Clase 12 Soluciones NCERT – Matemáticas Parte I – Capítulo 3 Arrays – Ejercicio 3.4 | Serie 1

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Usando transformaciones elementales, encuentre la inversa de cada una de las arrays, si existe en los ejercicios 1 a 6.

Pregunta 1.\left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right]

Solución:

\begin{aligned} &\text { Since we know that, } A= IA \Rightarrow\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text { A }\\ &\Rightarrow\left[\begin{array}{cc} 1 & -1 \\ 0 & 5 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ -2 & 1 \end{array}\right] \mathrm{A}\left[\mathrm{R}{2} \rightarrow \mathrm{R}{2}-2 \mathrm{R}_{1}\right]\\ &\Rightarrow\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ -2 / 5 & 1 / 5 \end{array}\right] \mathrm{A}\left[\mathrm{R}{2} \rightarrow \frac{1}{5} \mathrm{R}{2}\right]\\ &\left.\Rightarrow\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 3 / 5 & 1 / 5 \\ -2 / 5 & 1 / 5 \end{array}\right] \text { A }\left[R_{1} \rightarrow R_{1}+R_{2}\right.\\ \end{aligned}

Therefore,A^{-1}=\left[\begin{array}{cc} 3 / 5 & 1 / 5 \\ -2 / 5 & 1 / 5 \end{array}\right]

Pregunta 2.\left[\begin{array}{ll} 2 & 1 \\ 1 & 1 \end{array}\right]

Solución:

Sea A=\left[\begin{array}{ll} 2 & 1 \\ 1 & 1 \end{array}\right]

\begin{array}{l} We Know That, A=IA ⇒\left[\begin{array}{ll} 2 & 1 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] A \\ \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{lr} 1 & -1 \\ 0 & 1 \end{array}\right] A \\ \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{lr} 1 & -1 \\ -1 & 2 \end{array}\right] A \\ Therefore, A^{-1}=\left[\begin{array}{lr} 1 & -1 \\ -1 & 2 \end{array}\right] \end{array} [Tex]\begin{array}{l} \left(R_{1} \rightarrow R_{1}-R_{2}\right) \\ \left(R_{2} \rightarrow R_{2}-R_{ 1}\right) \end{array}[/Tex]

Pregunta 3.\left[\begin{array}{ll} 1 & 3 \\ 2 & 7 \end{array}\right]

Solución:

\begin{aligned} &\text { Let } A=\left[\begin{array}{ll} 1 & 3 \\ 2 & 7 \end{array}\right]\\ &\text { W.K.T., } A=IA \Rightarrow\left[\begin{array}{ll} 1 & 3 \\ 2 & 7 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text { A }\\ &\Rightarrow\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{rc} 1 & 0 \\ -2 & 1 \end{array}\right] \text { A }\left[R_{2} \rightarrow R_{2}-2 R_{1}\right]\\ &\left.\left.\Rightarrow \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\begin{array}{cc} - & -3 \\ -2 & 1 \end{array}\right] \text { A }\left[R_{1} \rightarrow R_{1}-3 R_{2}\right]\\ &Therefore,A^{-1}⇒\left[\begin{array}{cc} 7 & -3 \\ -2 & 1 \end{array}\right] \end{aligned}

Pregunta 4.\left[\begin{array}{ll} 2 & 3 \\ 5 & 7 \end{array}\right]

Solución:

\begin{aligned} &\text { Let } A=\left[\begin{array}{ll} 2 & 3 \\ 5 & 7 \end{array}\right]\\ &\text { W.K.T. , } A=I A \Rightarrow\left[\begin{array}{ll} 2 & 3 \\ 5 & 7 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text { A }\\ &\Rightarrow\left[\begin{array}{ll} 2 & 3 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{rc} 1 & 0 \\ -2 & 1 \end{array}\right] \text { A }\left[R_{2} \rightarrow R_{2}-2 R_{1}\right]\\ &⇒\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]=\left[\begin{array}{cc} -2 & -1 \\ 1 & 0 \end{array}\right] A\left[R_{1} \leftrightarrow R_{i}\right] \end{aligned}

\begin{array}{l} \Rightarrow\left[\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} -2 & 1 \\ 5 & -2 \end{array}\right] \text { A }\left[R_{2} \rightarrow R_{2}-2 R_{1}\right] \\ \Rightarrow\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} -7 & 3 \\ 5 & -2 \end{array}\right] \mathrm{A}\left[\mathrm{R}{1} \rightarrow \mathrm{R}{1}-\mathrm{R}_{2}\right] \\ Therefore, \mathrm{A}^{-1}=\left[\begin{array}{cc} -7 & 3 \\ 5 & -2 \end{array}\right] \end{array}

