Comprehensive Exercises on Matrices

Approach: Use the definition of matrix addition, adding corresponding entries.

Steps:

$A + B = \begin{pmatrix} 1+2 & 2+1 & 3+0 \\ 4+1 & 5+2 & 6+3 \end{pmatrix} = \begin{pmatrix} 3 & 3 & 3 \\ 5 & 7 & 9 \end{pmatrix}$

Answer: $A + B = \begin{pmatrix} 3 & 3 & 3 \\ 5 & 7 & 9 \end{pmatrix}$

Approach: First compute the scalar multiples, then carry out the subtraction.

Steps:

$2A = \begin{pmatrix} 2 & 4 \\ 6 & 8 \end{pmatrix}, \quad 3B = \begin{pmatrix} 0 & 3 \\ 3 & 0 \end{pmatrix}$

$2A - 3B = \begin{pmatrix} 2 & 4 \\ 6 & 8 \end{pmatrix} - \begin{pmatrix} 0 & 3 \\ 3 & 0 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 3 & 8 \end{pmatrix}$

Answer: $2A - 3B = \begin{pmatrix} 2 & 1 \\ 3 & 8 \end{pmatrix}$

Approach: Use the definition of matrix multiplication, and verify that matrix multiplication is not commutative.

Steps:

$AB = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix} = \begin{pmatrix} 4 & 6 \\ 1 & 3 \end{pmatrix}$

$BA = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 4 \\ 1 & 5 \end{pmatrix}$

Answer: $AB = \begin{pmatrix} 4 & 6 \\ 1 & 3 \end{pmatrix}$, $BA = \begin{pmatrix} 2 & 4 \\ 1 & 5 \end{pmatrix}$

Note that $AB \neq BA$, which verifies that matrix multiplication is not commutative.

Approach: Interchange the rows and the columns of the matrix.

Steps:

$A^T = \begin{pmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{pmatrix}$

Answer: $A^T = \begin{pmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{pmatrix}$

Approach: First compute the determinant, then find the inverse matrix.

Steps:

1. $|A| = 1 \times 4 - 2 \times 3 = -2 \neq 0$, so it is invertible

2. Compute the algebraic cofactors:

   • $A_{11} = 4$, $A_{12} = -3$
   • $A_{21} = -2$, $A_{22} = 1$

3. $A^* = \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$

4. $A^{-1} = \frac{1}{-2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}$

Answer: $A$ is invertible, and $A^{-1} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}$

Approach: Use the adjugate matrix method to find the inverse matrix.

Steps:

1. $|A| = 1 \neq 0$, so it is invertible

2. Compute the algebraic cofactors:

   • $A_{11} = 1$, $A_{12} = 0$, $A_{13} = 0$
   • $A_{21} = -2$, $A_{22} = 1$, $A_{23} = 0$
   • $A_{31} = 1$, $A_{32} = -2$, $A_{33} = 1$

3. $A^* = \begin{pmatrix} 1 & -2 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{pmatrix}$

4. $A^{-1} = \begin{pmatrix} 1 & -2 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{pmatrix}$

Answer: $A^{-1} = \begin{pmatrix} 1 & -2 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{pmatrix}$

Approach: Use the elementary row operation method to find the inverse matrix.

Steps:

1. Form the augmented matrix: $\begin{pmatrix} 2 & 1 & | & 1 & 0 \\ 1 & 1 & | & 0 & 1 \end{pmatrix}$

2. $R_1 \leftrightarrow R_2$: $\begin{pmatrix} 1 & 1 & | & 0 & 1 \\ 2 & 1 & | & 1 & 0 \end{pmatrix}$

3. $R_2 - 2R_1$: $\begin{pmatrix} 1 & 1 & | & 0 & 1 \\ 0 & -1 & | & 1 & -2 \end{pmatrix}$

4. $R_2 \times (-1)$: $\begin{pmatrix} 1 & 1 & | & 0 & 1 \\ 0 & 1 & | & -1 & 2 \end{pmatrix}$

5. $R_1 - R_2$: $\begin{pmatrix} 1 & 0 & | & 1 & -1 \\ 0 & 1 & | & -1 & 2 \end{pmatrix}$

Answer: $A^{-1} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$

Approach: Compute the algebraic cofactor of each entry.

