What is the value of c if value of determinant is 16 - 0 -1 -2 -3 1 0 -5 -6 2 5 0 -c 3 6 c 0

The correct options are A 1 C 7 Consider, 0 amp; 1 amp; 2 amp; 3 1 amp; 0 amp; 5 amp; 6 2 amp; 5 amp; ...

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Byju's Answer

Standard X

Mathematics

Addition and Subtraction of a Matrix

What is the v...

Question

What is the value of $c$ if value of determinant is 16 ?
$∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & - 1 & - 2 & - 3 1 & 0 & - 5 & - 6 2 & 5 & 0 & - c 3 & 6 & c & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$

\frac{11}{3}

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Solution

The correct options are
A $1$
C $- 7$
Consider, $∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & - 1 & - 2 & - 3 1 & 0 & - 5 & - 6 2 & 5 & 0 & - c 3 & 6 & c & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$
$R_{3} \to R_{3} - 2 R_{2}, R_{4} \to R_{4} - {3 R}_{2}$
$= ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & - 1 & - 2 & - 3 1 & 0 & - 5 & - 6 0 & 5 & 10 & - c + 12 0 & 6 & c + 15 & 18 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$
$= (- 1) ∣ ∣ ∣ ∣ \begin{matrix} - 1 & - 2 & - 3 5 & 10 & - c + 12 6 & c + 15 & 18 \end{matrix} ∣ ∣ ∣ ∣$
$C_{2} \to C_{2} - 2 C_{1}$
$∣ ∣ ∣ ∣ \begin{matrix} - 1 & 0 & - 3 5 & 0 & - c + 12 6 & c + 3 & 18 \end{matrix} ∣ ∣ ∣ ∣$
$= {(c + 3)}^{2}$
Now, ${(c + 3)}^{2} = 16$
$\Rightarrow c + 3 = \pm 4$
$\Rightarrow c = 1, - 7$

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What is the value of c if value of determinant is 16 ?∣∣ ∣ ∣ ∣∣0−1−2−310−5−6250−c36c0∣∣ ∣ ∣ ∣∣

What is the value of $c$ if value of determinant is 16 ?
$∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & - 1 & - 2 & - 3 1 & 0 & - 5 & - 6 2 & 5 & 0 & - c 3 & 6 & c & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$