1. Let $$ \mathrm{A}=\left(\begin{array}{ll}1 & 1 \ 0 & 2\end{array} ight), \mathrm{B}=\left(\begin{array}{cc}1 & 1 \ -1 & 0\end{array} ight) $$ and $$ \mathrm{C}=\left(\begin{array}{ll}2 & 1 \ 1 & 3\end{array} ight) $$. Then find: $$ \begin{array}{lll} ext { a. } \mathrm{AB} & ext { b. } \mathrm{BA} & ext { c. }(\mathrm{AB}) \mathrm{C} \ ext { d. } \mathrm{A}(\mathrm{BC}) & ext { e. } \mathrm{I}_{2} \mathrm{~A} & ext { f. } \mathrm{BI}_{2}\end{array} $$

Question

1. Let $$ \mathrm{A}=\left(\begin{array}{ll}1 & 1 \ 0 & 2\end{array}ight), \mathrm{B}=\left(\begin{array}{cc}1 & 1 \ -1 & 0\end{array}ight) $$ and $$ \mathrm{C}=\left(\begin{array}{ll}2 & 1 \ 1 & 3\end{array}ight) $$. Then find: $$ \begin{array}{lll}	ext { a. } \mathrm{AB} & 	ext { b. } \mathrm{BA} & 	ext { c. }(\mathrm{AB}) \mathrm{C} \ 	ext { d. } \mathrm{A}(\mathrm{BC}) & 	ext { e. } \mathrm{I}_{2} \mathrm{~A} & 	ext { f. } \mathrm{BI}_{2}\end{array} $$

UpStudy AI · Accepted Answer

Let's compute each part step-by-step using the given matrices:

$$
\mathrm{A} = \begin{pmatrix} 1 & 1 \ 0 & 2 \end{pmatrix}, \quad
\mathrm{B} = \begin{pmatrix} 1 & 1 \ -1 & 0 \end{pmatrix}, \quad
\mathrm{C} = \begin{pmatrix} 2 & 1 \ 1 & 3 \end{pmatrix}, \quad
\mathrm{I}_2 = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}
$$

### a. $$ \mathrm{AB} $$

Multiply matrix $$\mathrm{A}$$ by matrix $$\mathrm{B}$$:

$$
\mathrm{AB} = \begin{pmatrix} 1 & 1 \ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 1 \ -1 & 0 \end{pmatrix} = \begin{pmatrix}
1 \cdot 1 + 1 \cdot (-1) & 1 \cdot 1 + 1 \cdot 0 \
0 \cdot 1 + 2 \cdot (-1) & 0 \cdot 1 + 2 \cdot 0
\end{pmatrix} = \begin{pmatrix}
0 & 1 \
-2 & 0
\end{pmatrix}
$$

### b. $$ \mathrm{BA} $$

Multiply matrix $$\mathrm{B}$$ by matrix $$\mathrm{A}$$:

$$
\mathrm{BA} = \begin{pmatrix} 1 & 1 \ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 1 \ 0 & 2 \end{pmatrix} = \begin{pmatrix}
1 \cdot 1 + 1 \cdot 0 & 1 \cdot 1 + 1 \cdot 2 \
-1 \cdot 1 + 0 \cdot 0 & -1 \cdot 1 + 0 \cdot 2
\end{pmatrix} = \begin{pmatrix}
1 & 3 \
-1 & -1
\end{pmatrix}
$$

### c. $$ (\mathrm{AB})\mathrm{C} $$

Using $$ \mathrm{AB} = \begin{pmatrix} 0 & 1 \ -2 & 0 \end{pmatrix} $$:

$$
(\mathrm{AB})\mathrm{C} = \begin{pmatrix} 0 & 1 \ -2 & 0 \end{pmatrix} \begin{pmatrix} 2 & 1 \ 1 & 3 \end{pmatrix} = \begin{pmatrix}
0 \cdot 2 + 1 \cdot 1 & 0 \cdot 1 + 1 \cdot 3 \
-2 \cdot 2 + 0 \cdot 1 & -2 \cdot 1 + 0 \cdot 3
\end{pmatrix} = \begin{pmatrix}
1 & 3 \
-4 & -2
\end{pmatrix}
$$

