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  • There are two values of a which makes determinant, ,then sum of these number is A. 4

There are two values of a which makes determinant, \Delta=\left|\begin{array}{ccc} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2 a \end{array}\right|=86 ,then sum of these number is
A. 4
B. 5
C. -4
D. 9

Answers (1)

C)

\\ \begin{aligned} &\text { We have: }\\ &\Delta=\left|\begin{array}{ccc} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2 a \end{array}\right|=86\\ &1\left|\begin{array}{cc} \mathrm{a} & -1 \\ 4 & 2 \mathrm{a} \end{array}\right|-(-2)\left|\begin{array}{cc} 2 & -1 \\ 0 & 2 \mathrm{a} \end{array}\right|+5\left|\begin{array}{cc} 2 & \mathrm{a} \\ 0 & 4 \end{array}\right|=86 \end{aligned}

\\\vspace{\baselineskip} 1 [2a\textsuperscript{2} -(-4)] + 2 [4a -0] + 5 [8 -0] = 86\\ \\ \vspace{\baselineskip} 1 [2a\textsuperscript{2} + 4] + 2 [4a] + 5 [8] = 86\\ \\ \vspace{\baselineskip} 2a\textsuperscript{2} + 4 + 8a + 40 = 86\\ \\ \vspace{\baselineskip} 2a\textsuperscript{2} + 8a + 44 = 86\\ \\ \vspace{\baselineskip}2a\textsuperscript{2} + 8a = 42\\ \\ \vspace{\baselineskip}2 (a\textsuperscript{2} + 4a) = 42\\


\\ \vspace{\baselineskip} (a^{2} + 4a) = 21\\ \vspace{\baselineskip} \Rightarrow a^{2} + 4a -21 = 0\\ \vspace{\baselineskip} \Rightarrow (a + 7)(a -3) = 0\\ \vspace{\baselineskip} \therefore a = -7 \ \text{or} \ 3\\ \vspace{\baselineskip}

The sum of -7 and 3 = -4

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