Question 100If A = [−1320],B=[1−203],C=[1−4] and D=[41]\begin{bmatrix*}[r] -1 & 3 \ 2 & 0 \end{bmatrix*}, B = \begin{bmatrix*}[r] 1 & -2 \ 0 & 3 \end{bmatrix*}, C = \begin{bmatrix*}[r] 1 & -4 \end{bmatrix*} ext{ and } D = \begin{bmatrix*}[r] 4 \ 1 \end{bmatrix*}[−1230],B=[10−23],C=[1−4] and D=[41](a) Is the product AC possible? Justify your answer.(b) Find the matrix X, such that X = AB + B2 − DC

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Question 100If A = [−1320],B=[1−203],C=[1−4] and D=[41]\begin{bmatrix*}[r] -1 & 3 \ 2 & 0 \end{bmatrix*}, B = \begin{bmatrix*}[r] 1 & -2 \ 0 & 3 \end{bmatrix*}, C = \begin{bmatrix*}[r] 1 & -4 \end{bmatrix*} 	ext{ and } D = \begin{bmatrix*}[r] 4 \ 1 \end{bmatrix*}[−12​30​],B=[10​−23​],C=[1​−4​] and D=[41​](a) Is the product AC possible? Justify your answer.(b) Find the matrix X, such that X = AB + B2 − DC

Bright Tutorials · Accepted Answer

(a) Order of matrix A = 2 × 2, Order of matrix C = 1 × 2The product AC is not possible as the no. of columns in A is not equal to the no. of rows in C.Hence, product AC is not possible.(b) Given,⇒X=AB+B2−DC⇒X=[−1320][1−203]+[1−203][1−203]−[41][1−4]⇒X=[−1×1+3×0−1×−2+3×32×1+0×02×−2+0×3]+[1×1+−2×01×−2+−2×30×1+3×00×−2+3×3]−[4×14×−41×11×−4]⇒X=[−1+02+92+0−4+0]+[1+0−2−60+00+9]−[4−161−4]⇒X=[−1112−4]+[1−809]−[4−161−4]⇒X=[−1+1−411+(−8)−(−16)2+0−1(−4)+9−(−4)]⇒X=[−411−8+161−4+9+4]⇒X=[−41919]\Rightarrow X = AB + B^2 - DC \[1em] \Rightarrow X = \begin{bmatrix*}[r] -1 & 3 \ 2 & 0 \end{bmatrix*}\begin{bmatrix*}[r] 1 & -2 \ 0 & 3 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & -2 \ 0 & 3 \end{bmatrix*}\begin{bmatrix*}[r] 1 & -2 \ 0 & 3 \end{bmatrix*} - \begin{bmatrix*}[r] 4 \ 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & -4 \end{bmatrix*} \[1em] \Rightarrow X = \begin{bmatrix*}[r] -1 \times 1 + 3 \times 0 & -1 \times -2 + 3 \times 3 \ 2 \times 1 + 0 \times 0 & 2 \times -2 + 0 \times 3 \end{bmatrix*} + \begin{bmatrix*}[r] 1 \times 1 + -2 \times 0 & 1 \times -2 + -2 \times 3 \ 0 \times 1 + 3 \times 0 & 0 \times -2 + 3 \times 3 \end{bmatrix*} - \begin{bmatrix*}[r] 4 \times 1 & 4 \times -4 \ 1 \times 1 & 1 \times -4 \end{bmatrix*} \[1em] \Rightarrow X = \begin{bmatrix*}[r] -1 + 0 & 2 + 9 \ 2 + 0 & -4 + 0 \end{bmatrix*} + \begin{bmatrix*}[r] 1 + 0 & -2 - 6 \ 0 + 0 & 0 + 9 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & -16 \ 1 & -4 \end{bmatrix*} \[1em] \Rightarrow X = \begin{bmatrix*}[r] -1 & 11 \ 2 & -4 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & -8 \ 0 & 9 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & -16 \ 1 & -4 \end{bmatrix*} \[1em] \Rightarrow X = \begin{bmatrix*}[r] -1 + 1 - 4 & 11 + (-8) - (-16) \ 2 + 0 - 1 & (-4) + 9 - (-4) \end{bmatrix*} \[1em] \Rightarrow X = \begin{bmatrix*}[r] -4 & 11 - 8 + 16 \ 1 & -4 + 9 + 4 \end{bmatrix*} \[1em] \Rightarrow X = \begin{bmatrix*}[r] -4 & 19 \ 1 & 9 \end{bmatrix*}⇒X=AB+B2−DC⇒X=[−12​30​][10​−23​]+[10​−23​][10​−23​]−[41​][1​−4​]⇒X=[−1×1+3×02×1+0×0​−1×−2+3×32×−2+0×3​]+[1×1+−2×00×1+3×0​1×−2+−2×30×−2+3×3​]−[4×11×1​4×−41×−4​]⇒X=[−1+02+0​2+9−4+0​]+[1+00+0​−2−60+9​]−[41​−16−4​]⇒X=[−12​11−4​]+[10​−89​]−[41​−16−4​]⇒X=[−1+1−42+0−1​11+(−8)−(−16)(−4)+9−(−4)​]⇒X=[−41​11−8+16−4+9+4​]⇒X=[−41​199​]Hence, X = [−41919].\begin{bmatrix*}[r] -4 & 19 \ 1 & 9 \end{bmatrix*}.[−41​199​].

Short Answer Questions 2 — Question 23

Question 100