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Ex 4.4
Ex 4.4, 2
Ex 4.4, 3 Important
Ex 4.4, 4 Important
Ex 4.4, 5
Ex 4.4, 6 Important
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Ex 4.4, 8
Ex 4.4, 9
Ex 4.4, 10 Important
Ex 4.4, 11 Important
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Ex 4.4, 13
Ex 4.4, 14 Important
Ex 4.4, 15 Important
Ex 4.4, 16
Ex 4.4, 17 (MCQ) Important You are here
Ex 4.4, 18 (MCQ) Important
Last updated at May 29, 2023 by Teachoo
Ex 4.4, 17 (Method 1) Let A be a nonsingular square matrix of order 3 × 3. Then |adj A| is equal to A. |A| B. |A|2 C. |A|3 D. 3 |A| We know that |𝑎𝑑𝑗 𝐴| = |𝐴|^(𝑛 − 1) where n is the order of Matrix A Here, n = 3 |𝑎𝑑𝑗 𝐴| = |𝐴|^(3 − 1) = |𝐴|^2 Hence, B is the correct answer Nonsingular: Where |𝐴|≠ 0 Ex 4.4, 17 (Method 2) Let A be a nonsingular square matrix of order 3 × 3. Then |adj A| is equal to A. |A| B. |A|2 C. |A|3 D. 3 |A| We know that A (adj A) = |A|I Taking determinants both sides |A (ad jA)| = ||A|I| Solving |A (adj (A))| |A (adj (A))| = |A| |adj (A)| (|𝐴𝐵|=|𝐴||𝐵|) Solving ||A|I| ||A|I| = |A|3|I| = |A|3 Now, |A (ad jA)| = ||A|I| Putting values |A| |adj (A)| = |A|3 |adj (A)| = |A|3/|A| |adj (A)| = |A|2 Thus, B is the correct answer Since |A| is Constant Using Property |kA| = kn |A| where n is order of A