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BÀI THI GIỮA KỲ ĐSTT 
Ngày 14/01/2025-Thời gian: 60 phút-Đề 4 
Câu 1a 
$\begin{align} 
&nbsp; &amp; {{B}^{2}}=B.B=\left( \begin{matrix} 
&nbsp;&nbsp; 1 &amp; 0 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 1 &amp; 2 &amp; 1&nbsp; \\ 
&nbsp;&nbsp; 1 &amp; 0 &amp; -2&nbsp; \\ 
\end{matrix} \right).\left( \begin{matrix} 
&nbsp;&nbsp; 1 &amp; 0 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 1 &amp; 2 &amp; 1&nbsp; \\ 
&nbsp;&nbsp; 1 &amp; 0 &amp; -2&nbsp; \\ 
\end{matrix} \right) \\ 
&nbsp;&amp; =\left( \begin{matrix} 
&nbsp;&nbsp; 1\cdot 1+0\cdot 1+0\cdot 1 &amp; 1\cdot 0+0\cdot 2+0\cdot 0 &amp; 1\cdot 0+0\cdot 1+0\cdot (-2)&nbsp; \\ 
&nbsp;&nbsp; 1\cdot 1+2\cdot 1+1\cdot 1 &amp; 1\cdot 0+2\cdot 2+1\cdot 0 &amp; 1\cdot 0+2\cdot 1+1\cdot (-2)&nbsp; \\ 
&nbsp;&nbsp; 1\cdot 1+0\cdot 1-2\cdot 1 &amp; 1\cdot 0+0\cdot 2-2\cdot 0 &amp; 1\cdot 0+0\cdot 1-2\cdot (-2)&nbsp; \\ 
\end{matrix} \right)=\left( \begin{matrix} 
&nbsp;&nbsp; 1 &amp; 0 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 4 &amp; 4 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; -1 &amp; 0 &amp; 4&nbsp; \\ 
\end{matrix} \right) \\ 
\end{align}$ 
Câu 1b 
$B+{{I}_{3}}=\left( \begin{matrix} 
&nbsp;&nbsp; 1 &amp; 0 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 1 &amp; 2 &amp; 1&nbsp; \\ 
&nbsp;&nbsp; 1 &amp; 0 &amp; -2&nbsp; \\ 
\end{matrix} \right)+\left( \begin{array}{*{35}{l}} 
&nbsp;&nbsp; 1 &amp; 0 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 0 &amp; 1 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 0 &amp; 0 &amp; 1&nbsp; \\ 
\end{array} \right)=~\left( \begin{array}{*{35}{l}} 
&nbsp;&nbsp; 2 &amp; 0 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 1 &amp; 3 &amp; 1&nbsp; \\ 
&nbsp;&nbsp; 1 &amp; 0 &amp; -1&nbsp; \\ 
\end{array} \right)\xrightarrow{\begin{array}{*{35}{l}} 
&nbsp;&nbsp; 2{{H}_{2}}-{{H}_{1}}\to {{H}_{2}}&nbsp; \\ 
&nbsp;&nbsp; 2{{H}_{3}}-{{H}_{1}}\to {{H}_{3}}&nbsp; \\ 
\end{array}}\left( \begin{array}{*{35}{l}} 
&nbsp;&nbsp; 2 &amp; 0 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 0 &amp; 6 &amp; 2&nbsp; \\ 
&nbsp;&nbsp; 0 &amp; 0 &amp; -2&nbsp; \\ 
\end{array} \right)$ 
Þ $r(B+{{I}_{3}})=3$ 
Câu 1.c) 
$\begin{align} 
&nbsp; &amp; \det (B)=\left| \begin{array}{*{35}{l}} 
&nbsp;&nbsp; 1 &amp; 0 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 1 &amp; 2 &amp; 1&nbsp; \\ 
&nbsp;&nbsp; 1 &amp; 0 &amp; -2&nbsp; \\ 
\end{array} \right|=1\cdot {{\left( -1 \right)}^{1+1}}\left| \begin{array}{*{35}{l}} 
&nbsp;&nbsp; 2 &amp; 1&nbsp; \\ 
&nbsp;&nbsp; 0 &amp; -2&nbsp; \\ 
\end{array} \right|+0+0 \\ 
&nbsp;&amp; =1\cdot \left( -4-0 \right)=-4\ne 0&nbsp; 
\end{align}$ 
Þ tồn tại ${{B}^{\,-1}}$. &nbsp;Ta có: 
&nbsp;$\begin{array}{*{35}{l}} 
&nbsp;&nbsp; {{B}_{11}}={{(-1)}^{1+1}}.\left| \begin{array}{*{35}{l}} 
&nbsp;&nbsp; 2 &amp; 1&nbsp; \\ 
&nbsp;&nbsp; 0 &amp; -2&nbsp; \\ 
\end{array} \right|=-4;&nbsp; \\ 
&nbsp;&nbsp; {{B}_{12}}={{(-1)}^{1+2}}.\left| \begin{array}{*{35}{l}} 
&nbsp;&nbsp; 1 &amp; 1&nbsp; \\ 
&nbsp;&nbsp; 1 &amp; -2&nbsp; \\ 
\end{array} \right|=3;&nbsp; \\ 
