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Solve the matrix equation for $a,\ b,\ c,$a, b, c,​ and $d$d​ .$\frac{1}{2}$12​​ $\begin{bmatrix}-6a&-2\\3&14\\-10&-8\end{bmatrix}+\begin{bmatrix}5&3\\-\frac{5}{2}&b-3\\6&-1\end{bmatrix}=\begin{bmatrix}9&4d\\-1&11\\c+4&-5\end{bmatrix}$[−6a −23 14−10 −8]+[5 3−52​ b−36 −1]=[9 4d−1 11c+4 −5]​$a=$a=​ ,  $b=$b=​  ,  $c=$c=​  ,  $d=$d=​  Skip to navigationPrevious25, Fully Attempted.26, Fully Attempted.27, Fully Attempted.28, Fully Attempted.29, Fully Attempted.30, Unattempted. Flagged31, Unattempted.32, Unattempted.33, Unattempted.34, Unattempted.Next

Question

Solve the matrix equation for a, b, c,a,\ b,\ c,a, b, c,​ and ddd​ .12\frac{1}{2}12​​ [−6a−2314−10−8]+[53−52b−36−1]=[94d−111c+4−5]\begin{bmatrix}-6a&-2\\3&14\\-10&-8\end{bmatrix}+\begin{bmatrix}5&3\\-\frac{5}{2}&b-3\\6&-1\end{bmatrix}=\begin{bmatrix}9&4d\\-1&11\\c+4&-5\end{bmatrix}[−6a −23 14−10 −8]+[5 3−52​ b−36 −1]=[9 4d−1 11c+4 −5]​a=a=a=​ ,  b=b=b=​  ,  c=c=c=​  ,  d=d=d=​  Skip to navigationPrevious25, Fully Attempted.26, Fully Attempted.27, Fully Attempted.28, Fully Attempted.29, Fully Attempted.30, Unattempted. Flagged31, Unattempted.32, Unattempted.33, Unattempted.34, Unattempted.Next

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Solution

The given matrix equation is:

12\frac{1}{2} [−6a−2314−10−8]+[53−52b−36−1]=[94d−111c+4−5]\begin{bmatrix}-6a&-2\\3&14\\-10&-8\end{bmatrix}+\begin{bmatrix}5&3\\-\frac{5}{2}&b-3\\6&-1\end{bmatrix}=\begin{bmatrix}9&4d\\-1&11\\c+4&-5\end{bmatrix}

First, we can distribute the 12\frac{1}{2} to the first matrix:

[−3a−11.57−5−4]+[53−52b−36−1]=[94d−111c+4−5]\begin{bmatrix}-3a&-1\\1.5&7\\-5&-4\end{bmatrix}+\begin{bmatrix}5&3\\-\frac{5}{2}&b-3\\6&-1\end{bmatrix}=\begin{bmatrix}9&4d\\-1&11\\c+4&-5\end{bmatrix}

Next, we can add the two matrices on the left side of the equation:

[−3a+5−1+31.5−527+b−3−5+6−4−1]=[94d−111c+4−5]\begin{bmatrix}-3a+5&-1+3\\1.5-\frac{5}{2}&7+b-3\\-5+6&-4-1\end{bmatrix}=\begin{bmatrix}9&4d\\-1&11\\c+4&-5\end{bmatrix}

Simplify the left side:

[−3a+52−1b+41−5]=[94d−111c+4−5]\begin{bmatrix}-3a+5&2\\-1&b+4\\1&-5\end{bmatrix}=\begin{bmatrix}9&4d\\-1&11\\c+4&-5\end{bmatrix}

Now, we can equate the corresponding elements in the two matrices to solve for aa, bb, cc, and dd:

For aa:

−3a+5=9-3a+5=9

−3a=4-3a=4

a=−43a=-\frac{4}{3}

For bb:

b+4=11b+4=11

b=7b=7

For cc:

1=c+41=c+4

c=−3c=-3

For dd:

2=4d2=4d

d=12d=\frac{1}{2}

So, the solution is a=−43a=-\frac{4}{3}, b=7b=7, c=−3c=-3, and d=12d=\frac{1}{2}.

This problem has been solved

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Solve the matrix equation for $a,\ b,\ c,$a, b, c,​ and $d$d​ .$\frac{1}{2}$12​​ $\begin{bmatrix}-6a&-2\\3&14\\-10&-8\end{bmatrix}+\begin{bmatrix}5&3\\-\frac{5}{2}&b-3\\6&-1\end{bmatrix}=\begin{bmatrix}9&4d\\-1&11\\c+4&-5\end{bmatrix}$[−6a −23 14−10 −8]+[5 3−52​ b−36 −1]=[9 4d−1 11c+4 −5]​$a=$a=​ ,  $b=$b=​  ,  $c=$c=​  ,  $d=$d=​  Skip to navigationPrevious25, Fully Attempted.26, Fully Attempted.27, Fully Attempted.28, Fully Attempted.29, Fully Attempted.30, Unattempted. Flagged31, Unattempted.32, Unattempted.33, Unattempted.34, Unattempted.Next

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