If A - 5 x-2 2x-3 x-1 is symmetric- then x -

The correct option is C 5 A= 5 amp;x 2 2x 3 amp;x+1 A^T= 5 amp;2x 3 x 2 amp;x+1 Since A is symmetric, A=A^T ...

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Standard XII

Mathematics

Symmetric Matrix

If A = 5 ...

Question

If $A = ∣ ∣ ∣ \begin{matrix} 5 & x - 2 2 x + 3 & x + 1 \end{matrix} ∣ ∣ ∣$ is symmetric, then $x =$ _____

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Solution

The correct option is C $- 5$
A= $∣ ∣ ∣ \begin{matrix} 5 & x - 2 2 x - 3 & x + 1 \end{matrix} ∣ ∣ ∣$
$⟹ A^{T} =$ $∣ ∣ ∣ \begin{matrix} 5 & 2 x - 3 x - 2 & x + 1 \end{matrix} ∣ ∣ ∣$
Since $A$ is symmetric, $⟹ A = A^{T}$
$⟹ x - 2 = 2 x + 3$
$⟹ x = - 5$

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$(i) {(\frac{1}{4})}^{- 4} \times {(\frac{1}{4})}^{- 8} = {(\frac{1}{4})}^{- 4 x} (i i) {(\frac{- 1}{2})}^{- 19} \div {(\frac{- 1}{2})}^{8} = {(\frac{- 1}{2})}^{- 2 x + 1} (i i i) {(\frac{3}{2})}^{- 3} \div {(\frac{3}{2})}^{5} = {(\frac{3}{2})}^{2 x + 1} (i v) {(\frac{2}{5})}^{- 3} \times {(\frac{2}{5})}^{15} = {(\frac{2}{5})}^{2 + 3 x} (v) {(\frac{5}{4})}^{- x} \div {(\frac{5}{4})}^{- 4} = {(\frac{5}{4})}^{5} (v i) {(\frac{8}{3})}^{2 x + 1} \times {(\frac{8}{3})}^{5} = {(\frac{8}{3})}^{x + 2}$

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If A=∣∣∣5x−22x+3x+1∣∣∣ is symmetric, then x=_____

The correct option is C −5A=∣∣∣5x−22x−3x+1∣∣∣⟹AT=∣∣∣52x−3x−2x+1∣∣∣Since A is symmetric, ⟹A=AT⟹x−2=2x+3⟹x=−5

If $A = ∣ ∣ ∣ \begin{matrix} 5 & x - 2 2 x + 3 & x + 1 \end{matrix} ∣ ∣ ∣$ is symmetric, then $x =$ _____

The correct option is C $- 5$
A= $∣ ∣ ∣ \begin{matrix} 5 & x - 2 2 x - 3 & x + 1 \end{matrix} ∣ ∣ ∣$
$⟹ A^{T} =$ $∣ ∣ ∣ \begin{matrix} 5 & 2 x - 3 x - 2 & x + 1 \end{matrix} ∣ ∣ ∣$
Since $A$ is symmetric, $⟹ A = A^{T}$
$⟹ x - 2 = 2 x + 3$
$⟹ x = - 5$