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Worked Examples · Example 5

Q.Find the values of a,b,c,a, b, c, and dd from the following equation: [2a+ba−2b5c−d4c+3d]=[4−31124]\begin{bmatrix} 2a+b & a-2b \\ 5c-d & 4c+3d \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ 11 & 24 \end{bmatrix}

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Two matrices are equal if and only if their corresponding entries are equal. This gives four simple linear equations in a,b,c,da, b, c, d. Solving them yields a=1a = 1, b=2b = 2, c=3c = 3, d=4d = 4.

The core idea here is vector equality applied to matrices. When two matrices are declared equal, every entry in the same position must match exactly. That’s not a suggestion — it’s the definition. So this single matrix equation is really a compact way of writing four separate equations, one for each position.

Let’s unpack it.

We have:

[2a+ba−2b5c−d4c+3d]=[4−31124]\begin{bmatrix} 2a+b & a-2b \\ 5c-d & 4c+3d \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ 11 & 24 \end{bmatrix}

Since the matrices are 2×22 \times 2, equality means:

  1. Top-left entry: 2a+b=42a + b = 4
  2. Top-right entry: a−2b=−3a - 2b = -3
  3. Bottom-left entry: 5c−d=115c - d = 11
  4. Bottom-right entry: 4c+3d=244c + 3d = 24

Notice something beautiful: the equations for a,ba, b are completely separate from those for c,dc, d. They don’t mix. So we can solve them as two independent pairs.


Solving for aa and bb

We have:

{2a+b=4(1)a−2b=−3(2)\begin{cases} 2a + b = 4 \quad (1) \\ a - 2b = -3 \quad (2) \end{cases}

From (2), we get a=2b−3a = 2b - 3. Substitute into (1):

2(2b−3)+b=4  ⟹  4b−6+b=4  ⟹  5b=10  ⟹  b=22(2b - 3) + b = 4 \implies 4b - 6 + b = 4 \implies 5b = 10 \implies b = 2

Then a=2(2)−3=1a = 2(2) - 3 = 1. …

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