bmatrix}\) = \(\begin {bmatrix} 2 & -1\\ 1 & 0 \end {bmatrix}\)
By comparing
2b = 2
a - b = -1
2d = 1 and
c - d = 0
∴ b = \(^2/_2\) = 1
a - b = -1
⇒ a - 1 = -1
∴ a = 0
∴ d = \(^1/_2\)
⇒ c = d
∴ c = \(^1/_2\)
∴The matrice A = \(\begin {bmatrix} 0 & 1\\^1/_2 & ^1/_2 \end {bmatrix}\)
There is an explanation video available .