QuestionCari nilai $x+y$ jika matriks songsang bagi $\left(\begin{array}{cc}1 & 4 \\ -3 & -7\end{array}\right)$ ialah $\frac{1}{8}\left(\begin{array}{rr}-7 & -4 \\ 3 & y\end{array}\right)$ .

Studdy Solution

STEP 1

Assumptions1. The given matrix is $\left(\begin{array}{cc}1 &4 \\ -3 & -7\end{array}\right)$ . The inverse of the given matrix is $\frac{1}{8}\left(\begin{array}{rr}-7 & -4 \\3 & y\end{array}\right)$ 3. We need to find the value of $x+y$ , where $x$ is the element at the second row and first column of the inverse matrix, and $y$ is the element at the second row and second column of the inverse matrix.

STEP 2

The formula for the inverse of a2x2 matrix is given by $\left(\begin{array}{cc}a & b \\ c & d\end{array}\right)^{-1} = \frac{1}{ad-bc}\left(\begin{array}{cc}d & -b \\ -c & a\end{array}\right)$

STEP 3

Now, plug in the values from the given matrix into the formula to find the inverse matrix.
$\left(\begin{array}{cc}1 & \\ -3 & -7\end{array}\right)^{-1} = \frac{1}{1(-7)-(-3)}\left(\begin{array}{cc}-7 & - \\3 &1\end{array}\right)$

STEP 4

Calculate the determinant $ad-bc$ .
$ad-bc =1(-7)-4(-3) = -7+12 =$

STEP 5

Substitute the determinant into the formula.
$\left(\begin{array}{cc}1 &4 \\ -3 & -7\end{array}\right)^{-1} = \frac{1}{5}\left(\begin{array}{cc}-7 & -4 \\3 &1\end{array}\right)$

STEP 6

Multiply the matrix by the scalar $\frac{1}{5}$ .
$\left(\begin{array}{cc}1 &4 \\ -3 & -\end{array}\right)^{-1} = \left(\begin{array}{cc}-\frac{}{5} & -\frac{4}{5} \\ \frac{3}{5} & \frac{1}{5}\end{array}\right)$

STEP 7

Now, compare this matrix with the given inverse matrix $\frac{1}{}\left(\begin{array}{rr}-7 & -4 \\3 & y\end{array}\right)$ .

STEP 8

By comparing the elements at the second row and second column of both matrices, we can find the value of $y$ .
$\frac{1}{5} = \frac{y}{8}$

STEP 9

olve the equation for $y$ .
$y =8*\frac{}{5} = \frac{8}{5}$

STEP 10

Now that we have the value of $y$ , we can find the value of $x+y$ .
$x+y =3 + \frac{8}{5}$

STEP 11

Calculate the value of $x+y$ .
$x+y =3 + \frac{8}{5} = \frac{23}{5}$ So, the value of $x+y$ is $\frac{23}{5}$ .

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QuestionCari nilai $x+y$ jika matriks songsang bagi $\left(\begin{array}{cc}1 & 4 \\ -3 & -7\end{array}\right)$ ialah $\frac{1}{8}\left(\begin{array}{rr}-7 & -4 \\ 3 & y\end{array}\right)$ .

Studdy Solution

STEP 1

STEP 2

The formula for the inverse of a2x2 matrix is given by $\left(\begin{array}{cc}a & b \\ c & d\end{array}\right)^{-1} = \frac{1}{ad-bc}\left(\begin{array}{cc}d & -b \\ -c & a\end{array}\right)$

STEP 3

Now, plug in the values from the given matrix into the formula to find the inverse matrix.
$\left(\begin{array}{cc}1 & \\ -3 & -7\end{array}\right)^{-1} = \frac{1}{1(-7)-(-3)}\left(\begin{array}{cc}-7 & - \\3 &1\end{array}\right)$

STEP 4

Calculate the determinant $ad-bc$ .
$ad-bc =1(-7)-4(-3) = -7+12 =$

STEP 5

Substitute the determinant into the formula.
$\left(\begin{array}{cc}1 &4 \\ -3 & -7\end{array}\right)^{-1} = \frac{1}{5}\left(\begin{array}{cc}-7 & -4 \\3 &1\end{array}\right)$

STEP 6

Multiply the matrix by the scalar $\frac{1}{5}$ .
$\left(\begin{array}{cc}1 &4 \\ -3 & -\end{array}\right)^{-1} = \left(\begin{array}{cc}-\frac{}{5} & -\frac{4}{5} \\ \frac{3}{5} & \frac{1}{5}\end{array}\right)$

STEP 7

Now, compare this matrix with the given inverse matrix $\frac{1}{}\left(\begin{array}{rr}-7 & -4 \\3 & y\end{array}\right)$ .

STEP 8

By comparing the elements at the second row and second column of both matrices, we can find the value of $y$ .
$\frac{1}{5} = \frac{y}{8}$

STEP 9

olve the equation for $y$ .
$y =8*\frac{}{5} = \frac{8}{5}$

STEP 10

Now that we have the value of $y$ , we can find the value of $x+y$ .
$x+y =3 + \frac{8}{5}$

STEP 11

Calculate the value of $x+y$ .
$x+y =3 + \frac{8}{5} = \frac{23}{5}$ So, the value of $x+y$ is $\frac{23}{5}$ .

