Matrix A = [ * 20 c x&3&2 1&y&4 2&2&z ], xyz = 60 and 8x + 4y + 3… - Vidyadip

MCQ

Matrix $A = \left[ {\begin{array}{*{20}{c}}
x&3&2 \\
1&y&4 \\
2&2&z
\end{array}} \right]$, $xyz = 60$ and $8x + 4y + 3z = 20$, then $A.(Adj A)$ is equal to

✓
$\left[ {\begin{array}{*{20}{c}}
{68}&0&0 \\
0&{68}&0 \\
0&0&{68}
\end{array}} \right]$
B
$\left[ {\begin{array}{*{20}{c}}
{64}&0&0 \\
0&{64}&0 \\
0&0&{64}
\end{array}} \right]$
C
$\left[ {\begin{array}{*{20}{c}}
{34}&0&0 \\
0&{34}&0 \\
0&0&{34}
\end{array}} \right]$
D
$\left[ {\begin{array}{*{20}{c}}
{88}&0&0 \\
0&{88}&0 \\
0&0&{88}
\end{array}} \right]$

✓

Answer

Correct option: A.

$\left[ {\begin{array}{*{20}{c}}
{68}&0&0 \\
0&{68}&0 \\
0&0&{68}
\end{array}} \right]$

a
$A \cdot(A d j A)=|A| I$

$|A|=\left|\begin{array}{lll}{x} & {3} & {2} \\ {1} & {y} & {4} \\ {2} & {2} & {z}\end{array}\right| $

$ =x(y z-8)-3(z-8)+2(2-2 y) $

$=x y z-(8 x+4 y+3 z)+28 $

$=60-20+28=68 $

$\therefore \,\,\,\,\,A \cdot (Adj\,A) = 68I = \left[ {\begin{array}{*{20}{c}}
{68}&0&0\\
0&{68}&0\\
0&0&{68}
\end{array}} \right]$

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