Question 2, Exercise 2.3
Solutions of Question 2 of Exercise 2.3 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
Question 2(i)
Find the inverse of the matrix by using elementary row operation. [4−25210−123]
Solution
Let A=[4−25210−123]. Then |A|=|4−25210−123|=4(3)+2(6)+5(4+1)=49≠0. This gives, A is non-singular and A−1 exists. Now [4−25210−123|100010001]R∼[1414056−123|103012001] by R1+3R3 and R2+2R3R∼[14140560617|103012104] by R3+R1R∼[141401110617|1031−12104]by−R2+R3R∼[10−30011100−49|−34−51−12−56−8] by R3−6R2 and R1−4R2R∼[10−300111001|−34−51−12549−649849]by−149R3R∼[100010001|3491649−549−64917491049549−649849] by R1+30R3 and R2−11R3 Thus we have \begin{align} A^{-1}&=[3491649−549−64917491049549−649849]\\ \implies \quad A^{-1}&=\dfrac{1}{49}[316−5−617105−68] \end{align}
Question 2(ii)
Find the inverse of the matrix by using elementary row operation. [3−16134−151]
Solution
Let A=[3−16134−151]. Then |A|=|3−16134−151|=3(3−20)+1(1+4)+6(5+3)=−51+4+48=1≠0. This gives, A is non-singular and A−1 exists. Now [3−16134−151|100010001]R∼[1343−16−151|010100001] by R1↔R1R∼[1340−10−6085|0101−30011] by R2−3R1 and R3+R1R∼[1340−2−1085|0101−21011] by R2+R3R∼[1340112085|010−121−12011] by −12R2R∼[1340112001|010121−124−75] by R3−8R2R∼[134010001|010−5292−34−75] by R2−12R2R∼[104010001|152−2529−5292−34−75] by R1−3R2R∼[100010001|−172312−11−5292−34−75] by R1−4R3. Thus A−1=[−172312−11−5292−34−75]
Question 2(iii)
Find the inverse of the matrix by using elementary row operation. [12−30−20−2−22]
Solution
Let A=[12−30−20−2−22]. Then |A|=|12−30−20−2−22|=1(−4)−0−3(−4)=8≠0. This gives, A is non-singular and A−1 exists. Now [12−30−20−2−22|100010001]R∼[12−30−2002−4|100010201] by R3+2R1R∼[1010−2002−4|−10−1010201] by R1−R3R∼[10101001−2|−10−10−1201012] by −12R2 and 12R3R∼[10101000−2|−10−10−12011212] by R3−R2R∼[101010001|−10−10−120−12−14−14] by −12R3R∼[100010001|−1214−340−120−12−14−14] by R1−R3 Thus A−1=[−1214−340−120−12−14−14]
Question 2(iv)
Find the inverse of the matrix by using elementary row operation. [12−10−13102]
Solution
Let A=[12−10−13102]. Then |A|=|12−10−13102|=−2+6−1=3≠0. This gives, A is non-singular and A−1 exists. Now [12−10−13102|100010001]R∼[12−10−130−23|100010−101] by R3−R1R∼[1020−130−23|001010−101]byR1+R3R∼[1020−1300−3|001010−1−21] by R3−2R2R∼[1020−13001|0010101323−13] by −13R3R∼[1000−10001|−23−4353−1−111323−13] by R1−2R3 and R2−3R3R∼[100010001|−23−435311−11323−13] by −R2. Thus A−1=[−23−435311−11323−13].
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