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Exercise 3.1 · Q7

Q.Find the value of a,b,ca, b, c and dd from the equation: [a−b2a+c2a−b3c+d]=[−15013]\begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}

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Two matrices are equal if and only if their corresponding entries are equal. Equating the four entries gives a system of four linear equations in a,b,c,da,b,c,d; solving it yields a=1a=1, b=2b=2, c=3c=3, d=4d=4.

When two matrices are written as equal, every entry in the same position must be identical. This is the definition of matrix equality — no shortcuts, no tricks. So the given 2×22\times 2 matrix equation is really a compact way of writing four separate equations.

Let’s unpack them.

  1. Equate the (1,1) entries (top-left):

    a−b=−1a - b = -1

    This is our first equation.

  2. Equate the (1,2) entries (top-right):

    2a+c=52a + c = 5

    Second equation.

  3. Equate the (2,1) entries (bottom-left):

    2a−b=02a - b = 0

    Third equation.

  4. Equate the (2,2) entries (bottom-right):

    3c+d=133c + d = 13

    Fourth equation.

Now we have a system:

{a−b=−12a+c=52a−b=03c+d=13\begin{cases} a - b = -1 \\ 2a + c = 5 \\ 2a - b = 0 \\ 3c + d = 13 \end{cases}

Notice that the first and third equations both involve aa and bb only. Solve those together first.

From 2a−b=02a - b = 0, we get b=2ab = 2a.

Substitute into a−b=−1a - b = -1:

a−2a=−1  ⟹  −a=−1  ⟹  a=1a - 2a = -1 \implies -a = -1 \implies a = 1.

Then b=2(1)=2b = 2(1) = 2.

Now aa is known. Use the second equation: 2a+c=5  ⟹  2(1)+c=5  ⟹  c=32a + c = 5 \implies 2(1) + c = 5 \implies c = 3. …

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