&nbsp;&nbsp; {{B}_{13}}={{(-1)}^{1+3}}.\left| \begin{array}{*{35}{l}} 
&nbsp;&nbsp; 1 &amp; 2&nbsp; \\ 
&nbsp;&nbsp; 1 &amp; 0&nbsp; \\ 
\end{array} \right|=-2;&nbsp; \\ 
\end{array}$ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; $\begin{array}{*{35}{l}} 
&nbsp;&nbsp; {{B}_{21}}=-\left| \begin{array}{*{35}{l}} 
&nbsp;&nbsp; 0 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 0 &amp; -2&nbsp; \\ 
\end{array} \right|=0&nbsp; \\ 
&nbsp;&nbsp; {{B}_{22}}=+\left| \begin{array}{*{35}{l}} 
&nbsp;&nbsp; 1 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 1 &amp; -2&nbsp; \\ 
\end{array} \right|=-2&nbsp; \\ 
&nbsp;&nbsp; {{B}_{23}}=-\left| \begin{array}{*{35}{l}} 
&nbsp;&nbsp; 1 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 1 &amp; 0&nbsp; \\ 
\end{array} \right|=0&nbsp; \\ 
\end{array}$&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp; $\begin{array}{*{35}{l}} 
&nbsp;&nbsp; {{B}_{31}}=+\left| \begin{array}{*{35}{l}} 
&nbsp;&nbsp; 0 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 2 &amp; 1&nbsp; \\ 
\end{array} \right|=0&nbsp; \\ 
&nbsp;&nbsp; {{B}_{32}}=-\left| \begin{array}{*{35}{l}} 
&nbsp;&nbsp; 1 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 1 &amp; 1&nbsp; \\ 
\end{array} \right|=-1&nbsp; \\ 
&nbsp;&nbsp; {{B}_{33}}=+\left| \begin{array}{*{35}{l}} 
&nbsp;&nbsp; 1 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 1 &amp; 2&nbsp; \\ 
\end{array} \right|=2&nbsp; \\ 
\end{array}$ 
\[\begin{align} 
&nbsp; &amp; \Rightarrow {{B}^{-1}}=\frac{1}{\det (B)}\left( \begin{array}{*{35}{l}} 
&nbsp;&nbsp; -4 &amp; 0 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 3 &amp; -2 &amp; -1&nbsp; \\ 
&nbsp;&nbsp; -2 &amp; 0 &amp; 2&nbsp; \\ 
\end{array} \right)=\frac{1}{-4}\left( \begin{array}{*{35}{l}} 
&nbsp;&nbsp; -4 &amp; 0 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 3 &amp; -2 &amp; -1&nbsp; \\ 
&nbsp;&nbsp; -2 &amp; 0 &amp; 2&nbsp; \\ 
\end{array} \right)\Rightarrow {{B}^{-1}}=\left( \begin{array}{*{35}{l}} 
&nbsp;&nbsp; 1 &amp; 0 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; \frac{-3}{4} &amp; \frac{1}{2} &amp; \frac{1}{4}&nbsp; \\ 
&nbsp;&nbsp; \frac{1}{2} &amp; 0 &amp; \frac{-1}{2}&nbsp; \\ 
\end{array} \right) \\ 
&nbsp;&amp; \Rightarrow {{(3B)}^{-1}}=\frac{1}{3}{{B}^{-1}}=\frac{1}{3}\left( \begin{array}{*{35}{l}} 
&nbsp;&nbsp; 1 &amp; 0 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; \frac{-3}{4} &amp; \frac{1}{2} &amp; \frac{1}{4}&nbsp; \\ 
&nbsp;&nbsp; \frac{1}{2} &amp; 0 &amp; \frac{-1}{2}&nbsp; \\ 
\end{array} \right)=~\left( \begin{matrix} 
&nbsp;&nbsp; \frac{1}{3} &amp; 0 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; -\frac{1}{4} &amp; \frac{1}{6} &amp; \frac{1}{12}&nbsp; \\ 
&nbsp;&nbsp; \frac{1}{6} &amp; 0 &amp; -\frac{1}{6}&nbsp; \\ 
\end{matrix} \right) \\ 
\end{align}\] 
Câu 1.d) 
Giả sử tồn tại Y thoả mãn đề bài.$B$ là ma trận $3\times 3$ 
Þ ${{B}^{2025}}$ là ma trận $3\times 3$ 
Þ ${{B}^{2025}}Y$ là ma trận $3\times m$, với m là số cột của Y 
Mà ${{\theta }_{4}}$ có kích thước là $4\times 1$ 
Þ chắc chắn ${{B}^{2025}}Y\ne {{\theta }_{4}}$ 
Þ không tồn tại ma trận Y thoả mãn đề bài. 
Câu 2. 
