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where \( O \) denotes \( 2 \times 2 \) null matrix is f then surn of dights of th.plz help give perfect soln |
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Answer» M = \(\begin{bmatrix}a & -360 \\b & c \\\end{bmatrix}\) M2 = \(\begin{bmatrix} a & -360 \\ b & c \\ \end{bmatrix}\) \(\begin{bmatrix} a & -360 \\ b & c \\ \end{bmatrix}\) = \(\begin{bmatrix} a^2&-360 b \,\,\,-360(a+c)\\ b(a+c) &\,\, c^2 -360\,b\end{bmatrix}\) -360 (a+c) = 0 ⇒ a = -c a2 - 360 b = 0 ⇒ b = \(\frac{a^2}{360}\) a,b,c are integer so , for smallest positive value of b , we have a = 1 or a = -1 ∴ b = 1/360 is smallest positive value of b where o2 is null matrix . only needed condition is c = -a |
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