First slide

Applications of determinant

Question

$If A = [\begin{array}{cc} 5 a & - b \\ 3 & 2 \end{array}] and A adj A = A A^{T}, then 5 a + b is equal to$

Moderate

A

-1
B

5
C

4
D

13

Solution

$\underline{\underline{Given,}} A = [\begin{array}{cc} 5 a & - b \\ 3 & 2 \end{array}] and A adj A = A A^{T}$
$\underline{\underline{Clearly}} A (adj A) = A ∣ I_{2}$
$[∵ if A is square matrix of order n, then A (adj A) = (adj A) \cdot A = A ∣ I_{n}]$
$\begin{array}{l} A (adj A) = |\begin{array}{cc} 5 a & - b \\ 3 & 2 \end{array}| I_{2} \\ A (adj A) = (10 a + 3 b) I_{2} \\ A (adj A) = (10 a + 3 b) [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] \\ A (adj A) = [\begin{array}{cc} 10 a + 3 b & 0 \\ 0 & 10 a + 3 b \end{array}] - - - - - (1) \\ And A A^{T} = [\begin{array}{cc} 5 a & - b \\ 3 & 2 \end{array}] [\begin{array}{cc} 5 a & 3 \\ - b & 2 \end{array}] \end{array}$
$\begin{array}{l} A A^{T} = [\begin{array}{cc} 25 a^{2} + b^{2} & 15 a - 2 b \\ 15 a - 2 b & 13 \end{array}] \\ ∵ A (adj A) = A A^{T} \\ ∴ [\begin{array}{cc} 10 a + 3 b & 0 \\ 0 & 10 a + 3 b \end{array}] = [\begin{array}{cc} 25 a^{2} + b^{2} & 15 a - 2 b \\ 15 a - 2 b & 13 \end{array}] \end{array}$
[ using Eqs. (i) and (ii)]
$\begin{aligned} \Rightarrow 15 a - 2 b = 0 \\ \Rightarrow a = \frac{2 b}{15} \dots \dots \dots . . . (2) \end{aligned}$
$And 10 a + 3 b = 13 . - - - (3)$
On substituting the value of ‘a’ from Eq.(iii) in Eq.(iv), we get
$\begin{aligned} 10 (\frac{2 b}{15}) + 3 b = 13 \\ \Rightarrow \frac{20 b + 45 b}{15} = 13 \\ \Rightarrow \frac{65 b}{15} = 13 \\ \Rightarrow b = 3 \end{aligned}$
Now , substituting the value of b in Eq. (iii) , we get $5 a = 2$
Hence, 5a+b=2+3=5

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