$\begin{align} 
&nbsp; &amp; \bar{A}=\left( \begin{array}{*{35}{l}} 
&nbsp;&nbsp; 3 &amp; 4 &amp; 1 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 1 &amp; 1 &amp; 1 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 6 &amp; 7 &amp; {{m}^{2}}+5 &amp; 0&nbsp; \\ 
\end{array} \right)\xrightarrow{\begin{array}{*{35}{l}} 
&nbsp;&nbsp; 3{{H}_{2}}-{{H}_{1}}\to {{H}_{2}}&nbsp; \\ 
&nbsp;&nbsp; {{H}_{3}}-2{{H}_{1}}\to {{H}_{3}}&nbsp; \\ 
\end{array}}\left( \begin{array}{*{35}{l}} 
&nbsp;&nbsp; 3 &amp; 4 &amp; 1 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 0 &amp; -1 &amp; 2 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 0 &amp; -1 &amp; {{m}^{2}}+3 &amp; 0&nbsp; \\ 
\end{array} \right) \\ 
&nbsp;&amp; \xrightarrow{\begin{array}{*{35}{l}} 
&nbsp;&nbsp; {{H}_{3}}-{{H}_{2}}\to {{H}_{3}}&nbsp; \\ 
\end{array}}\left( \begin{array}{*{35}{l}} 
&nbsp;&nbsp; 3 &amp; 4 &amp; 1 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 0 &amp; -1 &amp; 2 &amp; 0&nbsp; \\ 
&nbsp;&nbsp; 0 &amp; 0 &amp; {{m}^{2}}+1 &amp; 0&nbsp; \\ 
\end{array} \right) \\ 
\end{align}$ 
Vì ${{m}^{2}}+1&gt;0,\text{ }\forall m\in \mathbb{R}\Rightarrow r(A)=r(\bar{A})=3=$ số ẩn 
Þ hệ phương trình có nghiệm duy nhất với mọi m 
$\left\{ \begin{matrix} 
&nbsp;&nbsp; 3{{x}_{1}}+4{{x}_{2}}+{{x}_{3}}=0&nbsp; \\ 
&nbsp;&nbsp; -{{x}_{2}}+2{{x}_{3}}=0&nbsp; \\ 
&nbsp;&nbsp; ({{m}^{2}}+1){{x}_{3}}=0&nbsp; \\ 
\end{matrix} \right.\Leftrightarrow \left\{ \begin{matrix} 
&nbsp;&nbsp; {{x}_{3}}=0&nbsp; \\ 
&nbsp;&nbsp; {{x}_{2}}=0&nbsp; \\ 
&nbsp;&nbsp; {{x}_{1}}=0&nbsp; \\ 
\end{matrix} \right.$ 
Vậy nghiệm đó là $\left( 0;\text{ }0;\text{ }0 \right)$ 
Câu 3. 
\[\begin{align} 
&nbsp; &amp; \left| \begin{array}{*{35}{l}} 
&nbsp;&nbsp; x &amp; 5 &amp; 3 &amp; 1&nbsp; \\ 
&nbsp;&nbsp; 0 &amp; x+5 &amp; 3 &amp; 1&nbsp; \\ 
&nbsp;&nbsp; 0 &amp; 5 &amp; x+3 &amp; 1&nbsp; \\ 
&nbsp;&nbsp; 0 &amp; 5 &amp; 3 &amp; x+1&nbsp; \\ 
\end{array} \right|=0\Leftrightarrow {{\left( -1 \right)}^{1+1}}\cdot x\cdot \left| \begin{array}{*{35}{l}} 
&nbsp;&nbsp; x+5 &amp; 3 &amp; 1&nbsp; \\ 
&nbsp;&nbsp; 5 &amp; x+3 &amp; 1&nbsp; \\ 
&nbsp;&nbsp; 5 &amp; 3 &amp; x+1&nbsp; \\ 
\end{array} \right|=0 \\ 
&nbsp;&amp; \Leftrightarrow x\cdot \left| \begin{array}{*{35}{l}} 
&nbsp;&nbsp; x+5 &amp; 3 &amp; 1&nbsp; \\ 
&nbsp;&nbsp; 5 &amp; x+3 &amp; 1&nbsp; \\ 
&nbsp;&nbsp; 5 &amp; 3 &amp; x+1&nbsp; \\ 
\end{array} \right|=0 \\ 
&nbsp;&amp; \Leftrightarrow x\left( \left[ \left( x+5 \right)\left( x+3 \right)\left( x+1 \right)+15+15 \right]-\left[ -5\left( x+3 \right)-3\left( x+5 \right)-15\left( x+1 \right) \right] \right) \\ 
&nbsp;&amp; \Leftrightarrow x({{x}^{3}}+9{{x}^{2}})=0\Leftrightarrow {{x}^{3}}(x+9)=0\Leftrightarrow \left[ \begin{matrix} 
&nbsp;&nbsp; x=0&nbsp; \\ 
&nbsp;&nbsp; x=-9&nbsp; \\ 
\end{matrix} \right. \\ 
\end{align}\] 
Vậy có 2 số thực thoả mãn là $x=0$ và $x=-9$ 
&nbsp;

ĐỀ SỐ 22: KTRA NGÀY 14/01/2025-ĐỀ 4

Đề thi gồm 3 câu hỏi tự luận có hướng dẫn giải chi tiết