Let begin pmatrix 1 amp 0 0 amp 1 end pmatrix p... - JAMB Mathematics 1998 Question
Let = \(\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\) p = \(\begin{pmatrix} 2 & 3 \ 4 & 5 \end{pmatrix}\) Q = \(\begin{pmatrix} u & 4+u \ -2v & v \end{pmatrix}\) be 2 x 2 matrices such that PQ = 1. Find (u, v)
A
(-\(\frac{5}{2}\) - 1)
B
(-\(\frac{5}{2}\) - \(\frac{3}{2}\))
C
(-\(\frac{5}{6}\) - 1)
D
(\(\frac{5}{2}\) - \(\frac{3}{2}\))
correct option: a
PQ = \(\begin{pmatrix} 2 & 3 \ 4 & 5 \end{pmatrix}\)\(\begin{pmatrix} u & 4+u \ -2v & v \end{pmatrix}\)
= \(\begin{pmatrix} (2u-6v & 2(4+u) +3v)\ 4u-10v & 4(4+u)+5v \end{pmatrix}\)
= \(\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\)
2u - 6v = 1.....(i)
4u - 10v = 0.......(ii)
2(4 + u) + 3v = 0......(iii)
4(4 + u) + 5v = 1......(iv)
2u - 6v = 1 .....(i) x 2
4u - 10v = 0......(ii) x 1
\(\frac{\text{4u - 12v = 0}}{\text{-4u - 10v = 0}}\)
-2v = 2 = v = -1
2u - 6(-1) = 1 = 2u = 5
u = -\(\frac{5}{2}\)
∴ (U, V) = (-\(\frac{5}{2}\) - 1)
= \(\begin{pmatrix} (2u-6v & 2(4+u) +3v)\ 4u-10v & 4(4+u)+5v \end{pmatrix}\)
= \(\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\)
2u - 6v = 1.....(i)
4u - 10v = 0.......(ii)
2(4 + u) + 3v = 0......(iii)
4(4 + u) + 5v = 1......(iv)
2u - 6v = 1 .....(i) x 2
4u - 10v = 0......(ii) x 1
\(\frac{\text{4u - 12v = 0}}{\text{-4u - 10v = 0}}\)
-2v = 2 = v = -1
2u - 6(-1) = 1 = 2u = 5
u = -\(\frac{5}{2}\)
∴ (U, V) = (-\(\frac{5}{2}\) - 1)
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