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SAT Math › Matrices

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If $\begin{bmatrix} a&b\ c&d \end{bmatrix} + \begin{bmatrix} 3&14\ 12&4 \end{bmatrix} = \begin{bmatrix} 19&2\ 4&5 \end{bmatrix}$ , what is ?

Explanation

You can treat matrices just like you treat other members of an equation. Therefore, you can subtract the matrix

$\begin{bmatrix} 3&14\ 12&4 \end{bmatrix}$

from both sides of the equation. This gives you:

$\begin{bmatrix} a&b\ c&d \end{bmatrix}=\begin{bmatrix} 19&2\ 4&5 \end{bmatrix}-\begin{bmatrix} 3&14\ 12&4 \end{bmatrix}$

Now, matrix subtraction is simple. You merely subtract each element, matching the correlative spaces with each other:

$\begin{bmatrix} a&b\ c&d \end{bmatrix} = \begin{bmatrix} 19-3&2-14\ 4-12&5-4 \end{bmatrix}$

Then, you simplify:

$\begin{bmatrix} 19-3&2-14\ 4-12&5-4 \end{bmatrix} = \begin{bmatrix} 16&-12\ -8&1 \end{bmatrix}$

Therefore,

If $\begin{bmatrix} a&b\ c&d \end{bmatrix} + \begin{bmatrix} 3&14\ 12&4 \end{bmatrix} = \begin{bmatrix} 19&2\ 4&5 \end{bmatrix}$ , what is ?

Explanation

You can treat matrices just like you treat other members of an equation. Therefore, you can subtract the matrix

$\begin{bmatrix} 3&14\ 12&4 \end{bmatrix}$

from both sides of the equation. This gives you:

$\begin{bmatrix} a&b\ c&d \end{bmatrix}=\begin{bmatrix} 19&2\ 4&5 \end{bmatrix}-\begin{bmatrix} 3&14\ 12&4 \end{bmatrix}$

Now, matrix subtraction is simple. You merely subtract each element, matching the correlative spaces with each other:

$\begin{bmatrix} a&b\ c&d \end{bmatrix} = \begin{bmatrix} 19-3&2-14\ 4-12&5-4 \end{bmatrix}$

Then, you simplify:

$\begin{bmatrix} 19-3&2-14\ 4-12&5-4 \end{bmatrix} = \begin{bmatrix} 16&-12\ -8&1 \end{bmatrix}$

Therefore,

Let $A = \begin{bmatrix} 6&4 \ 8&d \end{bmatrix}$

Which of the following values of makes a matrix without an inverse?

$d = 5 \frac{1}{3}$

$d = 5 \frac{1}{2}$

None of these

Explanation

A matrix $A = \begin{bmatrix} a& b\ c& d \end{bmatrix}$ lacks an inverse if and only if its determinant is equal to zero. The determinant of $A = \begin{bmatrix} 6&4 \ 8&d \end{bmatrix}$ is

$D = 6\cdot d - 4 \cdot 8 = 6d - 32$ .

We seek the value of that sets this quantity equal to 0. Setting it as such then solving for :

$6d \div 6 = 32 \div 6$

$d = 5 \frac{1}{3}$ ,

the correct response.

$5\begin{bmatrix} a \ b \end{bmatrix} = 14\begin{bmatrix} 20 \ 12 \end{bmatrix}$

What is ?

$\frac{448}{5}$

$\frac{41}{5}$

$\frac{191}{4}$

$\frac{148}{5}$

Explanation

You can begin by treating this equation just like it was:

That is, you can divide both sides by :

$\begin{bmatrix} a \ b \end{bmatrix} = \frac{14}{5}\begin{bmatrix} 20 \ 12 \end{bmatrix}$

Now, for scalar multiplication of matrices, you merely need to multiply the scalar by each component:

$\begin{bmatrix} a \ b \end{bmatrix} = \begin{bmatrix} \frac{14}{5}*20 \ \frac{14}{5}*12 \end{bmatrix}$

Then, simplify:

$\begin{bmatrix} a \ b \end{bmatrix} = \begin{bmatrix} 56 \ \frac{168}{5} \end{bmatrix}$

Therefore, $a+b=56+\frac{168}{5}=\frac{280}{5}+\frac{168}{5}=\frac{448}{5}$

$5\begin{bmatrix} a \ b \end{bmatrix} = 14\begin{bmatrix} 20 \ 12 \end{bmatrix}$

What is ?

$\frac{448}{5}$

$\frac{41}{5}$

$\frac{191}{4}$

$\frac{148}{5}$

Explanation

You can begin by treating this equation just like it was:

That is, you can divide both sides by :

$\begin{bmatrix} a \ b \end{bmatrix} = \frac{14}{5}\begin{bmatrix} 20 \ 12 \end{bmatrix}$

Now, for scalar multiplication of matrices, you merely need to multiply the scalar by each component:

$\begin{bmatrix} a \ b \end{bmatrix} = \begin{bmatrix} \frac{14}{5}*20 \ \frac{14}{5}*12 \end{bmatrix}$

Then, simplify:

$\begin{bmatrix} a \ b \end{bmatrix} = \begin{bmatrix} 56 \ \frac{168}{5} \end{bmatrix}$

Therefore, $a+b=56+\frac{168}{5}=\frac{280}{5}+\frac{168}{5}=\frac{448}{5}$

Define $A = \begin{bmatrix} 5& 10\ -6 & -12 \end{bmatrix}$ .

