Animated Solution for Mathematics - Matrices and Determinants: Comprehension Passage

We are given the matrix $A=\left[\begin{array}{ccc}1& 0& 0\\ 2& 1& 0\\ 3& 2& 1\end{array}\right]$ and the products $A{U}_1,A{U}_2,$ and $A{U}_3$. If we define $U=[U_1{U}_2{U}_3]$, then by the fundamental definition of matrix multiplication, $AU=[A{U}_{1}A{U}_{2}A{U}_3]$.

This is our 'Aha!' moment. We do not need to solve for $U_1,U_2,$ and $U_3$ individually. We can construct the matrix $AU$ directly:

Now, observe the structure of these matrices. $A$ is a lower triangular matrix, and $AU$ is an upper triangular matrix.

The determinant of any triangular matrix is simply the product of its diagonal elements. For $A$, the determinant is:

For $AU$, the determinant is:

Using the property $|AU|=|A|\cdot |U|$, we immediately find that $3=1\cdot |U|$, which means $|U|=3$.

To find the sum of the elements of $U^{-1}$, we first determine $U$ using $U=A^{-1}(AU)$. Since $A$ is lower triangular, its inverse is also lower triangular:

Performing the multiplication $U=A^{-1}(AU)$, we obtain:

We use the formula $U^{-1}=\frac{1}{|U|}\text{adj}(U)$. With $|U|=3$, we calculate the cofactor matrix, transpose it, and multiply by $1/3$:

The sum of the elements is the sum of the entries of the adjoint matrix divided by 3. Summing the entries: $(-1-2+0-7-5-3+9+6+3)=-18+18=0$. Thus, the sum of the elements of $U^{-1}$ is $0$.

Finally, we evaluate the expression $\left[\begin{array}{ccc}3& 2& 0\end{array}\right]U\left[\begin{array}{c}3\\ 2\\ 0\end{array}\right]$. Let $X=\left[\begin{array}{c}3\\ 2\\ 0\end{array}\right]$. We need to compute $X^{T}UX$.

First, compute $UX$:

Then, multiply by $X^T$:

The final value of the quadratic form is $5$.