SATHEE: Matrices and Determinants 1 Question 5

Matrices and Determinants 1 Question 5

5. Let $A = [\begin{array}{ccc} 0 & 2 q & r \\ p & q & - r \\ p & - q & r \end{array}]$ . If $A A^{T} = I_{3}$ , then $| p |$ is

(a) $\frac{1}{\sqrt{5}}$

(b) $\frac{1}{\sqrt{2}}$

(d) $\frac{1}{\sqrt{6}}$

(2019 Main, 11 Jan I)

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Answer:

Correct Answer: 5. (b)

Solution:

$\begin{aligned} Given, A A^{T} = I \\ \Rightarrow [\begin{array}{ccc} 0 & 2 q & r \\ p & q & - r \\ p & - q & r \end{array}] [\begin{array}{ccc} 0 & p & p \\ 2 q & q & - q \\ r & - r & r \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ \Rightarrow [\begin{array}{ccc} 0 + 4 q^{2} + r^{2} & 0 + 2 q^{2} - r^{2} & 0 - 2 q^{2} + r^{2} \\ 0 + 2 q^{2} - r^{2} & p^{2} + q^{2} + r^{2} & p^{2} - q^{2} - r^{2} \\ 0 - 2 q^{2} + r^{2} & p^{2} - q^{2} - r^{2} & p^{2} + q^{2} + r^{2} \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \end{aligned}$

We know that, if two matrices are equal, then corresponding elements are also equal, so

$4 q^{2} + r^{2} = 1 = p^{2} + q^{2} + r^{2},$ …(i)

$2 q^{2} - r^{2} = 0 \Rightarrow r^{2} = 2 q^{2}$ …(ii)

and

$p^{2} - q^{2} - r^{2} = 0$ …(iii)

Using Eqs. (ii) and (iii), we get

$p^{2} = 3 q^{2}$

Using Eqs. (ii) and (iv) in Eq. (i), we get

$4 q^{2} + 2 q^{2} = 1$

$\Rightarrow 6 q^{2} = 1$

$\Rightarrow 2 p^{2} = 1$ [using Eq. (iv)]

$p^{2} = \frac{1}{2} \Rightarrow | p | = \frac{1}{\sqrt{2}}$

Matrices and Determinants 1 Question 5

5. Let $A = [\begin{array}{ccc} 0 & 2 q & r \\ p & q & - r \\ p & - q & r \end{array}]$ . If $A A^{T} = I_{3}$ , then $| p |$ is

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Matrices and Determinants 1 Question 5

5. Let A=[02qrpq−rp−qr]. If AAT=I3, then |p| is

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5. Let $A = [\begin{array}{ccc} 0 & 2 q & r \\ p & q & - r \\ p & - q & r \end{array}]$ . If $A A^{T} = I_{3}$ , then $| p |$ is