Give $A^{-1}$ .

$A^{-1}$ is not defined.

$A^{-1} = \begin{bmatrix} 5& -10\ 6 & -12 \end{bmatrix}$

$A^{-1} = \begin{bmatrix} -12& -10\ 6 & -5\end{bmatrix}$

$A^{-1} = \begin{bmatrix} -\frac{1}{10}& - \frac{1}{12}\ \frac{1}{20} & -\frac{1}{24}\end{bmatrix}$

$A^{-1} = \begin{bmatrix} -\frac{1}{24}& - \frac{1}{12}\ \frac{1}{20} & -\frac{1}{10 }\end{bmatrix}$

Explanation

The inverse of a 2 x 2 matrix $\begin{bmatrix} A& B\ C& D \end{bmatrix}$ , if it exists, is the matrix

$\frac{\begin{bmatrix} D& -B\ -C&A \end{bmatrix}}{\begin{vmatrix} A& B\ C& D \end{vmatrix}}= \frac{\begin{bmatrix} D& -B\ -C&A \end{bmatrix}}{AD - BC}$

First, we need to establish that the inverse is defined, which it is if and only if determinant .

Set , and check:

$AD - BC = 5 \cdot \left (-12 \right ) - 10 \cdot \left ( -6\right ) =-60 - (-60 )= -60 + 60 = 0$

The determinant is equal to 0, so does not have an inverse.

$\textup{Let }A = \begin{bmatrix} -2 &x \ 5 & x- 7 \end{bmatrix}$ .

$\textup{Which of the following values of }x\textup{ makes a matrix without an inverse?}$

$A \textup{ has an inverse regardless of the value of }x$

Explanation

A matrix $A = \begin{bmatrix} a& b\ c& d \end{bmatrix}$ lacks an inverse if and only if its determinant is equal to zero. The determinant of $A = \begin{bmatrix} -2 &x \ 5 & x- 7 \end{bmatrix}$ is

Set this equal to 0 and solve for :

$-7x\div (-7) = - 14 \div (-7)$

the correct response.

Let equal the following:

$A = \begin{bmatrix} 4 &x \ x&-64\end{bmatrix}$ .

Which of the following real values of makes a matrix without an inverse?

has an inverse for all real values of

There are two such values: or

There is one such value:

Explanation

A matrix $A = \begin{bmatrix} a& b\ c& d \end{bmatrix}$ lacks an inverse if and only if its determinant is equal to zero. The determinant of $A = \begin{bmatrix} 4 &x \ x&-64\end{bmatrix}$ is

$D = 4 \cdot 64 - x (-x) = 256 +x^{2}$ , so

$256 +x^{2} = 0$

$256 +x^{2} - 256 = 0 - 256$

$x^{2} = - 256$

Since the square of all real numbers is nonnegative, this equation has no real solution. It follows that the determinant cannot be 0 for any real value of , and that must have an inverse for all real .

If $\frac{1}{2}$\bigl(\begin{smallmatrix} 2a\4b \end{smallmatrix} \bigr)$=$\bigl(\begin{smallmatrix} 53 \ 98 \end{smallmatrix} \bigr)$$ , what is ?

$\frac{103}{2}$

$\frac{99}{2}$

Explanation

Begin by distributing the fraction through the matrix on the left side of the equation. This will simplify the contents, given that they are factors of :

$\frac{1}{2}\begin{bmatrix} 2a\4b \end{bmatrix}=\begin{bmatrix} a\2b \end{bmatrix}$

Now, this means that your equation looks like:
$\begin{bmatrix} a\2b \end{bmatrix}=\begin{bmatrix} 53 \ 98 \end{bmatrix}$

This simply means:

and

Therefore,

If $\frac{1}{2}$\bigl(\begin{smallmatrix} 2a\4b \end{smallmatrix} \bigr)$=$\bigl(\begin{smallmatrix} 53 \ 98 \end{smallmatrix} \bigr)$$ , what is ?

$\frac{103}{2}$

$\frac{99}{2}$

Explanation

Begin by distributing the fraction through the matrix on the left side of the equation. This will simplify the contents, given that they are factors of :

$\frac{1}{2}\begin{bmatrix} 2a\4b \end{bmatrix}=\begin{bmatrix} a\2b \end{bmatrix}$

Now, this means that your equation looks like:
$\begin{bmatrix} a\2b \end{bmatrix}=\begin{bmatrix} 53 \ 98 \end{bmatrix}$

This simply means:

and

Therefore,

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