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Exercise 3.2 · Q9

Q.Find xx and yy, if 2[130x]+[y012]=[5618]2 \begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}.

Rajasthan RbseTextbookSubjective· 2mImportance★★★★★
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Matrix addition is element-wise, so we equate corresponding entries after scalar multiplication. Solving the two independent equations gives x=3x = 3 and y=3y = 3.

Matrix addition works exactly like adding two grids of numbers — you add the entry in row 1, column 1 of the first matrix to the entry in row 1, column 1 of the second, and so on. When a matrix is multiplied by a scalar (like the 22 here), every entry inside gets multiplied by that scalar first. So the left-hand side becomes a single matrix whose entries are simple expressions in xx and yy. The right-hand side is a known matrix. Since two matrices are equal only when every corresponding entry matches, we get a set of equations — one per position — and solve them.

Let’s go through it step by step.

  1. Multiply the first matrix by the scalar 22.

2[130x]=[2⋅12⋅32⋅02⋅x]=[2602x]2 \begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} = \begin{bmatrix} 2 \cdot 1 & 2 \cdot 3 \\ 2 \cdot 0 & 2 \cdot x \end{bmatrix} = \begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix}

  1. Add the second matrix to this result.

[2602x]+[y012]=[2+y6+00+12x+2]=[2+y612x+2]\begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2 + y & 6 + 0 \\ 0 + 1 & 2x + 2 \end{bmatrix} = \begin{bmatrix} 2 + y & 6 \\ 1 & 2x + 2 \end{bmatrix}

  1. Set this equal to the given right-hand side.

[2+y612x+2]=[5618]\begin{bmatrix} 2 + y & 6 \\ 1 & 2x + 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}

  1. Equate corresponding entries.
    • Top-left: 2+y=52 + y = 5
    • Top-right: 6=66 = 6 (already satisfied, gives no new info)
    • Bottom-left: 1=11 = 1 (also automatically satisfied)
    • Bottom-right: 2x+2=82x + 2 = 8 …

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