### d. $$ \mathrm{A}(\mathrm{BC}) $$

First, compute $$ \mathrm{BC} $$:

$$
\mathrm{BC} = \begin{pmatrix} 1 & 1 \ -1 & 0 \end{pmatrix} \begin{pmatrix} 2 & 1 \ 1 & 3 \end{pmatrix} = \begin{pmatrix}
1 \cdot 2 + 1 \cdot 1 & 1 \cdot 1 + 1 \cdot 3 \
-1 \cdot 2 + 0 \cdot 1 & -1 \cdot 1 + 0 \cdot 3
\end{pmatrix} = \begin{pmatrix}
3 & 4 \
-2 & -1
\end{pmatrix}
$$

Now, multiply by $$\mathrm{A}$$:

$$
\mathrm{A}(\mathrm{BC}) = \begin{pmatrix} 1 & 1 \ 0 & 2 \end{pmatrix} \begin{pmatrix} 3 & 4 \ -2 & -1 \end{pmatrix} = \begin{pmatrix}
1 \cdot 3 + 1 \cdot (-2) & 1 \cdot 4 + 1 \cdot (-1) \
0 \cdot 3 + 2 \cdot (-2) & 0 \cdot 4 + 2 \cdot (-1)
\end{pmatrix} = \begin{pmatrix}
1 & 3 \
-4 & -2
\end{pmatrix}
$$

### e. $$ \mathrm{I}_2 \mathrm{A} $$

Multiply the identity matrix by $$\mathrm{A}$$:

$$
\mathrm{I}_2 \mathrm{A} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \ 0 & 2 \end{pmatrix} = \begin{pmatrix}
1 \cdot 1 + 0 \cdot 0 & 1 \cdot 1 + 0 \cdot 2 \
0 \cdot 1 + 1 \cdot 0 & 0 \cdot 1 + 1 \cdot 2
\end{pmatrix} = \begin{pmatrix}
1 & 1 \
0 & 2
\end{pmatrix} = \mathrm{A}
$$

### f. $$ \mathrm{B}\mathrm{I}_2 $$

Multiply $$\mathrm{B}$$ by the identity matrix:

$$
\mathrm{B}\mathrm{I}_2 = \begin{pmatrix} 1 & 1 \ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = \begin{pmatrix}
1 \cdot 1 + 1 \cdot 0 & 1 \cdot 0 + 1 \cdot 1 \
-1 \cdot 1 + 0 \cdot 0 & -1 \cdot 0 + 0 \cdot 1
\end{pmatrix} = \begin{pmatrix}
1 & 1 \
-1 & 0
\end{pmatrix} = \mathrm{B}
$$

### Summary of Results

a. $$ \mathrm{AB} = \begin{pmatrix} 0 & 1 \ -2 & 0 \end{pmatrix} $$

b. $$ \mathrm{BA} = \begin{pmatrix} 1 & 3 \ -1 & -1 \end{pmatrix} $$

c. $$ (\mathrm{AB})\mathrm{C} = \begin{pmatrix} 1 & 3 \ -4 & -2 \end{pmatrix} $$

d. $$ \mathrm{A}(\mathrm{BC}) = \begin{pmatrix} 1 & 3 \ -4 & -2 \end{pmatrix} $$

e. $$ \mathrm{I}_2 \mathrm{A} = \mathrm{A} = \begin{pmatrix} 1 & 1 \ 0 & 2 \end{pmatrix} $$

f. $$ \mathrm{B}\mathrm{I}_2 = \mathrm{B} = \begin{pmatrix} 1 & 1 \ -1 & 0 \end{pmatrix} $$

Answer for part a:

$$
\mathrm{AB} = \begin{pmatrix} 0 & 1 \ -2 & 0 \end{pmatrix}
$$
a. $$ \mathrm{AB} = \begin{pmatrix} 0 & 1 \ -2 & 0 \end{pmatrix} $$

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