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Exercise 4.5 · Q111

Q.If A=[1−253]A = \begin{bmatrix} 1 & -2 \\ 5 & 3 \end{bmatrix}, B=[1−34−7]B = \begin{bmatrix} 1 & -3 \\ 4 & -7 \end{bmatrix}, then find the matrix A−2B+6IA - 2B + 6I, where II is the unit matrix of order 2.

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✓ Free question

Given A=[1−253]A = \begin{bmatrix} 1 & -2 \\ 5 & 3 \end{bmatrix}, B=[1−34−7]B = \begin{bmatrix} 1 & -3 \\ 4 & -7 \end{bmatrix}, I=[1001]I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.

2B=[2−68−14]2B = \begin{bmatrix} 2 & -6 \\ 8 & -14 \end{bmatrix}, 6I=[6006]\quad 6I = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}

A−2B=[1−2−2−(−6)5−83−(−14)]=[−14−317]A - 2B = \begin{bmatrix} 1-2 & -2-(-6) \\ 5-8 & 3-(-14) \end{bmatrix} = \begin{bmatrix} -1 & 4 \\ -3 & 17 \end{bmatrix}

A−2B+6I=[−1+64+0−3+017+6]=[54−323]A - 2B + 6I = \begin{bmatrix} -1+6 & 4+0 \\ -3+0 & 17+6 \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ -3 & 23 \end{bmatrix}

✓Final answer

A−2B+6I=[54−323]A - 2B + 6I = \begin{bmatrix} 5 & 4 \\ -3 & 23 \end{bmatrix}

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