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Exercise 3.3 · Q2

Q.If A=[−123579−211]A = \begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix} and B=[−41−5120131]B = \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix}, then verify that

(i) (A+B)′=A′+B′(A + B)' = A' + B',
(ii) (A−B)′=A′−B′(A - B)' = A' - B'
Uttarakhand UbseTextbookSubjective· 3mImportance★★★★★
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✓ Free question

The transpose of a sum (or difference) of matrices equals the sum (or difference) of their transposes. For the given matrices AA and BB, we verify that (A+B)′=A′+B′(A+B)' = A' + B' and (A−B)′=A′−B′(A-B)' = A' - B' by computing both sides and showing they match.

Why This Works: The Intuition

Matrix transpose is a simple operation — it flips rows into columns and columns into rows. When you add two matrices, you add corresponding entries. So if you transpose after adding, you're just flipping the positions of those sums. But if you transpose each matrix first and then add, you're adding the flipped entries — which is exactly the same thing, because addition is commutative and works entry-by-entry. The same logic holds for subtraction. This property is called linearity of the transpose: it respects addition and scalar multiplication.

Let's verify it concretely with the given 3×33 \times 3 matrices.


Step-by-Step Verification

1. Compute A+BA + B and A−BA - B

First, add and subtract the matrices entry-wise.

A+B=[−1+(−4)2+13+(−5)5+17+29+0−2+11+31+1]=[−53−2699−142]A + B = \begin{bmatrix} -1 + (-4) & 2 + 1 & 3 + (-5) \\ 5 + 1 & 7 + 2 & 9 + 0 \\ -2 + 1 & 1 + 3 & 1 + 1 \end{bmatrix} = \begin{bmatrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{bmatrix}

A−B=[−1−(−4)2−13−(−5)5−17−29−0−2−11−31−1]=[318459−3−20]A - B = \begin{bmatrix} -1 - (-4) & 2 - 1 & 3 - (-5) \\ 5 - 1 & 7 - 2 & 9 - 0 \\ -2 - 1 & 1 - 3 & 1 - 1 \end{bmatrix} = \begin{bmatrix} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \end{bmatrix}

2. Transpose A+BA + B and A−BA - B

Transpose means: row ii becomes column ii.

(A+B)′=[−56−1394−292](A + B)' = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}

(A−B)′=[34−315−2890](A - B)' = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}

3. Compute A′A' and B′B' individually

Transpose AA and BB:

A′=[−15−2271391]A' = \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix}

B′=[−411123−501]B' = \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix}

4. Add and subtract the transposes

Now add A′A' and B′B':

A′+B′=[−1+(−4)5+1−2+12+17+21+33+(−5)9+01+1]=[−56−1394−292]A' + B' = \begin{bmatrix} -1 + (-4) & 5 + 1 & -2 + 1 \\ 2 + 1 & 7 + 2 & 1 + 3 \\ 3 + (-5) & 9 + 0 & 1 + 1 \end{bmatrix} = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}

This matches (A+B)′(A + B)' exactly.

Now subtract B′B' from A′A':

A′−B′=[−1−(−4)5−1−2−12−17−21−33−(−5)9−01−1]=[34−315−2890]A' - B' = \begin{bmatrix} -1 - (-4) & 5 - 1 & -2 - 1 \\ 2 - 1 & 7 - 2 & 1 - 3 \\ 3 - (-5) & 9 - 0 & 1 - 1 \end{bmatrix} = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}

This matches (A−B)′(A - B)' exactly.

Watch out

A common mistake is to forget that the transpose flips the order of rows and columns. When adding or subtracting, make sure you're comparing corresponding entries after transposition — not just looking at the shape. Also, note that (A−B)′=A′−B′(A - B)' = A' - B' works because subtraction is just addition of a scalar multiple (−1-1 times BB), and the transpose is linear.

Tip

You never actually need to compute both sides fully to verify these properties — they hold for any matrices of the same size. But doing the concrete check builds confidence and catches arithmetic errors. A quick mental check: the (i,j)(i,j) entry of (A+B)′(A+B)' is aji+bjia_{ji} + b_{ji}, which is exactly the (i,j)(i,j) entry of A′+B′A' + B'.


✓Final answer

Both identities are verified: (A+B)′=A′+B′(A+B)' = A' + B' and (A−B)′=A′−B′(A-B)' = A' - B' hold for the given matrices.

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