Matrices

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SAT Subject Test in Math II › 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 ?

CORRECT

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,

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

What is ?

$\frac{448}{5}$

CORRECT

$\frac{41}{5}$

$\frac{191}{4}$

$\frac{148}{5}$

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

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

CORRECT

$\frac{103}{2}$

$\frac{99}{2}$

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,

Let $A = \begin{bmatrix} 5& -4\ -2& -7 \end{bmatrix}$ and be the 2 x 2 identity matrix.

Let .

Which of the following is equal to ?

$B = \begin{bmatrix} 4& -4 \ -2& -8\end{bmatrix}$

CORRECT

$B = \begin{bmatrix} 4& -5 \ -3& -8\end{bmatrix}$

$B = \begin{bmatrix} 4& -3 \ -1& -8\end{bmatrix}$

$B = \begin{bmatrix} 5& -5 \ -3& -7\end{bmatrix}$

$B = \begin{bmatrix} 3& -5 \ -3& -9\end{bmatrix}$

Explanation

The 2 x 2 identity matrix is $I =\begin{bmatrix} 1& 0\ 0& 1 \end{bmatrix}$ .

, or, equivalently,

$B = \begin{bmatrix} 5& -4\ -2& -7 \end{bmatrix} - \begin{bmatrix} 1& 0\ 0& 1 \end{bmatrix}$

Subtract the elements in the corresponding positions:

$B = \begin{bmatrix} 5-1& -4-0\ -2-0& -7 -1\end{bmatrix}$

$B = \begin{bmatrix} 4& -4 \ -2& -8\end{bmatrix}$

Simplify:

$\begin{bmatrix} 13 & 4\ 71 &3 \end{bmatrix} + 4\begin{bmatrix} 5 & 3\ 11 &22 \end{bmatrix}$

$\begin{bmatrix} 33 & 16\ 115 & 91\end{bmatrix}$

CORRECT

$\begin{bmatrix} 49\ 206\end{bmatrix}$

$\begin{bmatrix} 49 & 206\end{bmatrix}$

$\begin{bmatrix} 20 & 12\ 44 &88 \end{bmatrix}$

$\begin{bmatrix} 31 & 93\ 3 &21 \end{bmatrix}$

Explanation

Scalar multiplication and addition of matrices are both very easy. Just like regular scalar values, you do multiplication first:

$\begin{bmatrix} 13 & 4\ 71 &3 \end{bmatrix} + 4\begin{bmatrix} 5 & 3\ 11 &22 \end{bmatrix}= \begin{bmatrix} 13 & 4\ 71 &3 \end{bmatrix}+\begin{bmatrix} 20 & 12\ 44 &88 \end{bmatrix}$

The addition of matrices is very easy. You merely need to add them directly together, correlating the spaces directly.

$\begin{bmatrix} 13 & 4\ 71 &3 \end{bmatrix} + \begin{bmatrix} 20 & 12\ 44 &88 \end{bmatrix} = \begin{bmatrix} 13 +20 & 4+12\ 71+44 &3 +88\end{bmatrix}$

$\begin{bmatrix} 13 +20 & 4+12\ 71+44 &3 +88\end{bmatrix}= \begin{bmatrix} 33 & 16\ 115 & 91\end{bmatrix}$

$A = \begin{bmatrix} x+8& 7\ 5& x - 7 \end{bmatrix}$

$B = \begin{bmatrix} 4&x-6 \ x-7 & 2 \end{bmatrix}$

Calculate:

$\begin{bmatrix} x+12& x+1\ x-2 & x - 5 \end{bmatrix}$

CORRECT

$\begin{bmatrix} x+12& x+1\ x-2 & x - 9 \end{bmatrix}$

$\begin{bmatrix} x+12& x+1\ x-12 & x - 9 \end{bmatrix}$

$\begin{bmatrix} x+12& x-13\ x-12 & x - 9 \end{bmatrix}$

$\begin{bmatrix} x+12& x+1\ x-12 & x - 9 \end{bmatrix}$

Explanation

To add two matrices, add the elements in corresponding positions:

$A + B = \begin{bmatrix} x+8& 7\ 5& x - 7 \end{bmatrix} + \begin{bmatrix} 4&x-6 \ x-7 & 2 \end{bmatrix}$

$= \begin{bmatrix} x+8 + 4& 7+ \left ( x-6 \right )\ 5+\left ( x - 7 \right ) & x - 7 +2 \end{bmatrix}$

$= \begin{bmatrix} x+12& x+1\ x-2 & x - 5 \end{bmatrix}$

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

Give $A^{-1}$ .

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

CORRECT

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

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

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

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

Explanation

has three rows and two columns; since the number of rows is not equal to the number of columns, is not a square matrix, and, therefore, it does not have an inverse.

$A = \begin{bmatrix} 8& x\ x& 4 \end{bmatrix}$

For which of the following real values of does have determinant of sixteen?

CORRECT

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 $A = \begin{bmatrix} 8& x\ x& 4 \end{bmatrix}$ is

$D = 8 \cdot 4 - x \cdot x = 32 - x^{2}$

We seek the value of that sets this quantity equal to 16. Setting it as such then solving for :

$32 - x^{2} = 16$

$32 - x^{2} - 32 = 16 - 32$

$- x^{2} = -16$

$x^{2} = 16$

$x = \pm 4$

Therefore, either or .

