First slide

Applications of determinant

Question

$If A = [\begin{array}{lll} 0 1 2 \\ 1 2 3 \\ 3 a 1 \end{array}] and A^{- 1} = [\begin{array}{ccc} \frac{1}{2} & \frac{- 1}{2} & \frac{1}{2} \\ - 4 & 3 & c \\ \frac{5}{2} & \frac{- 3}{2} & \frac{1}{2} \end{array}], then the values of a and c are respectively is equal to$

Easy

A

$1, 1$
B

$1, - 1$
C

$1, 2$
D

$- 1, 1$

Solution

$\begin{array}{l} Given that A = [\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{array}] and A^{- 1} = [\begin{array}{ccc} \frac{1}{2} & - \frac{1}{2} & \frac{1}{2} \\ - 4 & 3 & c \\ \frac{5}{2} & - \frac{3}{2} & \frac{1}{2} \end{array}] \\ we have I = A A^{- 1} \\ = \frac{1}{2} [\begin{matrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{matrix}] [\begin{array}{ccc} 1 & - 1 & 1 \\ - 8 & 6 & 2 c \\ 5 & - 3 & 1 \end{array}] \\ = \frac{1}{2} [\begin{array}{ccl} 0 - 8 + 10 & 0 + 6 - 6 & 0 + 2 c + 2 \\ 1 - 16 + 15 & - 1 + 12 - 9 & 1 + 4 c + 3 \\ 3 - 8 a + 5 & - 3 + 6 a - 3 & 3 + 2 a c + 1 \end{array}] \end{array}$
$= [\begin{array}{ccc} 1 & 0 & c + 1 \\ 0 & 1 & 2 (c + 1) \\ 4 (1 - a) & 3 (a - 1) & 2 + a c \end{array}]$
$Comparing the elements of A A^{- 1} with those of I = [\begin{array}{lll} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}],$
$we have c + 1 = 0 or c = - 1$
$a - 1 = 0 or a = 1$

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