Pregunta 5.\left[\begin{array}{ll} 2 & 1 \\ -7 & 4 \end{array}\right]

Solución:

\begin{aligned} &\text { Let } A=\left[\begin{array}{ll} 2 & 1 \\ 7 & 4 \end{array}\right]\\ &\text { W.K.T. , } A=IA \Rightarrow\left[\begin{array}{ll} 2 & 1 \\ 7 & 4 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text { A }\\ &\Rightarrow\left[\begin{array}{ll} 2 & 1 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ -3 & 1 \end{array}\right] \text { A }\left[R_{2} \rightarrow R_{2}-3 R_{1}\right]\\ &⇒\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{cc} 4 & -1 \\ -3 & 1 \end{array}\right] A\left[R_{1} \rightarrow R_{1}-R_{2}\right]\\ &⇒\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 4 & -1 \\ -7 & 2 \end{array}\right] A \quad\left[R, \rightarrow R,-R_{1}\right]\\ &Therefore,A=\left[\begin{array}{cc} 4 & -1 \\ -7 & 2 \end{array}\right] \end{aligned}

Pregunta 6.\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right]

Solución:

\begin{aligned} &\text { Let } A=\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right]\\ &\text { W.K.T. , } A=IA \Rightarrow\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text { A }\\ &⇒\left[\begin{array}{ll} 1 & 3 \\ 2 & 5 \end{array}\right]=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] A \quad\left[\mathrm{R}{1} \leftrightarrow \mathrm{R}{2}\right]\\ &\Rightarrow\left[\begin{array}{cc} 1 & 3 \\ 0 & -1 \end{array}\right]=\left[\begin{array}{cc} 0 & 1 \\ 1 & -2 \end{array}\right] \text { A }\left[R_{2} \rightarrow R_{2}-2 R_{1}\right] \end{aligned}

\begin{array}{l} \Rightarrow\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 0 & 1 \\ -1 & 2 \end{array}\right] \text { A }\left[\mathrm{R}{2} \rightarrow(-1) \mathrm{R}{2}\right] \\ \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 3 & -5 \\ -1 & 2 \end{array}\right] \mathrm{A} \quad\left[\mathrm{R}{1} \rightarrow \mathrm{R}{1}-3 \mathrm{R}_{2}\right] \\ Therefore,\mathrm{A}^{-1}=\left[\begin{array}{rr} 3 & -5 \\ -1 & 2 \end{array}\right] \end{array}

Usando transformación elemental, encuentre la inversa de cada una de las arrays, si existe en los ejercicios 7 a 14.

Pregunta 7.\left[\begin{array}{ll} 3 & 1 \\ 5 & 2 \end{array}\right]

Solución:

\begin{array}{l} \text { Ans. Let } A=\left[\begin{array}{ll} 3 & 1 \\ 5 & 2 \end{array}\right] \\ \text { W.K.T. , } A=IA \Rightarrow\left[\begin{array}{ll} 3 & 1 \\ 5 & 2 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] A \\ \Rightarrow\left[\begin{array}{ll} 6 & 2 \\ 5 & 2 \end{array}\right]=\left[\begin{array}{ll} 2 & 0 \\ 0 & 1 \end{array}\right] \text { A }\left[R_{1} \rightarrow 2 R_{1}\right] \\ \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 5 & 2 \end{array}\right]=\left[\begin{array}{cc} 2 & -1 \\ 0 & 1 \end{array}\right] \text { A }\left[R_{1} \rightarrow R_{1}-R_{2}\right] \end{array}

\begin{array}{l} \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 2 \end{array}\right]=\left[\begin{array}{cc} 2 & -1 \\ -10 & 6 \end{array}\right] \mathrm{A}\left[R_{2} \rightarrow R_{2}-5 R_{1}\right] \\ \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 2 & -1 \\ -5 & 3 \end{array}\right] \mathrm{A} \quad\left[\mathrm{R}{2} \rightarrow \frac{1}{2} \mathrm{R}{2}\right] \\ Therefore, \quad \mathrm{A}^{-1}=\left[\begin{array}{rr} 2 & -1 \\ -5 & 3 \end{array}\right] \end{array}

