First Principles AI

Accedmychevron_right11thchevron_rightmathchevron_rightMatrices And Determinantschevron_rightExercise 4.2

Solution

Given:

A=\begin{bmatrix} 1&2&1\\ -5&0&4\\ 5&4&0 \end{bmatrix}

Step 1: Compute $|A|$

Expand along the first row:

\begin{aligned} |A| &=1\left|\begin{matrix}0&4\\4&0\end{matrix}\right|-2\left|\begin{matrix}-5&4\\5&0\end{matrix}\right|+1\left|\begin{matrix}-5&0\\5&4\end{matrix}\right|\\ &=1(0-16)-2(0-20)+1(-20-0)\\ &=-16+40-20\\ &=4 \end{aligned}

So $A$ is invertible.

Step 2: Find the cofactor matrix and adjoint

The adjoint is $\operatorname{adj}(A)=(\text{cofactor matrix})^T$ .

From cofactor computation:

\operatorname{adj}(A)= \begin{bmatrix} -16&4&8\\ 20&-5&-9\\ -20&6&10 \end{bmatrix}

Step 3: Use $A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A)$

A^{-1}=\frac{1}{4} \begin{bmatrix} -16&4&8\\ 20&-5&-9\\ -20&6&10 \end{bmatrix} = \begin{bmatrix} -4&1&2\\ 5&-\frac{5}{4}&-\frac{9}{4}\\ -5&\frac{3}{2}&\frac{5}{2} \end{bmatrix}

Step 4: Verify $A^{-1}A=I_3$

Multiplying (or equivalently checking $AA^{-1}$ ) gives the identity matrix $I_3$ .

\boxed{A^{-1}=\begin{bmatrix} -4&1&2\\ 5&-\frac{5}{4}&-\frac{9}{4}\\ -5&\frac{3}{2}&\frac{5}{2} \end{bmatrix}}

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