第12类NCERT解决方案-数学第I部分–第3章矩阵–练习3.4 |套装1

如果练习1至6中存在矩阵，则使用基本转换查找每个矩阵的逆。

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

解决方案：

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

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

问题2。 $\left[\begin{array}{ll} 2 & 1 \\ 1 & 1 \end{array}\right]$

解决方案：

Let 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]

问题3。 $\left[\begin{array}{ll} 1 & 3 \\ 2 & 7 \end{array}\right]$

解决方案：

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

问题4。 $\left[\begin{array}{ll} 2 & 3 \\ 5 & 7 \end{array}\right]$

解决方案：

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

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

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

解决方案：

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

问题6。 $\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right]$

解决方案：

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

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

如果练习7至练习14中存在矩阵，则使用基本变换查找每个矩阵的逆。

问题7。 $\left[\begin{array}{ll} 3 & 1 \\ 5 & 2 \end{array}\right]$

解决方案：

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

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

问题8。 $\begin{bmatrix}4 & 5 \\3 & 4 \\\end{bmatrix}$

解决方案：

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

问题9。 $\begin{bmatrix}3 & 10 \\2 & 7 \\\end{bmatrix}$

解决方案：

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

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

问题10。 $\begin{bmatrix}3& -1 \\-4 & 2 \\\end{bmatrix}$

解决方案：

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

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

第3章矩阵–练习3.4 |套装2