Home/Boards/Maharashtra Msbshse/Class 11 Science/Mathematics/Ch 4 — Determinants and Matrices/If A = 1 -2 \5 3 , B = 1 -3 \4 -7 , then find the matrix A -…Exercise 4.5 · Q111Q.If A=[1−253]A = \begin{bmatrix} 1 & -2 \\ 5 & 3 \end{bmatrix}A=[15−23], B=[1−34−7]B = \begin{bmatrix} 1 & -3 \\ 4 & -7 \end{bmatrix}B=[14−3−7], then find the matrix A−2B+6IA - 2B + 6IA−2B+6I, where III is the unit matrix of order 2.Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est33% · 69/212 QuestionsConceptShort answerLong answerMethodsCommon mistakesRelated✓ Free questionConcept understanding — Operations on MatricesMatrices of the same order can be checked for equality (A=BA=BA=B iff every corresponding entry matches), added (A+B=[aij+bij]A+B=[a_{ij}+b_{ij}]A+B=[aij+bij], entry-by-entry, undefined for mismatched orders), subtracted (A−B=A+(−B)A-B=A+(-B)A−B=A+(−B)), and scaled by a number (kA=[k aij]kA=[k\,a_{ij}]kA=[kaij], every entry multiplied by kkk). These operations obey familiar-looking laws — addition is commutative (A+B=B+AA+B=B+AA+B=B+A) and associative ((A+B)+C=A+(B+C)(A+B)+C=A+(B+C)(A+B)+C=A+(B+C)), the zero matrix is the additive identity, −A-A−A is the additive inverse, and scalar multiplication distributes over both matrix sums and scalar sums. The most common use is solving a linear matrix equation for an unknown matrix XXX (e.g. 3A−4B+5X=C3A-4B+5X=C3A−4B+5X=C): treat XXX exactly as you would an unknown number, isolating it algebraically, then compute the resulting combination of known matrices entry-by-entry. These "sum-and-scale" operations are the foundation on top of which matrix multiplication (a genuinely different, non-commutative operation) is built. Compute 2B2B2B and 6I6I6I separately, then combine with AAA entry-wise. ✓Final answer A−2B+6I=[54−323]A - 2B + 6I = \begin{bmatrix} 5 & 4 \\ -3 & 23 \end{bmatrix}A−2B+6I=[5−3423] Given A=[1−253]A = \begin{bmatrix} 1 & -2 \\ 5 & 3 \end{bmatrix}A=[15−23], B=[1−34−7]B = \begin{bmatrix} 1 & -3 \\ 4 & -7 \end{bmatrix}B=[14−3−7], I=[1001]I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}I=[1001]. 2B=[2−68−14]2B = \begin{bmatrix} 2 & -6 \\ 8 & -14 \end{bmatrix}2B=[28−6−14], 6I=[6006]\quad 6I = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}6I=[6006] 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=[1−25−8−2−(−6)3−(−14)]=[−1−3417] 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}A−2B+6I=[−1+6−3+04+017+6]=[5−3423] ✓Final answer A−2B+6I=[54−323]A - 2B + 6I = \begin{bmatrix} 5 & 4 \\ -3 & 23 \end{bmatrix}A−2B+6I=[5−3423] Scale B by 2 and I by 6, then combine A − 2B + 6I entry-by-entry. Forgetting that 6I6I6I only adds 6 to the diagonal entries, not to every entry of the matrix. Sign errors subtracting the negative entries of 2B2B2B. Exercise 4.5If A=[2−35−4−61]A = \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix}A=25−6−3−41, B=[−122203]B = \begin{bmatrix} -1 & 2 \\ 2 & 2 \\ 0 & 3 \end{bmatrix}B=−120223 and C=[43−14−21]C = \begin{bmatrix} 4 & 3 \\ -1 & 4 \\ -2 & 1 \end{bmatrix}C=4−1−2341, showView this question →Exercise 4.5If A=[2−35−4−61]A = \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix}A=25−6−3−41, B=[−122203]B = \begin{bmatrix} -1 & 2 \\ 2 & 2 \\ 0 & 3 \end{bmatrix}B=−120223 and C=[43−14−21]C = \begin{bmatrix} 4 & 3 \\ -1 & 4 \\ -2 & 1 \end{bmatrix}C=4−1−2341, showView this question →Exercise 4.5If A=[12−3−37−80−61]A = \begin{bmatrix} 1 & 2 & -3 \\ -3 & 7 & -8 \\ 0 & -6 & 1 \end{bmatrix}A=1−3027−6−3−81, B=[9−12−42540−3]B = \begin{bmatrix} 9 & -1 & 2 \\ -4 & 2 & 5 \\ 4 & 0 & -3 \end{bmatrix}B=9−44−12025−3, then find the matrix CCC such that A+B+CA+B+CA+B+C iView this question →Exercise 4.5If A=[1−23−5−60]A = \begin{bmatrix} 1 & -2 \\ 3 & -5 \\ -6 & 0 \end{bmatrix}A=13−6−2−50, B=[−1−24215]B = \begin{bmatrix} -1 & -2 \\ 4 & 2 \\ 1 & 5 \end{bmatrix}B=−141−225 and C=[24−1−4−36]C = \begin{bmatrix} 2 & 4 \\ -1 & -4 \\ -3 & 6 \end{bmatrix}C=2−1−34−46, finView this question →Exercise 4.5Solve the following equations for XXX and YYY, if 3X−Y=[1−1−11]3X - Y = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}3X−Y=[1−1−11] and X−3Y=[0−10−1]X - 3Y = \begin{bmatrix} 0 & -1 \\ 0 & -1 \end{bmatrix}X−3Y=[00−1−1].View this question →Exercise 4.5Find matrices AAA and BBB, if 2A−B=[6−60−421]2A - B = \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix}2A−B=[6−4−6201] and A−2B=[328−21−7]A - 2B = \begin{bmatrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{bmatrix}A−2B=[3−2218−7].View this question →🎓Unlock everything free for 14 days✓Full step-by-step solutions✓Concept-first explanations✓Methods, shortcuts & mistakes✓PYQ mapping + timed mock testsStart 14-day free trial →See Plans & Pricing →Full access for 14 days. No credit card required.←Previous question · 4.5Q110 — If A = 2 -3 \5 -4 \-6 1 , B = -1 2 \2 2 \0 3 and C = 4 3 \-1 4 \-2 1 ,…→Next question · 4.5Q112 — If A = 1 2 -3 \-3 7 -8 \0 -6 1 , B = 9 -1 2 \-4 2 5 \4 0 -3 , then fin…