Linear Algebra Mid-Term Prediction (Solutions)

Problem 1 Solution

a) Finding eigenvalues and eigenvectors of $B$:

Given matrix $B=\left(\begin{array}{ccc}2& -1& 0\\ 1& 2& -2\\ 1& 0& 1\end{array}\right)$.

First, find the eigenvalues by solving $\det (B-\lambda I)=0$.

$B-\lambda I=\left(\begin{array}{ccc}2-\lambda & -1& 0\\ 1& 2-\lambda & -2\\ 1& 0& 1-\lambda \end{array}\right)$

$\det (B-\lambda I)=(2-\lambda )[(2-\lambda )(1-\lambda )]-(-1)(-2)-(0)=0$

Solving this characteristic polynomial will give us the eigenvalues $\lambda$.

b) Diagonalisability of $B$:

For $B$ to be diagonalisable, the number of linearly independent eigenvectors must equal the multiplicity of each eigenvalue. We find these eigenvectors by solving $(B-\lambda I)\mathbf{v}=0$ for each $\lambda$.

If we find 3 linearly independent eigenvectors, then $B$ is diagonalisable, and we can construct matrix $P$ from these eigenvectors. $D$ is a diagonal matrix with eigenvalues on the diagonal.

Detailed calculations for the eigenvalues, eigenvectors, and the diagonalisation process will require solving the characteristic polynomial and system of equations for each eigenvalue.

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Problem 2 Solution

a) Solving the system using Gauss-Jordan elimination:

Given matrix $C$ and system $C\mathbf{x}=0$,

$C=\left(\begin{array}{ccc}1& 2& -1\\ -2& 4& 2\\ 3& 6& -3\end{array}\right)$

We apply Gauss-Jordan elimination to reduce $C$ to its reduced row echelon form (RREF), then solve for $\mathbf{x}$.

b) Solution set as a vector subspace and its basis:

The solution set will be all vectors $\mathbf{x}$ that satisfy the RREF form of $C\mathbf{x}=0$. We find the basis by identifying the free variables and expressing the solution set in terms of these variables, showing it forms a subspace.

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Problem 3 Solution

a) Finding the inverse of $M$ using Gauss-Jordan elimination:

Given $M=\left(\begin{array}{ccc}0& 1& 2\\ 2& 0& -2\\ 1& -1& 1\end{array}\right)$,

To find $M^{-1}$, augment $M$ with the identity matrix $I$ and apply Gauss-Jordan elimination until the left side is $I$. The right side will then be $M^{-1}$.

b) Showing $N=M^{-1}$:

If $NM=I$, then by definition, $N$ is the inverse of $M$. Since we’ve found $M^{-1}$, we can show $N$ and $M^{-1}$ are the same by comparison.

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Problem 4 Solution

a) Proving $\{v_1,v_2,v_3\}$ forms a basis for ${\mathbb{R}}^3$:

Vectors form a basis if they are linearly independent and span ${\mathbb{R}}^3$. We can show linear independence by showing the determinant of the matrix formed by these vectors is non-zero.

b) Coordinates of $w$ in the basis $\{v_1,v_2,v_3\}$:

Solve the linear equation $\alpha v_1+\beta v_2+\gamma v_3=w$ for coefficients $\alpha ,\beta ,\gamma$. These coefficients are the coordinates of $w$ in the given basis.