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Solve the matrix equation for $a,\ b,\ c,$a, b, c,​ and $d$d​ .$2\left(\begin{bmatrix}3a&11\\4&6\end{bmatrix}+\begin{bmatrix}-7&-6\\11&5d\end{bmatrix}\right)=\begin{bmatrix}22&b\\c&52\end{bmatrix}$2([3a 114 6]+[−7 −611 5d])=[22 bc 52]​$a=$a=​ ,  $b=$b=​  ,  $c=$c=​  ,  $d=$d=​

Question

Solve the matrix equation for a, b, c,a,\ b,\ c,a, b, c,​ and ddd​ .2([3a1146]+[−7−6115d])=[22bc52]2\left(\begin{bmatrix}3a&11\\4&6\end{bmatrix}+\begin{bmatrix}-7&-6\\11&5d\end{bmatrix}\right)=\begin{bmatrix}22&b\\c&52\end{bmatrix}2([3a 114 6]+[−7 −611 5d])=[22 bc 52]​a=a=a=​ ,  b=b=b=​  ,  c=c=c=​  ,  d=d=d=​

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Solution

To solve the matrix equation, we first need to distribute the 2 into both matrices on the left side of the equation. This gives us:

2([3a1146])+2([−7−6115d])=[22bc52]2\left(\begin{bmatrix}3a&11\\4&6\end{bmatrix}\right) + 2\left(\begin{bmatrix}-7&-6\\11&5d\end{bmatrix}\right) = \begin{bmatrix}22&b\\c&52\end{bmatrix}

This simplifies to:

[6a22812]+[−14−122210d]=[22bc52]\begin{bmatrix}6a&22\\8&12\end{bmatrix} + \begin{bmatrix}-14&-12\\22&10d\end{bmatrix} = \begin{bmatrix}22&b\\c&52\end{bmatrix}

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

[6a−1422−128+2212+10d]=[22bc52]\begin{bmatrix}6a-14&22-12\\8+22&12+10d\end{bmatrix} = \begin{bmatrix}22&b\\c&52\end{bmatrix}

This simplifies to:

[6a−14103012+10d]=[22bc52]\begin{bmatrix}6a-14&10\\30&12+10d\end{bmatrix} = \begin{bmatrix}22&b\\c&52\end{bmatrix}

Now, we can equate the corresponding elements in the two matrices to form a system of equations:

6a−14=226a - 14 = 22

10=b10 = b

30=c30 = c

12+10d=5212 + 10d = 52

Solving these equations gives:

a=22+146=6a = \frac{22 + 14}{6} = 6

b=10b = 10

c=30c = 30

d=52−1210=4d = \frac{52 - 12}{10} = 4

So, the solution to the matrix equation is a=6a = 6, b=10b = 10, c=30c = 30, and d=4d = 4.

This problem has been solved

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