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Now it is time to put your new-found knowledge to the test through the practice questions on Linear Algebra listed below. The solutions are available at the end of the lesson but try your best to solve these practice questions by yourself.

1. For matrices $A = \begin{bmatrix} 4 & 7 & 3 \\ 3 & 7 & 0 \\ 4 & 6 & 5 \\ \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 & 5 \\ 3 & 8 & 3 \\ 0 & 1 & 8 \\ \end{bmatrix}$, find:

a. $A+B$

b. $A-B$

c. $A\cdot B$

d. $A \times B$

e. $B \times A$

2. For matrices $C = \begin{bmatrix} 5 & 1 \\ 0 & 8 \\ \end{bmatrix}$ and $D = \begin{bmatrix} 4 & 1 \\ 8 & 2 \\ \end{bmatrix}$, find where possible:

a. $det(C)$

b. $C^{-1}$

c. $det(D)$

d. $D^{-1}$

If the inverse does not exist, state why.

## Solutions

Were you able to solve all of these questions? If not, you might want to refer back to previous lessons. Nonetheless, here are the solutions:

1.a. $A + B = \begin{bmatrix} 5 & 9 & 8 \\ 6 & 15 & 3 \\ 4 & 7 & 13 \\ \end{bmatrix}$

1.b. $A-B = \begin{bmatrix} 3 & 5 & -2 \\ 0 & -1 & -3 \\ 4 & 5 & -3 \\ \end{bmatrix}$

1.c. $A \cdot B = 144$

1.d. $A \times B = \begin{bmatrix} 4 \times 1 + 7 \times 3 + 3 \times 0 & 4 \times 2 + 7 \times 8 + 3 \times 1 & 4 \times 5 + 7 \times 3 + 3 \times 8 \\ 3 \times 1 + 7 \times 3 + 0 \times 0 & 3 \times 2 + 7 \times 8 + 0 \times 1 & 3 \times 5 + 7 \times 3 + 0 \times 8 \\ 4 \times 1 + 6 \times 3 + 5 \times 0 & 4 \times 2 + 6 \times 8 + 5 \times 1 & 4 \times 5 + 6 \times 3 + 5 \times 8 \\ \end{bmatrix}$ $ = \begin{bmatrix} 25 & 67 & 65 \\ 24 & 62 & 36 \\ 22 & 61 & 78 \end{bmatrix}$

1.e . $B \times A = \begin{bmatrix} 1 \times 4 + 2 \times 3 + 5 \times 4 & 1 \times 7 + 2 \times 7 + 5 \times 6 & 1 \times 3 + 2 \times 0 + 5 \times 5 \\ 3 \times 4 + 8 \times 3 + 3 \times 4 & 3 \times 7 + 8 \times 7 + 3 \times 6 & 3 \times 3 + 8 \times 0 + 3 \times 5 \\ 0 \times 4 + 1 \times 3 + 8 \times 4 & 0 \times 7 + 1 \times 7 + 8 \times 6 & 0 \times 3 + 1 \times 0 + 8 \times 5 \\ \end{bmatrix}$ $= \begin{bmatrix} 30 & 51 & 28 \\ 48 & 95 & 24 \\ 35 & 55 & 40 \\\end{bmatrix}$

2.a. $det(C) = 5\times8 – 0 \times 1 = 40$

2.b. $C^{-1} = \dfrac{1}{40}\begin{bmatrix} 8 & -1 \\ 0 & 5 \\ \end{bmatrix}$ $= \begin{bmatrix} \dfrac{1}{5} & -\dfrac{1}{40} \\ 0 & \frac{1}{8} \end{bmatrix}$

2.c. $det(D)= 4 \times 2 – 8 \times 1 = 0 $

2.d. The inverse does not exist as the determinant is zero.

Hope you enjoyed solving those questions and are now more confident of your conceptual knowledge of Linear Algebra moving forward.

## End of the Course

With this, we have come to the end of our Linear Algebra for Data Science Mini-Course. We hope that this short course helped you as a stepping stone towards your journey in learning the concept of Linear Algebra and mathematics for Data Science. If you have any questions or feedback, please feel free to let us know in the comment section.

If you have any questions or feedback, feel free to let us know in the comment section.

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