Linear Algebra Examples

$[\begin{array}{c} a \\ b \end{array}] = [\begin{array}{c} d & g \\ h & j \end{array}] [\begin{array}{c} a \\ b \end{array}] + [\begin{array}{c} A^{- 1} \\ 0 \end{array}] q i$

Step 1

Simplify each term.

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Step 1.1

Multiply $[\begin{array}{c} d & g \\ h & j \end{array}] [\begin{array}{c} a \\ b \end{array}]$ .

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Step 1.1.1

Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. In this case, the first matrix is $2 \times 2$ and the second matrix is $2 \times 1$ .

Step 1.1.2

Multiply each row in the first matrix by each column in the second matrix.

$[\begin{array}{c} a \\ b \end{array}] = [\begin{array}{c} d a + g b \\ h a + j b \end{array}] + [\begin{array}{c} A^{- 1} \\ 0 \end{array}] q i$

Step 1.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$[\begin{array}{c} a \\ b \end{array}] = [\begin{array}{c} d a + g b \\ h a + j b \end{array}] + [\begin{array}{c} \frac{1}{A} \\ 0 \end{array}] q i$

Step 2

Reorder factors in $[\begin{array}{c} a \\ b \end{array}] = [\begin{array}{c} d a + g b \\ h a + j b \end{array}] + [\begin{array}{c} \frac{1}{A} \\ 0 \end{array}] q i$ .

$[\begin{array}{c} a \\ b \end{array}] = [\begin{array}{c} d a + g b \\ h a + j b \end{array}] + q i [\begin{array}{c} \frac{1}{A} \\ 0 \end{array}]$

Step 3

Simplify.

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Step 3.1

Simplify each term.

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Step 3.1.1

Multiply $q i$ by each element of the matrix.

$[\begin{array}{c} a \\ b \end{array}] = [\begin{array}{c} d a + g b \\ h a + j b \end{array}] + [\begin{array}{c} q i \frac{1}{A} \\ q i \cdot 0 \end{array}]$

Step 3.1.2

Simplify each element in the matrix.

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Step 3.1.2.1

Multiply $q i \frac{1}{A}$ .

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Step 3.1.2.1.1

Combine $\frac{1}{A}$ and $q$ .

$[\begin{array}{c} a \\ b \end{array}] = [\begin{array}{c} d a + g b \\ h a + j b \end{array}] + [\begin{array}{c} \frac{q}{A} i \\ q i \cdot 0 \end{array}]$

Step 3.1.2.1.2

Combine $\frac{q}{A}$ and $i$ .

$[\begin{array}{c} a \\ b \end{array}] = [\begin{array}{c} d a + g b \\ h a + j b \end{array}] + [\begin{array}{c} \frac{q i}{A} \\ q i \cdot 0 \end{array}]$

Step 3.1.2.2

Multiply $0$ by $q$ .

$[\begin{array}{c} a \\ b \end{array}] = [\begin{array}{c} d a + g b \\ h a + j b \end{array}] + [\begin{array}{c} \frac{q i}{A} \\ 0 i \end{array}]$

Step 3.1.2.3

Multiply $0$ by $i$ .

$[\begin{array}{c} a \\ b \end{array}] = [\begin{array}{c} d a + g b \\ h a + j b \end{array}] + [\begin{array}{c} \frac{q i}{A} \\ 0 \end{array}]$

Step 3.2

Add the corresponding elements.

$[\begin{array}{c} a \\ b \end{array}] = [\begin{array}{c} d a + g b + \frac{q i}{A} \\ h a + j b + 0 \end{array}]$

Step 3.3

Add $h a + j b$ and $0$ .

$[\begin{array}{c} a \\ b \end{array}] = [\begin{array}{c} d a + g b + \frac{q i}{A} \\ h a + j b \end{array}]$

Step 4

Move all terms containing a variable to the left side.

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Step 4.1

Move all terms containing variables to the left side of the equation.

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Step 4.1.1

Subtract $[\begin{array}{c} d a + g b + \frac{q i}{A} \\ h a + j b \end{array}]$ from both sides of the equation.

$[\begin{array}{c} a \\ b \end{array}] - [\begin{array}{c} d a + g b + \frac{q i}{A} \\ h a + j b \end{array}] = 0$

Step 4.1.2

Subtract the corresponding elements.

$[\begin{array}{c} a - (d a + g b + \frac{q i}{A}) \\ b - (h a + j b) \end{array}] = 0$

Step 4.1.3

Simplify each element.

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Step 4.1.3.1

Simplify each term.

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Step 4.1.3.1.1

Apply the distributive property.

$[\begin{array}{c} a - (d a) - (g b) - \frac{q i}{A} \\ b - (h a + j b) \end{array}] = 0$

Step 4.1.3.1.2

Remove parentheses.

$[\begin{array}{c} a - d a - g b - \frac{q i}{A} \\ b - (h a + j b) \end{array}] = 0$

Step 4.1.3.2

Simplify each term.

$[\begin{array}{c} a - d a - g b - \frac{q i}{A} \\ b - h a - j b \end{array}] = 0$