pregunta 8\begin{bmatrix}4 & 5 \\3 & 4 \\\end{bmatrix}

Solución:

\begin{aligned} &\text { Ans. Let } A=\left[\begin{array}{ll} 4 & 5 \\ 3 & 4 \end{array}\right]\\ &\text { W.K.T. , } A=IA \Rightarrow\left[\begin{array}{ll} 4 & 5 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text { A }\\ &\Rightarrow\left[\begin{array}{ll} 1 & 1 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right] \text { A } \quad\left[R_{1} \rightarrow R_{1}-R_{2}\right]\\ &\Rightarrow\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{lr} 1 & -1 \\ -3 & 4 \end{array}\right] \text { A }\left[R_{2} \rightarrow R_{2}-3 R_{1}\right]\\ &\Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 4 & -5 \\ -3 & 4 \end{array}\right] \text { A }\left[R_{1} \rightarrow R_{1}-R_{2}\right]\\ &Therefore,A^{-1}=\begin{bmatrix}4 & -5 \\-3 & 4 \\\end{bmatrix} \end{aligned}

Pregunta 9.\begin{bmatrix}3 & 10 \\2 & 7 \\\end{bmatrix}

Solución:

\begin{aligned} &\text { Let } A=\left[\begin{array}{cc} 3 & 10 \\ 2 & 7 \end{array}\right]\\ &\text { W.K.T. , } A=IA \Rightarrow\left[\begin{array}{cc} 3 & 10 \\ 2 & 7 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text { A }\\ &\Rightarrow\left[\begin{array}{ll} 1 & 3 \\ 2 & 7 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right] \mathrm{A} \quad\left[\mathrm{R}{1} \rightarrow \mathrm{R}{1}-\mathrm{R}_{2}\right]\\ &\Rightarrow\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ -2 & 3 \end{array}\right] \text { A }\left[R_{2} \rightarrow R_{2}-2 R_{1}\right]\\ & \end{aligned}

\begin{array}{l} \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 7 & -10 \\ -2 & 3 \end{array}\right] \mathrm{A} \quad\left[\mathrm{R}{1} \rightarrow \mathrm{R}{1}-3 \mathrm{R}_{2}\right] \\ Therefore, \quad \mathrm{A}^{-1}=\left[\begin{array}{rr} 7 & -10 \\ -2 & 3 \end{array}\right] \end{array}

Pregunta 10.\begin{bmatrix}3& -1 \\-4 & 2 \\\end{bmatrix}

Solución:

\begin{array}{l} \text { W.K.T. , } A=I A \Rightarrow\left[\begin{array}{rr} 3 & -1 \\ -4 & 2 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] \mathrm{A} \\ \Rightarrow\left[\begin{array}{cc} -1 & 1 \\ -4 & 2 \end{array}\right]=\left[\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right] \mathrm{A} \quad\left[\mathrm{R}{1} \rightarrow \mathrm{R}{1}+\mathrm{R}_{2}\right] \\ \Rightarrow\left[\begin{array}{rr} 1 & -1 \\ -4 & 2 \end{array}\right]=\left[\begin{array}{cc} -1 & -1 \\ 0 & 1 \end{array}\right] \mathrm{A}\left[R_{1} \rightarrow(-1) R_{1}\right] \end{array}

\begin{array}{l} \Rightarrow⇒\left[\begin{array}{cc} 1 & -1 \\ 0 & -2 \end{array}\right]=\left[\begin{array}{cc} -1 & -1 \\ -4 & 3 \end{array}\right] A\left[\mathrm{R}{2} \rightarrow \mathrm{R}{2}+4 \mathrm{R}_{1}\right] \\ \Rightarrow\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} -1 & -1 \\ 2 & 3 / 2 \end{array}\right] \mathrm{A}\left[\mathrm{R}{2} \rightarrow \frac{-1}{2} \mathrm{R}{2}\right] \\ \Rightarrow \quad\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & 1 / 2 \\ 2 & 3 / 2 \end{array}\right] \mathrm{A}\left[\mathrm{R}{1} \rightarrow \mathrm{R}{1}+\mathrm{R}_{2}\right] \\ \mathrm{Therefore,A}^{-1}= {\left[\begin{array}{cc} 1 & 1 / 2 \\ 2 & 3 / 2 \end{array}\right]} \end{array}

Capítulo 3 Arrays – Ejercicio 3.4 | conjunto 2

Publicación traducida automáticamente

Artículo escrito por kavyagupta0098 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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