First Principles AI

Accedmychevron_right11thchevron_rightmathchevron_rightMatrices And Determinantschevron_rightExercise 4.2

Solution

(i) $\left|\begin{matrix}2&1&x\\-1&-4&-3\\x&1&0\end{matrix}\right|=5$

Expand along the 3rd column:

\begin{aligned} \Delta&=x\left|\begin{matrix}-1&-4\\x&1\end{matrix}\right|-(-3)\left|\begin{matrix}2&1\\x&1\end{matrix}\right|+0\\ &=x((-1)(1)-(-4)x)+3(2\cdot 1-1\cdot x)\\ &=x(-1+4x)+3(2-x)\\ &=4x^2-x+6-3x\\ &=4x^2-4x+6 \end{aligned}

Given $\Delta=5$ :

4x^2-4x+6=5\Rightarrow 4x^2-4x+1=0\Rightarrow (2x-1)^2=0

\boxed{x=\frac{1}{2}}

(ii) $\left|\begin{matrix}1&x-1&3\\-1&x+1&2\\2&-3&x\end{matrix}\right|=9$

Compute the determinant by expansion:

\begin{aligned} \Delta &=1\left|\begin{matrix}x+1&2\\-3&x\end{matrix}\right|-(x-1)\left|\begin{matrix}-1&2\\2&x\end{matrix}\right|+3\left|\begin{matrix}-1&x+1\\2&-3\end{matrix}\right|\\ &=((x+1)x-2(-3))-(x-1)((-1)x-2\cdot 2)+3(((-1)(-3)-(x+1)\cdot 2))\\ &=(x^2+x+6)-(x-1)(-x-4)+3(3-2x-2)\\ &=(x^2+x+6)+(x-1)(x+4)+3(1-2x)\\ &=(x^2+x+6)+(x^2+3x-4)+3-6x\\ &=2x^2-2x+5 \end{aligned}

Given $\Delta=9$ :

2x^2-2x+5=9\Rightarrow 2x^2-2x-4=0\Rightarrow x^2-x-2=0

(x-2)(x+1)=0

\boxed{x=2\ \text{or}\ x=-1}

(iii) $\left|\begin{matrix}1&1&1\\2&x&2\\3&6&x\end{matrix}\right|=0$

Expand along the first row:

\begin{aligned} \Delta &=1\left|\begin{matrix}x&2\\6&x\end{matrix}\right|-1\left|\begin{matrix}2&2\\3&x\end{matrix}\right|+1\left|\begin{matrix}2&x\\3&6\end{matrix}\right|\\ &=(x^2-12)-(2x-6)+(12-3x)\\ &=x^2-12-2x+6+12-3x\\ &=x^2-5x+6 \end{aligned}

Given $\Delta=0$ :

x^2-5x+6=0\Rightarrow (x-2)(x-3)=0

\boxed{x=2\ \text{or}\ x=3}

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