First slide

Transpose of matrix

Question

If $3 A = [\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & - 2 \\ x & 2 & y \end{array}]$ and $A A^{'} = I,$ then $3 A = [\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & - 2 \\ x & 2 & y \end{array}] and A A^{'} = I, then x + y$ is equal to

Moderate

A

$- 3$
B

$- 2$
C

$- 1$
D

0

Solution

$(3 A)^{'} = [\begin{array}{ccc} 1 & 2 & x \\ 2 & 1 & 2 \\ 2 & - 2 & y \end{array}]$
Now, $9 I = (3 A) (3 A)^{'}$
$\begin{array}{l} = [\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & - 2 \\ x & 2 & y \end{array}] [\begin{array}{ccc} 1 & 2 & x \\ 2 & 1 & 2 \\ 2 & - 2 & y \end{array}] \\ = [\begin{array}{ccc} 9 & 0 & x + 4 + 2 y \\ 0 & 9 & 2 x + 2 - 2 y \\ x + 4 + 2 y & 2 x + 2 - 2 y & x^{2} + 4 + y^{2} \end{array}] \end{array}$
$\begin{array}{l} \Rightarrow x + 2 y + 4 = 0 \\ 2 x - 2 y + 2 = 0 \\ x^{2} + y^{2} + 4 = 9 \\ \Rightarrow x = - 2, y = - 1 \end{array}$

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