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Common Core: High School - Algebra : High School: Algebra

Study concepts, example questions & explanations for Common Core: High School - Algebra

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All Common Core: High School - Algebra Resources

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Example Questions

Example Question #601 : High School: Algebra

Does the following matrix have an inverse?

$A=\begin{bmatrix} 46 & -46 \\ 68 & -68 \end{bmatrix}$ $\displaystyle A=\begin{bmatrix} 46 & -46 \\ 68 & -68 \end{bmatrix}$

Possible Answers:

Yes

Correct answer:

Explanation:

In order to determine if a matrix has an inverse is to calculate the determinant.

$\det(A)=a\cdotc-b\cdot d$ $\displaystyle \det(A)=a\cdotc-b\cdot d$

Where $\uptext{a}, \uptext{b}, \uptext{c}$ $\displaystyle \uptext{a}, \uptext{b}, \uptext{c}$, and $\uptext{d}$ $\displaystyle \uptext{d}$ correspond to the entries in the following matrix.

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ $\displaystyle \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

If the determinant is not equal to zero, an inverse exists, and if it's equal to zero, no inverse exists.

Now let's calculate the determinant.

$\det(A)= 46 \cdot 68 - -46 \cdot -68$ $\displaystyle \det(A)= 46 \cdot 68 - -46 \cdot -68$

$\det(A)= 3128 - 3128$ $\displaystyle \det(A)= 3128 - 3128$

$\det(A)= 0$ $\displaystyle \det(A)= 0$

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Example Question #602 : High School: Algebra

Does the following matrix have an inverse?

$A=\begin{bmatrix} -59 & 43 \\ 36 & -89 \end{bmatrix}$ $\displaystyle A=\begin{bmatrix} -59 & 43 \\ 36 & -89 \end{bmatrix}$

Possible Answers:

Yes

Correct answer:

Yes

Explanation:

In order to determine if a matrix has an inverse is to calculate the determinant.

$\det(A)=a\cdotc-b\cdot d$ $\displaystyle \det(A)=a\cdotc-b\cdot d$

Where $\uptext{a}, \uptext{b}, \uptext{c},$ $\displaystyle \uptext{a}, \uptext{b}, \uptext{c},$ and $\uptext{d}$ $\displaystyle \uptext{d}$ correspond to the entries in the following matrix.

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ $\displaystyle \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

If the determinant is not equal to zero, an inverse exists, and if it's equal to zero, no inverse exists.

Now let's calculate the determinant.

$\det(A)= -59 \cdot 36 - 43 \cdot -89$ $\displaystyle \det(A)= -59 \cdot 36 - 43 \cdot -89$

$\det(A)= -2124 - -3827$ $\displaystyle \det(A)= -2124 - -3827$

$\det(A)= 1703$ $\displaystyle \det(A)= 1703$

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Example Question #603 : High School: Algebra

Does the following matrix have an inverse?

$A=\begin{bmatrix} -64 & -60 \\ -69 & 21 \end{bmatrix}$ $\displaystyle A=\begin{bmatrix} -64 & -60 \\ -69 & 21 \end{bmatrix}$

Possible Answers:

Yes

Correct answer:

Yes

Explanation:

In order to determine if a matrix has an inverse is to calculate the determinant.

$\det(A)=a\cdotc-b\cdot d$ $\displaystyle \det(A)=a\cdotc-b\cdot d$

Where $\uptext{a}, \uptext{b}, \uptext{c},$ $\displaystyle \uptext{a}, \uptext{b}, \uptext{c},$ and $\uptext{d}$ $\displaystyle \uptext{d}$ correspond to the entries in the following matrix.

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ $\displaystyle \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

If the determinant is not equal to zero, an inverse exists, and if it's equal to zero, no inverse exists.

Now let's calculate the determinant.

$\det(A)= -64 \cdot -69 - -60 \cdot 21$ $\displaystyle \det(A)= -64 \cdot -69 - -60 \cdot 21$

$\det(A)= 4416 - -1260$ $\displaystyle \det(A)= 4416 - -1260$

$\det(A)= 5676$ $\displaystyle \det(A)= 5676$

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Example Question #604 : High School: Algebra

Does the following matrix have an inverse?