Steps:

1. Compute the algebraic cofactors:

   • $A_{11} = 4$, $A_{12} = -3$
   • $A_{21} = -2$, $A_{22} = 1$

2. Arrange them in the transposed positions: $A^* = \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$

Answer: $A^* = \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$

Approach: Reduce to row echelon form by elementary transformations and count the nonzero rows.

Steps:

$\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1 \end{pmatrix}$

$R_2 - 2R_1$: $\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 1 & 1 & 1 \end{pmatrix}$

$R_3 - R_1$: $\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & -1 & -2 \end{pmatrix}$

$R_2 \leftrightarrow R_3$: $\begin{pmatrix} 1 & 2 & 3 \\ 0 & -1 & -2 \\ 0 & 0 & 0 \end{pmatrix}$

There are 2 nonzero rows, so $r(A) = 2$.

Answer: $r(A) = 2$

Approach: Reduce to row echelon form by elementary transformations and count the nonzero rows.

Steps:

$\begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 6 & 8 \\ 1 & 1 & 1 & 1 \end{pmatrix}$

$R_2 - 2R_1$: $\begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \end{pmatrix}$

$R_3 - R_1$: $\begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & -2 & -3 \end{pmatrix}$

$R_2 \leftrightarrow R_3$: $\begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & -1 & -2 & -3 \\ 0 & 0 & 0 & 0 \end{pmatrix}$

There are 2 nonzero rows, so $r(A) = 2$.

Answer: $r(A) = 2$

Approach: Compute the rank of each matrix and check whether they are equal.

Steps:

1. $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, $|A| = -2 \neq 0$, so $r(A) = 2$

2. $B = \begin{pmatrix} 2 & 4 \\ 6 & 8 \end{pmatrix}$, $B = 2A$, so $r(B) = r(A) = 2$

3. The two matrices have the same rank, so they are equivalent.

Answer: $A$ and $B$ are equivalent

Approach: Use the determinant property for block matrices.

Steps:

For the block matrix $\begin{pmatrix} A & B \\ O & C \end{pmatrix}$, we have $A_{12} = O$, so:

$|A| = |A_{11}| \cdot |A_{22}| = |A| \cdot |C|$

Answer: $\left|\begin{pmatrix} A & B \\ O & C \end{pmatrix}\right| = |A| \cdot |C|$

Approach: First find $A^{-1}$, then compute $X = A^{-1}B$.

Steps:

1. $A^{-1} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}$

2. $X = A^{-1}B = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix} \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix}$

3. $= \begin{pmatrix} -3 & -4 \\ 4 & 5 \end{pmatrix}$

Answer: $X = \begin{pmatrix} -3 & -4 \\ 4 & 5 \end{pmatrix}$

Approach: Write the system in matrix form and solve it with the inverse matrix.

Steps:

1. Write the system in matrix form: $AX = b$, where: $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \end{pmatrix}, \quad b = \begin{pmatrix} 5 \\ 7 \end{pmatrix}$

2. $A^{-1} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}$

3. $X = A^{-1}b = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix} \begin{pmatrix} 5 \\ 7 \end{pmatrix} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}$

Answer: $x = -3$, $y = 4$

Approach: Use the properties of idempotent matrices and the properties of the rank.

Steps:

1. Because $A^2 = A$, the matrix $A$ is idempotent

2. For an idempotent matrix, $r(A) = \text{tr}(A)$

3. This is because the eigenvalues of an idempotent matrix can only be 0 or 1, and the number of eigenvalues equal to 1 is the rank of the matrix

4. The trace of a matrix equals the sum of its eigenvalues, so $r(A) = \text{tr}(A)$

Answer: the proof is complete