First Principles AI

Accedmychevron_right11thchevron_rightmathchevron_rightMatrices And Determinantschevron_rightExercise 4.2

Solution

A matrix is singular iff its determinant is zero.

(i) $A=\begin{bmatrix}4&2&3\\7&\lambda&6\\2&3&1\end{bmatrix}$ is singular

Compute $|A|$ by expansion along the first row:

\begin{aligned} |A| &=4\left|\begin{matrix}\lambda&6\\3&1\end{matrix}\right|-2\left|\begin{matrix}7&6\\2&1\end{matrix}\right|+3\left|\begin{matrix}7&\lambda\\2&3\end{matrix}\right|\\ &=4(\lambda\cdot 1-6\cdot 3)-2(7\cdot 1-6\cdot 2)+3(7\cdot 3-\lambda\cdot 2)\\ &=4(\lambda-18)-2(7-12)+3(21-2\lambda)\\ &=4\lambda-72-2(-5)+63-6\lambda\\ &=1-2\lambda \end{aligned}

Singular means $|A|=0$ :

1-2\lambda=0\Rightarrow \boxed{\lambda=\frac{1}{2}}

(ii) $B=\begin{bmatrix}-2&4&5\\1&-2&1\\2&\lambda&0\end{bmatrix}$ is singular

Expand along the third column:

\begin{aligned} |B| &=5\left|\begin{matrix}1&-2\\2&\lambda\end{matrix}\right|-1\left|\begin{matrix}-2&4\\2&\lambda\end{matrix}\right|+0\\ &=5(1\cdot\lambda-(-2)\cdot 2)-( (-2)\lambda-4\cdot 2)\\ &=5(\lambda+4)-(-2\lambda-8)\\ &=5\lambda+20+2\lambda+8\\ &=7\lambda+28 \end{aligned}

Set $|B|=0$ :

7\lambda+28=0\Rightarrow \boxed{\lambda=-4}

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