$A=\begin{bmatrix} 0 & 0 \\ -93 & 93 \end{bmatrix}$ $\displaystyle A=\begin{bmatrix} 0 & 0 \\ -93 & 93 \end{bmatrix}$

Possible Answers:

Yes

Correct answer:

Explanation:

In order to determine if a matrix has an inverse is to calculate the determinant.

$\det(A)=a\cdotc-b\cdot d$ $\displaystyle \det(A)=a\cdotc-b\cdot d$

Where $\uptext{a}, \uptext{b}, \uptext{c},$ $\displaystyle \uptext{a}, \uptext{b}, \uptext{c},$ and $\uptext{d}$ $\displaystyle \uptext{d}$ correspond to the entries in the following matrix.

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ $\displaystyle \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

If the determinant is not equal to zero, an inverse exists, and if it's equal to zero, no inverse exists.

Now let's calculate the determinant.

$\det(A)= 0 \cdot -93 - 0 \cdot 93$ $\displaystyle \det(A)= 0 \cdot -93 - 0 \cdot 93$

$\det(A)= 0 - 0$ $\displaystyle \det(A)= 0 - 0$

$\det(A)= 0$ $\displaystyle \det(A)= 0$

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Example Question #605 : High School: Algebra

Does the following matrix have an inverse?

$A=\begin{bmatrix} 91 & -22 \\ 9 & 66 \end{bmatrix}$ $\displaystyle A=\begin{bmatrix} 91 & -22 \\ 9 & 66 \end{bmatrix}$

Possible Answers:

Yes

Correct answer:

Yes

Explanation:

In order to determine if a matrix has an inverse is to calculate the determinant.

$\det(A)=a\cdotc-b\cdot d$ $\displaystyle \det(A)=a\cdotc-b\cdot d$

Where $\uptext{a}, \uptext{b}, \uptext{c},$ $\displaystyle \uptext{a}, \uptext{b}, \uptext{c},$ and $\uptext{d}$ $\displaystyle \uptext{d}$ correspond to the entries in the following matrix.

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ $\displaystyle \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

If the determinant is not equal to zero, an inverse exists, and if it's equal to zero, no inverse exists.

Now let's calculate the determinant.

$\det(A)= 91 \cdot 9 - -22 \cdot 66$ $\displaystyle \det(A)= 91 \cdot 9 - -22 \cdot 66$

$\det(A)= 819 - -1452$ $\displaystyle \det(A)= 819 - -1452$

$\det(A)= 2271$ $\displaystyle \det(A)= 2271$

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Example Question #606 : High School: Algebra

Does the point $\left( -5 , 22 \right)$ $\displaystyle \left( -5 , 22 \right)$ exist on the line $\displaystyle y = x + 27?$

Possible Answers:

The point $\left( -5 , 22 \right)$ $\displaystyle \left( -5 , 22 \right)$ exists on the line.

The point $\left( -5 , 22 \right)$ $\displaystyle \left( -5 , 22 \right)$ does not exist on the line.

Correct answer:

The point $\left( -5 , 22 \right)$ $\displaystyle \left( -5 , 22 \right)$ exists on the line.

Explanation:

To see if this point exists on the line, we need to plug in the $\uptext{x}$ $\displaystyle \uptext{x}$ value into the equation and see if it equals the $\uptext{y}$ $\displaystyle \uptext{y}$ value.

$y = 1 \cdot -5 + 27$ $\displaystyle y = 1 \cdot -5 + 27$

$\displaystyle y = -5 + 27$

y = 22 $\displaystyle y = 22$

Since the $\displaystyle y$ values are equal, then the point $\left( -5 , 22 \right)$ $\displaystyle \left( -5 , 22 \right)$ exist on the line.

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Example Question #607 : High School: Algebra

Does the point $\left( -9 , 3 \right)$ $\displaystyle \left( -9 , 3 \right)$ exist on the line $\displaystyle y = - 24 x + 18?$

Possible Answers:

The point $\left( -9 , 3 \right)$ $\displaystyle \left( -9 , 3 \right)$ exists on the line.

The point $\left( -9 , 3 \right)$ $\displaystyle \left( -9 , 3 \right)$ does not exist on the line.

Correct answer:

The point $\left( -9 , 3 \right)$ $\displaystyle \left( -9 , 3 \right)$ does not exist on the line.

Explanation:

$y = -24 \cdot -9 + 18$ $\displaystyle y = -24 \cdot -9 + 18$

$\displaystyle y = 216 + 18$

y = 234 $\displaystyle y = 234$

Since the $\displaystyle y$ values are not equal, then the point $\left( -9 , 3 \right)$ $\displaystyle \left( -9 , 3 \right)$ does not exist on the line.

