If for a matrix A - A -6 and adj- A - beginbmatrix 1 -2 - 4 4 - 1 - 111-1 - k - 0lendbmatrix then k is equal toA- -1B- 2C- 1D- 0

The correct option is B 2 |adj A| = |A|^2 = 36 ⇒ 1 amp; 2 amp; 4 4 amp; 1 amp; 1 1 amp; k amp; 0 = 36 ⇒ ( 1) 2 amp; 4 1 am ...

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Standard XII

Mathematics

Symmetric Matrix

If for a matr...

Question

If for a matrix $A, | A | = 6$ and $a d j A = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 2 & 4 4 & 1 & 1 - 1 & k & 0 \end{matrix} ⎤ ⎥ ⎦$ then $k$ is equal to

-1

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Solution

The correct option is B
2

$| a d j A | = | A |^{2} = 36$

$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 1 & - 2 & 4 4 & 1 & 1 - 1 & k & 0 \end{matrix} ∣ ∣ ∣ ∣ = 36$

$\Rightarrow (- 1) ∣ ∣ ∣ \begin{matrix} - 2 & 4 1 & 1 \end{matrix} ∣ ∣ ∣ - k ∣ ∣ ∣ \begin{matrix} 1 & 4 4 & 1 \end{matrix} ∣ ∣ ∣ = 36$ [Expanding along $R_{3}$ ]

$\Rightarrow (- 1) (- 2 - 4) - k (1 - 16) = 36$

$\Rightarrow 6 + 15 k = 36 \Rightarrow k = 2$

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If for a matrix A, |A|=6 and adj A=⎡⎢⎣1−24411−1k0⎤⎥⎦ then k is equal to

The correct option is B 2 |adj A|=|A|2=36 ⇒ ∣∣ ∣∣1−24411−1k0∣∣ ∣∣=36 ⇒ (−1)∣∣∣−2411∣∣∣−k∣∣∣1441∣∣∣=36 [Expanding along R3] ⇒ (−1)(−2−4)−k(1−16)=36 ⇒ 6+15k=36⇒k=2

If for a matrix $A, | A | = 6$ and $a d j A = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 2 & 4 4 & 1 & 1 - 1 & k & 0 \end{matrix} ⎤ ⎥ ⎦$ then $k$ is equal to