$A = \begin{bmatrix} 0&0& 0 & 3\ 0& 0 & 3& 0\ 0& 3& 0& 0 \3& 0&0& 0 \end{bmatrix}$

Evaluate $A^{2}$ .

$A ^{2}= \begin{bmatrix} 9& 0&0& 0\ 0& 9& 0& 0\ 0& 0 & 9& 0\ 0&0& 0 & 9 \end{bmatrix}$

CORRECT

$A ^{2}= \begin{bmatrix}0&0& 0 & 9\0& 0 & 9& 0 \ 0& 9& 0& 0 \ 9 & 0&0& 0 \end{bmatrix}$

$A ^{2}= \begin{bmatrix} 0&0& 0 & 0 \ 0&0& 0 & 0 \ 0&0& 0 & 0\ 0&0& 0 & 0 \end{bmatrix}$

$A ^{2}= \begin{bmatrix}0& 0 & 9& 0 \ 0& 9& 0& 0 \ 9 & 0&0& 0 \0&0& 0 & 9 \end{bmatrix}$

$A ^{2}= \begin{bmatrix}9 & 0&0& 0 \0&0& 0 & 9 \ 0& 0 & 9& 0 \ 0& 9& 0& 0 \end{bmatrix}$

Explanation

$A^{2} = A \cdot A = \begin{bmatrix} 0&0& 0 & 3\ 0& 0 & 3& 0\ 0& 3& 0& 0 \3& 0&0& 0 \end{bmatrix}\begin{bmatrix} 0&0& 0 & 3\ 0& 0 & 3& 0\ 0& 3& 0& 0 \3& 0&0& 0 \end{bmatrix}$

The element in row , column , of $A^{2}$ can be found by multiplying row of by row of - that is, by multiplying elements in corresponding positions and adding the products. Therefore,

$A^{2} = \begin{bmatrix} 0 \cdot 0 + 0 \cdot 0 +0 \cdot 0 +3 \cdot 3 &0 \cdot 0 +0 \cdot 0 + 0 \cdot 3 +3 \cdot 0 &0 \cdot 0 + 0 \cdot 3 +0 \cdot 0 +3 \cdot 0&0 \cdot 3 + 0 \cdot 0 +0 \cdot 0 +3 \cdot 0 \ 0 \cdot 0 + 0 \cdot 0 +3 \cdot 0 +0 \cdot 3 &0 \cdot 0 +0 \cdot 0 + 3 \cdot 3 +0 \cdot 0 &0 \cdot 0 + 0 \cdot 3 +3 \cdot 0 +0 \cdot 0&0 \cdot 3 + 0 \cdot 0 +3 \cdot 0 +0 \cdot 0\ 0 \cdot 0 + 3 \cdot 0 +0 \cdot 0 +0 \cdot 3 &0 \cdot 0 +3 \cdot 0 + 0 \cdot 3 +0 \cdot 0 &0 \cdot 0 + 3 \cdot 3 +0 \cdot 0 +0\cdot 0&0 \cdot 3 + 3 \cdot 0 +0 \cdot 0 +0 \cdot 0\3 \cdot 0 + 0 \cdot 0 +0 \cdot 0 +0 \cdot 3 &3 \cdot 0 +0 \cdot 0 + 0 \cdot 3 +0 \cdot 0 &3 \cdot 0 + 0 \cdot 3 +0 \cdot 0 +0 \cdot 0&3 \cdot 3 + 0 \cdot 0 +0 \cdot 0 +0\cdot 0 \end{bmatrix}$

$= \begin{bmatrix} 0+0+0+9 & 0 +0+0+0 & 0 +0+0+0 & 0 +0+0+0 \ 0 +0+0+0 & 0+0+9+0 & 0 +0+0+0& 0 +0+0+0 \ 0 +0+0+0 & 0 +0+0+0& 0+9+0+0 & 0 +0+0+0 \0 +0+0+0 & 0 +0+0+0& 0 +0+0+0 & 9+0+0+0 \end{bmatrix}$

$= \begin{bmatrix} 9& 0&0& 0\ 0& 9& 0& 0\ 0& 0 & 9& 0\ 0&0& 0 & 9 \end{bmatrix}$

Multiply:

$\begin{bmatrix} x& 8\ -6& t \end{bmatrix}\begin{bmatrix} t& x \end{bmatrix}$

The matrices cannot be multiplied.

CORRECT

$\begin{bmatrix} tx +8x\xt-6t \end{bmatrix}$

$\begin{bmatrix} tx +6x\xt-8t \end{bmatrix}$

$\begin{bmatrix} tx +8x&xt-6t \end{bmatrix}$

$\begin{bmatrix} tx +6x& xt-8t \end{bmatrix}$

Explanation

Two matrices can be multiplied if and only if the number of columns in the first matrix and the number of rows in the second are equal. The first matrix has two columns; the second matrix has one row. This violates the condition, so they cannot be multiplied in this order.

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