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Example Question #3 : Plotted Solutions And The Graph Of Two Variable Equations: Ccss.Math.Content.Hsa Rei.D.10

Does the point $\left( -11 , 60 \right)$ $\displaystyle \left( -11 , 60 \right)$ exist on the line $\displaystyle y = - 2 x + 38?$

Possible Answers:

The point $\left( -11 , 60 \right)$ $\displaystyle \left( -11 , 60 \right)$ does not exist on the line.

The point $\left( -11 , 60 \right)$ $\displaystyle \left( -11 , 60 \right)$ exists on the line.

Correct answer:

The point $\left( -11 , 60 \right)$ $\displaystyle \left( -11 , 60 \right)$ exists on the line.

Explanation:

$y = -2 \cdot -11 + 38$ $\displaystyle y = -2 \cdot -11 + 38$

$\displaystyle y = 22 + 38$

y = 60 $\displaystyle y = 60$

Since the y values are equal, then the point $\left( -11 , 60 \right)$ $\displaystyle \left( -11 , 60 \right)$ exist on the line.

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Example Question #4 : Plotted Solutions And The Graph Of Two Variable Equations: Ccss.Math.Content.Hsa Rei.D.10

Does the point $\left( -45 , 16 \right)$ $\displaystyle \left( -45 , 16 \right)$ exist on the line $\displaystyle y = 9 x + 45?$

Possible Answers:

The point $\left( -45 , 16 \right)$ $\displaystyle \left( -45 , 16 \right)$ does not exist on the line.

The point $\left( -45 , 16 \right)$ $\displaystyle \left( -45 , 16 \right)$ exists on the line.

Correct answer:

The point $\left( -45 , 16 \right)$ $\displaystyle \left( -45 , 16 \right)$ does not exist on the line.

Explanation:

$y = 9 \cdot -45 + 45$ $\displaystyle y = 9 \cdot -45 + 45$

$\displaystyle y = -405 + 45$

$\displaystyle y = -360$

Since the $\displaystyle y$ values are not equal, then the point $\left( -45 , 16 \right)$ $\displaystyle \left( -45 , 16 \right)$ does not exist on the line.

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Example Question #4 : Plotted Solutions And The Graph Of Two Variable Equations: Ccss.Math.Content.Hsa Rei.D.10

Does the point $\left( -27 , 44 \right)$ $\displaystyle \left( -27 , 44 \right)$ exist on the line $\displaystyle y = - x + 17?$

Possible Answers:

The point $\left( -27 , 44 \right)$ $\displaystyle \left( -27 , 44 \right)$ exists on the line.

The point $\left( -27 , 44 \right)$ $\displaystyle \left( -27 , 44 \right)$ does not exist on the line.

Correct answer:

The point $\left( -27 , 44 \right)$ $\displaystyle \left( -27 , 44 \right)$ exists on the line.

Explanation:

$y = -1 \cdot -27 + 17$ $\displaystyle y = -1 \cdot -27 + 17$

$\displaystyle y = 27 + 17$

y = 44 $\displaystyle y = 44$

Since the $\displaystyle y$ values are equal, then the point $\left( -27 , 44 \right)$ $\displaystyle \left( -27 , 44 \right)$ exist on the line.

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All Common Core: High School - Algebra Resources

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Common Core: High School - Algebra : High School: Algebra

Study concepts, example questions & explanations for Common Core: High School - Algebra

All Common Core: High School - Algebra Resources

Example Questions

Example Question #601 : High School: Algebra

Example Question #602 : High School: Algebra

Example Question #603 : High School: Algebra

Example Question #604 : High School: Algebra

Example Question #605 : High School: Algebra

Example Question #606 : High School: Algebra

Example Question #607 : High School: Algebra

Example Question #3 : Plotted Solutions And The Graph Of Two Variable Equations: Ccss.Math.Content.Hsa Rei.D.10

Example Question #4 : Plotted Solutions And The Graph Of Two Variable Equations: Ccss.Math.Content.Hsa Rei.D.10

Example Question #4 : Plotted Solutions And The Graph Of Two Variable Equations: Ccss.Math.Content.Hsa Rei.D.10

All Common Core: High School - Algebra Resources

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