SATHEE: Matrices and Determinants 3 Question 5

Matrices and Determinants 3 Question 5

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

(2016 Main)

(a) -1

(b) 5

(d) 13

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Answer:

Correct Answer: 6. (b)

Solution:

Given, $A = [\begin{matrix} 5 a & - b \\ 3 & 2 \end{matrix}]$ and $A$ adj $A = A A^{T}$

Clearly, $A (adj A) = | A | I_{2}$

$[∵ if A is square matrix of order n$ then $A (adj A) = (adj A) \cdot A = | A | I_{n}]$

$= [\begin{matrix} 5 a & - b \\ 3 & 2 \end{matrix}]$ $I_{2} = (10 a + 3 b) I_{2}$

$= (10 a + 3 b)$ $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$[\begin{matrix} 10 a + 3 b & 0 \\ 0 & 10 a + 3 b \end{matrix}]$

and $A A^{T} = [\begin{matrix} 5 a & - b & 5 a & 3 \\ 3 & 2 & - b & 2 \end{matrix}]$

= $[\begin{matrix} 25 a^{2} + b^{2} & 15 a - 2 b \\ 15 a - 2 b & 13 \end{matrix}]$

$∵ A (adj A) = A A^{T}$

$∴ [\begin{matrix} 10 a + 3 b & 0 \\ 0 & 10 a + 3 b \end{matrix}]$ = $[\begin{matrix} 25 a^{2} + b^{2} & 15 a - 2 b \\ 15 a - 2 b & 13 \end{matrix}]$

$\Rightarrow 15 a - 2 b = 0$

$\Rightarrow a = \frac{2 b}{15}$

and $10 a + 3 b = 13$

On substituting the value of ’ $a$ ’ from Eq. (iii) in Eq. (iv), we get

$10 \cdot \frac{2 b}{15} + 3 b = 13$

$\Rightarrow \frac{20 b + 45 b}{15} = 13$

$\Rightarrow \frac{65 b}{15} = 13$

$\Rightarrow b = 3$

Now, substituting the value of $b$ in Eq. (iii), we get

$5 a = 2$

Hence, $5 a + b = 2 + 3 = 5$

Matrices and Determinants 3 Question 5

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

Answer:

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Matrices and Determinants 3 Question 5

6. If A=[5a−b32] and A adj A=AAT, then 5a+b is equal to

Answer:

Engineering Preparation

Medical Preparation

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