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QuestionCari nilai x+yx+yx+y jika matriks songsang bagi (14−3−7)\left(\begin{array}{cc}1 & 4 \\ -3 & -7\end{array}\right)(1−3​4−7​) ialah 18(−7−43y)\frac{1}{8}\left(\begin{array}{rr}-7 & -4 \\ 3 & y\end{array}\right)81​(−73​−4y​).

Studdy Solution

STEP 1

STEP 2

The formula for the inverse of a2x2 matrix is given by(abcd)−1=1ad−bc(d−b−ca)\left(\begin{array}{cc}a & b \\ c & d\end{array}\right)^{-1} = \frac{1}{ad-bc}\left(\begin{array}{cc}d & -b \\ -c & a\end{array}\right)(ac​bd​)−1=ad−bc1​(d−c​−ba​)

STEP 3

STEP 4

Calculate the determinant ad−bcad-bcad−bc.ad−bc=1∗(−7)−4∗(−3)=−7+12=ad-bc =1*(-7)-4*(-3) = -7+12 =ad−bc=1∗(−7)−4∗(−3)=−7+12=

STEP 5

Substitute the determinant into the formula.(14−3−7)−1=15(−7−431)\left(\begin{array}{cc}1 &4 \\ -3 & -7\end{array}\right)^{-1} = \frac{1}{5}\left(\begin{array}{cc}-7 & -4 \\3 &1\end{array}\right)(1−3​4−7​)−1=51​(−73​−41​)

STEP 6

Multiply the matrix by the scalar 15\frac{1}{5}51​.(14−3−)−1=(−5−453515)\left(\begin{array}{cc}1 &4 \\ -3 & -\end{array}\right)^{-1} = \left(\begin{array}{cc}-\frac{}{5} & -\frac{4}{5} \\ \frac{3}{5} & \frac{1}{5}\end{array}\right)(1−3​4−​)−1=(−5​53​​−54​51​​)

STEP 7

Now, compare this matrix with the given inverse matrix 1(−7−43y)\frac{1}{}\left(\begin{array}{rr}-7 & -4 \\3 & y\end{array}\right)1​(−73​−4y​).

STEP 8

By comparing the elements at the second row and second column of both matrices, we can find the value of yyy.15=y8\frac{1}{5} = \frac{y}{8}51​=8y​

STEP 9

olve the equation for yyy.y=8∗5=85y =8*\frac{}{5} = \frac{8}{5}y=8∗5​=58​

STEP 10

Now that we have the value of yyy, we can find the value of x+yx+yx+y.x+y=3+85x+y =3 + \frac{8}{5}x+y=3+58​

STEP 11

Calculate the value of x+yx+yx+y.x+y=3+85=235x+y =3 + \frac{8}{5} = \frac{23}{5}x+y=3+58​=523​So, the value of x+yx+yx+y is 235\frac{23}{5}523​.

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QuestionCari nilai $x+y$ jika matriks songsang bagi $\left(\begin{array}{cc}1 & 4 \\ -3 & -7\end{array}\right)$ ialah $\frac{1}{8}\left(\begin{array}{rr}-7 & -4 \\ 3 & y\end{array}\right)$ .

The formula for the inverse of a2x2 matrix is given by $\left(\begin{array}{cc}a & b \\ c & d\end{array}\right)^{-1} = \frac{1}{ad-bc}\left(\begin{array}{cc}d & -b \\ -c & a\end{array}\right)$

Calculate the determinant $ad-bc$ .
$ad-bc =1(-7)-4(-3) = -7+12 =$

Substitute the determinant into the formula.
$\left(\begin{array}{cc}1 &4 \\ -3 & -7\end{array}\right)^{-1} = \frac{1}{5}\left(\begin{array}{cc}-7 & -4 \\3 &1\end{array}\right)$

Multiply the matrix by the scalar $\frac{1}{5}$ .
$\left(\begin{array}{cc}1 &4 \\ -3 & -\end{array}\right)^{-1} = \left(\begin{array}{cc}-\frac{}{5} & -\frac{4}{5} \\ \frac{3}{5} & \frac{1}{5}\end{array}\right)$

Now, compare this matrix with the given inverse matrix $\frac{1}{}\left(\begin{array}{rr}-7 & -4 \\3 & y\end{array}\right)$ .

By comparing the elements at the second row and second column of both matrices, we can find the value of $y$ .
$\frac{1}{5} = \frac{y}{8}$

olve the equation for $y$ .
$y =8*\frac{}{5} = \frac{8}{5}$

Now that we have the value of $y$ , we can find the value of $x+y$ .
$x+y =3 + \frac{8}{5}$

Calculate the value of $x+y$ .
$x+y =3 + \frac{8}{5} = \frac{23}{5}$ So, the value of $x+y$ is $\frac{23}{5}$ .